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[ "Set the time Ian traveled as $I$ , and set Han's speed as $H$ . Therefore, Jan's speed is $H+5.$\nWe get the following equation for how much Han is ahead of Ian: $H+5I = 70.$\nThe expression for how much Jan is ahead of Ian is: $2(H+5)+10I.$\nThis simplifies to: $2H+10+10I.$\nHowever, this is just $2(H+5I)+10.$\nSubstitute, from the first equation, $H+5I$ as $70.$\nTherefore, the answer is $140 + 10$ , which is $150$ , or $\\boxed{150}$", "We let Ian's speed and time equal $I_s$ and $I_t$ , respectively. Similarly, let Han's and Jan's speed and time be $H_s$ $H_t$ $J_s$ $J_t$ . The problem gives us 5 equations\n\\begin{align} H_s&=I_s+5 \\\\ H_t&=I_t+1 \\\\ J_s&=I_s+10 \\\\ J_t&=I_t+2 \\\\ H_s \\cdot H_t & =I_s \\cdot I_t+70 \\end{align}\nSubstituting equations $(1)$ and $(2)$ into $(5)$ gives:\n\\[(I_s+5)(I_t+1)=I_s I_t+70 \\Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \\Longrightarrow I_s+5I_t=65 \\quad (*)\\]\nWe are asked the difference between Jan's and Ian's distances, or\n\\[J_s J_t-I_s I_t=x,\\]\nWhere $x$ is the difference between Jan's and Ian's distances and the answer to the problem. Substituting $(3)$ and $(4)$ equations into this equation gives:\n\\[(I_s+10)(I_t+2)-I_s I_t=x \\Longrightarrow I_s I_t+2I_s+10I_t+20-I_s I_t=x \\Longrightarrow\\]\n\\[2I_s+10I_t+20=x \\Longrightarrow 2(I_s+5I_t)+20=x\\]\nSubstituting $(*)$ into this equation gives:\n\\[2(65)+20=x \\Longrightarrow 130+20=x \\Longrightarrow 150=x\\]\nTherefore, the answer is $150$ miles or $\\boxed{150}$", "Let Ian drive $D$ miles, at a speed of $R$ , for some time $T$ (in hours). Hence, we have $D=RT$ . We can find a similar equation for Han, who drove $D + 70$ miles, at a rate of $R+5$ , for $T+1$ hours, giving us $D + 70 = (R + 5)(T + 1)$ . We can do the same for Jan, giving us $D + x = (R + 10)(T + 2)$ , where $x$ is how much further Jan traveled than Ian. We now have three equations: \\[D= RT\\] \\[D + 70 = (R+5)(T+1) = RT + R + 5T + 5\\] \\[D + x = (R + 10)(T + 2) = RT + 10 T + 2R + 20.\\] Substituting $RT$ for $D$ in the second and third equations and cancelling gives us: \\[70 = 5T + R + 5 \\Longrightarrow 5T + R = 65\\] \\[x = 10T + 2R + 20 \\Longrightarrow x = 2(5T + R ) + 20 \\Longrightarrow x= 2(65) + 20 = 150.\\] Since $x = 150$ , our answer is $\\boxed{150}$", "Let Ian drive $d$ miles, $t$ hours, and at speed $s$\nIan's Equation: $d=s \\cdot t.$\nHan drove 70 more miles, traveled 5 miles per hour faster and traveled 1 more hour than Ian.\nHan's Equation: $d+70=(s+5) \\cdot (t+1).$\nLet Jan have driven $m$ miles. Jan also has driven 10 miles per hour faster and traveled 2 more hours than Ian.\nJan's Equation: $m=(s+10) \\cdot (t+2).$\nLet's group the equations together:\n$(1) \\phantom{a} d=s \\cdot t$\n$(2) \\phantom{a} d+70=(s+5) \\cdot (t+1)$\n$(3) \\phantom{a} m=(s+10) \\cdot (t+2)$\nLet's see what we want to find: We want to find $n$ . The equation is $m=d+n$ where $n$ is the number of more miles traveled by Jan than Ian.\nOnto the calculating part.\nExpanding the second equation, we get\n$d+70=st+5t+s+5.$\nNote that $st=d$ by the first equation, so substituting we get\n$d+70=d+5t+s+5.$\nSimplifying gets us\n$(4) \\phantom{a} s+5t=65.$\n(Note for the above process: You could have substituted $d$ with $st$ but that would lead you to the same result since $d-d=st-st=0.$\nLet's look at the third equation. Expanding, we get\n$m=st+10t+2s+20.$\nSince we want $n$ we want the equation $m=d+n$ . We write the expanded third equation into this form since $st=d.$\n$(5) \\phantom{a} m=d+(10t+2s+20)$\nLet's take a closer look at the $n$ section of the equation:\n$(6) \\phantom{a} n=10t+2s+20$\nThis looks very similar to equation 4, if you multiply equation 4 by $2$ you get\n$(7) \\phantom{a} 10t+2s=130$\nPlugging equation 7 into equation 6 we have\n$n=(10t+2s)+20 \\Rightarrow n=130+20$\nCalculating gets us\n$(8) \\phantom{a} n=150.$\nSubstituting equation 8 into equation 5 gets us\n$m=d+n \\Rightarrow m=d+150.$\nThe $n$ term is $150$ which is what we want to find so the answer is $\\boxed{150}.$", "Since Han drove for $1$ hour and drove $70$ miles more than Ian during that hour, we know that Ian's speed is $65$ miles per hour since Han drove $5$ mph faster than him. Now Jan went $10$ mph faster than Ian for $2$ hours, so we can tell that she drove $75 \\cdot 2$ miles more than Ian, therefore the answer is $\\boxed{150}.$" ]

[ "There are $6$ people to split the $1500$ dollars among, so each person must raise $\\frac{1500}6=250$ dollars. $\\Rightarrow\\boxed{250}$" ]

[ "First, note that $n$ does not have a prime number larger than $61$ as one of its factors. Also, note that $n$ does not equal $1$\nTherefore, since the prime factorization of $n$ only has primes from $2$ to $59$ $n$ and $P$ share at least one common factor other than $1$ . Therefore $P+n$ is not prime for any $n$ , so the answer is $\\Rightarrow{\\boxed{0}$" ]

[ "First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with $5$ guesses, and one with $4$ . Since the problem is asking for the minimum number, the answer is $\\boxed{4}$", "Since the hidden rectangle can only hide two adjacent squares, we may think that we eliminate 8 squares and we're done, but think again. This is the AMC 10, so there must be a better solution (also note that every other solution choice is below 8 so we're probably not done) So, we think again, we notice that we haven't used the adjacent condition, and then it clicks. If we eliminate the four squares with only one edge on the boundary of the 9x9 square. We are left with 5 diagonal squares, since our rectangle cant be diagonal, we can ensure that we find it in 4 moves. So our answer is : $\\boxed{4}$", "We realize that every $2 \\times 1$ rectangle must contain an edge and no more than one edge. There are a total of four edges so the answer is $\\boxed{4.}$ .\n~darrenn.cp" ]

[ "The $3 \\times 3$ grid can be colored like a checkerboard with alternating black and white squares.\nLet the top left square be white, and the rest of the squares alternate colors.\nEach $2 \\times 1$ rectangle always covers $1$ white square and $1$ black square.\nYou can ensure that at least one of your guessed squares is covered by the rectangle by choosing either each of the white squares ( $5$ turns) or each of the black squares ( $4$ turns).\nSince it is ideal to be the most efficient with our turns, by choosing all the black squares, we guarantee that one of the $\\boxed{4}$ squares are of the $2 \\times 1$ rectangle." ]

[ "First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with $5$ guesses, and one with $4$ . Since the problem is asking for the minimum number, the answer is $\\boxed{4}$", "Since the hidden rectangle can only hide two adjacent squares, we may think that we eliminate 8 squares and we're done, but think again. This is the AMC 10, so there must be a better solution (also note that every other solution choice is below 8 so we're probably not done) So, we think again, we notice that we haven't used the adjacent condition, and then it clicks. If we eliminate the four squares with only one edge on the boundary of the 9x9 square. We are left with 5 diagonal squares, since our rectangle cant be diagonal, we can ensure that we find it in 4 moves. So our answer is : $\\boxed{4}$", "We realize that every $2 \\times 1$ rectangle must contain an edge and no more than one edge. There are a total of four edges so the answer is $\\boxed{4.}$ .\n~darrenn.cp" ]

[ "The $3 \\times 3$ grid can be colored like a checkerboard with alternating black and white squares.\nLet the top left square be white, and the rest of the squares alternate colors.\nEach $2 \\times 1$ rectangle always covers $1$ white square and $1$ black square.\nYou can ensure that at least one of your guessed squares is covered by the rectangle by choosing either each of the white squares ( $5$ turns) or each of the black squares ( $4$ turns).\nSince it is ideal to be the most efficient with our turns, by choosing all the black squares, we guarantee that one of the $\\boxed{4}$ squares are of the $2 \\times 1$ rectangle." ]

[ "Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins.\nYou must have $1$ more penny. If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes, and quarters. But you only have $5$ more coins to assign.\nNow you have $61 - 1 = 60$ cents remaining for $4$ coins, which may be nickels, quarters, or dimes. If you have only one more dime, that leaves $50$ cents in $3$ nickels or quarters, which is impossible. If you have two dimes, that leaves $40$ cents for $2$ nickels or quarters, which is again impossible. If you have three dimes, that leaves $30$ cents for $1$ nickel or quarter, which is still impossible. And all four remaining coins being dimes will not be enough.\nTherefore, you must have no more dimes to assign, and the $60$ cents in $4$ coins must be divided between the quarters and nickels. We quickly see that $2$ nickels and $2$ quarters work. Thus, the total count is $2$ quarters, $2$ nickels, $1$ penny, plus one more coin of each type that we originally subtracted. Double-checking, that gives a total $2 + 2 + 1 + 4 = 9$ coins, and a total of $2\\cdot 25 + 2\\cdot 5 + 1 + (1 + 5 + 10 + 25) = 102$ cents.\nThere is only $1$ dime in that combo, so the answer is $\\boxed{1}$", "We see that there must be 102 cents, so therefore there's at least 2 pennies. That leaves 7 coins. We assume that there are 3 quarters, leaving 25 cents with 4 coins left. If all 4 are nickels, that would only be 20 cents, missing 5. Therefore, one nickel must be changed into 1 dime, so the answer is $\\boxed{1}$", "It is clear that there should only be $2$ pennies; any more would take up too many coins, and the limit is $9$ . Now we have $ $1$ left, and $7$ coins to use. Looking at the quarters, we can make methodical guesses. If there is $1$ quarter, then we would have to make $ $0.75$ with $6$ coins. We take a few educated guesses for the nickel and dime combos, and see that $1$ quarter will not work. Trying values for $2$ quarters, we see that this will not work either. When we reach $3$ quarters, the remaining is $ $0.25$ made from $4$ coins. We try with $2$ dimes, which does not work (it only takes $3$ coins) and we try with $1$ dime. After trying $2$ pennies, $3$ quarters, $1$ dime, and $3$ nickels, it is evident that this combo works. Therefore, the answer is $\\boxed{1}$" ]

[ "Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$ , respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each other. If we place $S$ and $T$ next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. $ST\\square\\square, \\square ST\\square, \\square\\square ST$ ). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $TS\\square\\square, \\square TS\\square, \\square\\square TS$ ). Thus there are 6 ways of placing $S$ and $T$ directly next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$ . Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\\times 2=12$ ways to place $S$ and $T$ next to each other. Subtracting this from the total number of arrangements gives us $24-12=12$ total arrangements $\\implies\\boxed{12}$", "Let's try complementary counting. There $4!$ ways to arrange the 4 marbles. However, there are $2\\cdot3!$ arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, \\[4!-2\\cdot3!=\\boxed{12}\\]", "We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a \"super marble.\" There are $2!$ ways to arrange Steelie and Tiger within this \"super marble.\" Then there are $3!$ ways to arrange the \"super marble\" and Zara's two other marbles in a row. Since there are $4!$ ways to arrange the marbles without any restrictions, the answer is given by $4!-2!\\cdot 3!=\\boxed{12}$", "We will use the following\n$\\textbf{Georgeooga-Harryooga Theorem:}$ The Georgeooga-Harryooga Theorem states that if you have $a$ distinguishable objects and $b$ of them cannot be together, then there are $\\frac{(a-b)!(a-b+1)!}{b!}$ ways to arrange the objects.\n$\\textit{Proof. (Created by AoPS user RedFireTruck)}$\nLet our group of $a$ objects be represented like so $1$ $2$ $3$ , ..., $a-1$ $a$ . Let the last $b$ objects be the ones we can't have together.\nThen we can organize our objects like so $\\square1\\square2\\square3\\square...\\square a-b-1\\square a-b\\square$\nWe have $(a-b)!$ ways to arrange the objects in that list.\nNow we have $a-b+1$ blanks and $b$ other objects so we have $_{a-b+1}P_{b}=\\frac{(a-b+1)!}{(a-2b+1)!}$ ways to arrange the objects we can't put together.\nBy fundamental counting principle our answer is $\\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$\nProof by RedFireTruck talk ) 12:09, 1 February 2021 (EST)\nBack to the problem. By the Georgeooga-Harryooga Theorem , our answer is $\\frac{(4-2)!(4-2+1)!}{(4-2\\cdot2+1)!}=\\boxed{12}$" ]

[ "For the next five races, Zou wins four and loses one. Let $W$ and $L$ denote a win and a loss, respectively. There are five possible outcome sequences for Zou:\nWe proceed with casework:\nCase (1): Sequences #1-4, in which Zou does not lose the last race.\nThe probability that Zou loses a race is $\\frac13,$ and the probability that Zou wins the next race is $\\frac13.$ For each of the three other races, the probability that Zou wins is $\\frac23.$\nThere are four sequences in this case. The probability of one such sequence is $\\left(\\frac13\\right)^2\\left(\\frac23\\right)^3.$\nCase (2): Sequence #5, in which Zou loses the last race.\nThe probability that Zou loses a race is $\\frac13.$ For each of the four other races, the probability that Zou wins is $\\frac23.$\nThere is one sequence in this case. The probability is $\\left(\\frac13\\right)^1\\left(\\frac23\\right)^4.$\nAnswer\nThe requested probability is \\[4\\left(\\frac13\\right)^2\\left(\\frac23\\right)^3+\\left(\\frac13\\right)^1\\left(\\frac23\\right)^4=\\frac{32}{243}+\\frac{16}{243}=\\frac{48}{243}=\\frac{16}{81},\\] from which the answer is $16+81=\\boxed{097}.$", "We have $5$ cases, depending on which race Zou lost. Let $\\text{W}$ denote a won race, and $\\text{L}$ denote a lost race for Zou. The possible cases are $\\text{WWWWL, WWWLW, WWLWW, WLWWW, LWWWW}$ . The first case has probability $\\left(\\frac{2}{3} \\right)^4 \\cdot \\frac{1}{3} = \\frac{16}{3^5}$ . The second case has probability $\\left( \\frac{2}{3} \\right)^3 \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} = \\frac{8}{3^5}$ . The third has probability $\\left( \\frac{2}{3} \\right)^2 \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{2}{3} = \\frac{8}{3^5}$ . The fourth has probability $\\frac{2}{3} \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\left( \\frac{2}{3} \\right)^2 = \\frac{8}{3^5}$ . Lastly, the fifth has probability $\\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\left( \\frac{2}{3} \\right)^3 = \\frac{8}{3^5}$ . Adding these up, the total probability is $\\frac{16 + 8 \\cdot 4}{3^5} = \\frac{16 \\cdot 3}{3^5} = \\frac{16}{81}$ , so $m+n = \\boxed{097}$", "Case 1: Zou loses the first race\nIn this case, Zou must win the rest of the races. Thus, our probability is $\\frac{8}{243}$\nCase 2: Zou loses the last race\nThere is only one possibility for this, so our probability is $\\frac{16}{243}$\nCase 3: Neither happens\nThere are three ways that this happens. Each has one loss that is not the last race. Therefore, the probability that one happens is $\\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{2}{3} \\cdot \\frac{2}{3} \\cdot \\frac{2}{3} = \\frac{8}{243}$ . Thus, the total probability is $\\frac{8}{243} \\cdot 3 = \\frac{24}{243}$\nAdding these up, we get $\\frac{48}{243} = \\frac{16}{81}$ , so $16+81=\\boxed{097}$", "Note that Zou wins one race. The probability that he wins the last race is $\\left(\\frac{2}{3}\\right)^4\\left(\\frac{1}{3}\\right)=\\frac{16}{243}.$ Now, if he doesn't win the last race, then there must be two races where the winner of the previous race loses. We can choose any $4$ of the middle races for Zou to win. So the probability for this case is $4\\left(\\frac{2}{3}\\right)^3\\left(\\frac{1}{3}\\right)^2=\\frac{32}{243}.$ Thus, the answer is $\\frac{16}{243}+\\frac{32}{243}=\\frac{16}{81}\\implies\\boxed{097}.$" ]

[ "Small correction: The writer below has maximized the area of the rectangle (with sides parallel to the walls) that fits around the table, but there is a larger single dimension we can find in the table. The height or width is maximized when the diagonal of the table is horizontal or vertical. By the Pythagorean Theorem, this diagonal is $\\sqrt{8^2+10^2} = \\sqrt{164},$ which is between $\\sqrt{144}$ and $\\sqrt{169},$ so the answer is still $\\textbf{(C)}.$ -hailstone\nWe begin by thinking about the motion of the table. As it moves, the table will have it's maximum height and width when the rectangle's sides form $45$ degree angles relative to the sides of the square. Therefore, by the Pythagorean Theorem, we have that $S= 4\\sqrt{2}+5\\sqrt{2}$ , with $4\\sqrt{2}$ being the length of the leg formed by the side of the square with length $8$ and $5\\sqrt{2}$ being the length of the leg formed by the side of the square with length $10$ . Adding these up yields $9\\sqrt{2}$\nWe have that $\\sqrt{2}\\approx 1.414\\approx 1.4$ . That means that $9\\sqrt{2}\\approx 12.6$ , which rounds up to $\\boxed{13}$" ]

[ "Solution by e_power_pi_times_i\nSince the dimensions of $DEFG$ are half of the dimensions of $ABCD$ , the area of $DEFG$ is $\\dfrac{1}{2}\\cdot\\dfrac{1}{2}$ of $ABCD$ , so the area of $ABCD$ is $\\dfrac{1}{4}\\cdot72 = \\boxed{18}$" ]

[ "Solution by e_power_pi_times_i\nLet $F$ be the point such that $DF$ and $CF$ are parallel to $CE$ and $DE$ , respectively, and let $DE = x$ and $BE^2 = 169-x^2$ . Then, $[FDEC] = x(4+\\sqrt{169-x^2}) = [ABC] + [BED] + [ABD] + [AFD] = 6 + \\dfrac{x\\sqrt{169-x^2}}{2} + 30 + \\dfrac{(x-3)(4+\\sqrt{169-x^2})}{2}$ . So, $4x+x\\sqrt{169-x^2} = 60 + x\\sqrt{169-x^2} - 3\\sqrt{169-x^2}$ . Simplifying $3\\sqrt{169-x^2} = 60 - 4x$ , and $1521 - 9x^2 = 16x^2 - 480x + 3600$ . Therefore $25x^2 - 480x + 2079 = 0$ , and $x = \\dfrac{48\\pm15}{5}$ . Checking, $x = \\dfrac{63}{5}$ is the answer, so $\\dfrac{DE}{DB} = \\dfrac{\\dfrac{63}{5}}{13} = \\dfrac{63}{65}$ . The answer is $\\boxed{128}$", "Solution by Arjun Vikram\nExtend lines $AD$ and $CE$ to meet at a new point $F$ . Now, we see that $FAC\\sim FDE \\sim ACB$ . Using this relationship, we can see that $AF=\\frac{15}4$ , (so $FD=\\frac{63}4$ ), and the ratio of similarity between $FDE$ and $FAC$ is $\\frac{63}{15}$ . This ratio gives us that $\\frac{63}5$ . By the Pythagorean Theorem, $DB=13$ . Thus, $\\frac{DE}{DB}=\\frac{63}{65}$ , and the answer is $63+65=\\boxed{128}$" ]

[ "We first notice that the paper strips cover up part of the others. Since the width of the overlap is $1$ and the length of the overlap is $1$ , the area of \neach of the strips with the overlap is $(10\\cdot 1)-1=9$ . Since there are 4 strips, $4\\cdot 9=36 \\implies \\boxed{36}$" ]

[ "Because the arch has a height of $16$ inches, an equation that models the arch is $y = ax^2 + 16$ , where $x$ is the horizontal distance from the center and $y$ is the height. The arch has a span of $40$ inches, so the arch meets the ground $20$ inches from the center. That means $0 = 400a + 16$ , so $a = -\\frac{1}{25}$\nThus, the equation that models height based on distance from the center $y = -\\frac{1}{25}x^2 + 16$ , so the height of the arch $5$ inches from the center is $-\\frac{1}{25} \\cdot 5^2 + 16 = \\boxed{15}$ inches." ]

[ "Solution by e_power_pi_times_i\nNotice that $\\measuredangle DAE = 90^\\circ+60^\\circ = 150^\\circ$ and that $AD = AE$ . Then triangle $ADE$ is isosceles, so $\\measuredangle AED = \\dfrac{180^\\circ-150^\\circ}{2} = \\boxed{15}$", "WLOG, let the side length of the square and the equilateral triangle be $1$ $\\angle{DAE}=90^\\circ+60^\\circ=150^\\circ$ . Apply the law of cosines then the law of sines, we find that $\\angle{AED}=15^\\circ$ . Select $\\boxed{15}$" ]

[ "We know that triangle $RCQ$ is similar to triangle $ABC$ . We draw a line to point $D$ on hypotenuse $AB$ such that $\\angle QDR$ is $90 ^\\circ$ and that $RDQC$ is a rectangle. Since triangle $RCQ$ is similar to triangle $ABC$ , let $RC$ be $4x$ and $RD/CQ$ be $3x$ . Now we have line segment $AQ$ $8-3x$ , and line segment $RB$ $6-4x$ . Since $BD + DA = AB$ , we use simple algebra and Pythagorean Theorem to get $\\sqrt {(3x)^2 + (6-4x)^2}$ $\\sqrt {(4x)^2 + (8-3x)^2}$ $10$ . Expanding and simplifying gives us $\\sqrt {25x^2-48x+36}$ $\\sqrt {25x^2-48x+64}$ $10$\nSquaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by $\\sqrt {25x^2-48x+36}$ . Now, we can square both sides and simplify to get $0 = 72 - 20 \\sqrt{25x^2-48x+36}$ . Dividing both sides by $4$ , we get $18 - 5 \\sqrt {25x^2-48x+36}$ $0$ . We then add $5 \\sqrt {25x^2-48x+36}$ to both sides to get $18 = 5 \\sqrt {25x^2-48x+36}$ . Since this is very messy, let $25x^2 - 48x = y$ . Squaring both sides, we get $324 = 25y + 900, 25y = -576$ . Solving for $y$ , we have $y = -23.04$ . Plugging in $y$ as $25x^2-48x$ , we have $25x^2-48x+23.04 = 0$ . Using the quadratic equation, we get $\\frac {48+0}{50}$ . Therefore, $x = \\frac {48}{50}$\nRemember that our desired answer is the hypotenuse of the triangle $3x - 4x - 5x$ . Since $5x$ is the hypotenuse, our answer is $\\boxed{4.8}$" ]

[ "Solution by e_power_pi_times_i\nIf arcs $AB$ $BC$ , and $CD$ are congruent, then $\\measuredangle ACB = \\measuredangle BDC = \\measuredangle CBD = \\theta$ . Because $ABCD$ is cyclic, $\\measuredangle CAD = \\measuredangle CBD = \\theta$ , and $\\measuredangle ADB = \\measuredangle ACB = \\theta$ . Then, $\\measuredangle EAD = \\measuredangle EDA = \\dfrac{180^\\circ - 40^\\circ}{2} = 70^\\circ$ $\\theta = 55^\\circ$ $\\measuredangle ACD = 180^\\circ - 55^\\circ - 110^\\circ = \\boxed{15}$" ]

[ "Solution by e_power_pi_times_i\nBecause $AB = OD$ , triangles $ABO$ and $BOE$ are isosceles. Denote $\\measuredangle BAO = \\measuredangle AOB = \\theta$ . Then $\\measuredangle ABO = 180^\\circ-2\\theta$ , and $\\measuredangle EBO = \\measuredangle OEB = 2\\theta$ , so $\\measuredangle BOE = 180^\\circ-4\\theta$ . Notice that $\\measuredangle AOB + \\measuredangle BOE + 45^\\circ = 180^\\circ$ . Therefore $\\theta+180-4\\theta = 135^\\circ$ , and $\\theta = \\boxed{15}$", "Draw $BO$ . Let $y = \\angle BAO$ . Since $AB = OD = BO$ , triangle $ABO$ is isosceles, so $\\angle BOA = \\angle BAO = y$ . Angle $\\angle EBO$ is exterior to triangle $ABO$ , so $\\angle EBO = \\angle BAO + \\angle BOA = y + y = 2y$\nTriangle $BEO$ is isosceles, so $\\angle BEO = \\angle EBO = 2y$ . Then $\\angle EOD$ is external to triangle $AEO$ , so $\\angle EOD = \\angle EAO + \\angle AEO = y + 2y = 3y$ . But $\\angle EOD = 45^\\circ$ , so $\\angle BAO = y = 45^\\circ/3 = \\boxed{15}$" ]

[ "By the associative property , we can rearrange the numbers in the numerator and the denominator. [katex display=true]\\frac{3}{3}\\cdot \\frac{5}{5}\\cdot\\frac{7}{7}\\cdot\\frac{9}{9}\\cdot\\frac{11}{11}=1\\cdot1\\cdot1\\cdot1\\cdot1=\\boxed{1}[/katex]", "Notice that the $9 \\times 11$ in the denominator of the first fraction cancels with the same term in the second fraction, the $7$ s in the numerator and denominator of the second fraction cancel, and the $3 \\times 5$ in the numerator of the first fraction cancels with the same term in the denominator second fraction. Then everything in the expression cancels, leaving us with $\\boxed{1}$" ]

[ "Let $r$ be the length of the radius of the circle. A right triangle is formed by half of the chord, half of the radius (since the chord bisects it), and the radius. Thus, it is a $30^\\circ$ $60^\\circ$ $90^\\circ$ triangle , and the area of two such triangles is $2 \\cdot \\frac{1}{2} \\cdot \\frac{r}{2} \\cdot \\frac{r\\sqrt{3}}{2} = \\frac{r^2\\sqrt{3}}{4}$ . The central angle which contains the entire chord is $60 \\cdot 2 = 120$ degrees , so the area of the sector is $\\frac{1}{3}r^2\\pi$ ; the rest of the area of the circle is then equal to $\\frac{2}{3}r^2\\pi$\nThe smaller area cut off by the chord is equal to the area of the sector minus the area of the triangle. The larger area is equal to the area of the circle not within the sector and the area of the triangle. Thus, the desired ratio is $\\frac{\\frac{2}{3}r^2\\pi + \\frac{r^2\\sqrt{3}}{4}}{\\frac{1}{3}r^2\\pi - \\frac{r^2\\sqrt{3}}{4}} = \\frac{8\\pi + 3\\sqrt{3}}{4\\pi - 3\\sqrt{3}}$\nTherefore, $abcdef = 2^53^4 = 2592 \\Longrightarrow \\boxed{592}$" ]

[ "The two lines can both intersect the circle twice, and can intersect each other once, so $2+2+1= \\boxed{5}.$" ]

[ "Call the center of the circle $O$ . By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.\nThus, $\\angle{AOP}+\\angle{POB}+\\angle{COQ}+\\angle{QOD}=180$ , or $(\\arctan(\\tfrac{19}{r})+\\arctan(\\tfrac{26}{r}))+(\\arctan(\\tfrac{37}{r})+\\arctan(\\tfrac{23}{r}))=180$\nTake the $\\tan$ of both sides and use the identity for $\\tan(A+B)$ to get \\[\\tan(\\arctan(\\tfrac{19}{r})+\\arctan(\\tfrac{26}{r}))+\\tan(\\arctan(\\tfrac{37}{r})+\\arctan(\\tfrac{23}{r}))=n\\cdot0=0.\\]\nUse the identity for $\\tan(A+B)$ again to get \\[\\frac{\\tfrac{45}{r}}{1-19\\cdot\\tfrac{26}{r^2}}+\\frac{\\tfrac{60}{r}}{1-37\\cdot\\tfrac{23}{r^2}}=0.\\]\nSolving gives $r^2=\\boxed{647}$", "Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on ( $a, b, c,$ and $d$ are the tangent lengths, not the side lengths). \\[A = \\sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\\sqrt{647}\\] $r^2=\\frac{A^2}{(a+b+c+d)^2} = \\boxed{647}$", "Using the formulas established in solution 2, one notices: \\[r^2=\\frac{A^2}{(a+b+c+d)^2}\\] \\[r^2=\\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}\\] \\[r^2=\\frac{abc+bcd+acd+abd}{a+b+c+d}\\] \\[r^2=\\boxed{647}\\]" ]

[ "The location of the center of the circle must be in the $34 \\times 13$ rectangle that is one unit away from the sides of rectangle $ABCD$ . We want to find the area of the right triangle with hypotenuse one unit away from $\\overline{AC}$ . Let this triangle be $A'B'C'$\nNotice that $ABC$ and $A'B'C'$ share the same incenter ; this follows because the corresponding sides are parallel, and so the perpendicular inradii are concurrent, except that the inradii of $\\triangle ABC$ extend one unit farther than those of $\\triangle A'B'C'$ . From $A = rs$ , we note that $r_{ABC} = \\frac{[ABC]}{s_{ABC}} = \\frac{15 \\cdot 36 /2}{(15+36+39)/2} = 6$ . Thus $r_{A'B'C'} = r_{ABC} - 1 = 5$ , and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, $[A'B'C'] = [ABC] \\cdot \\left(\\frac{r_{A'B'C'}}{r_{ABC}}\\right)^2 = \\frac{15 \\times 36}{2} \\cdot \\frac{25}{36} = \\frac{375}{2}$\nThe probability is $\\frac{2[A'B'C']}{34 \\times 13} = \\frac{375}{442}$ , and $m + n = \\boxed{817}$", "2004 I AIME-10.png\nLet the bisector of $\\angle CAD$ be $AE$ , with $E$ on $CD$ . By the angle bisector theorem, $DE = 36/5$ . Since $\\triangle AOR \\sim \\triangle AED$ $O$ is the center of the circle), we find that $AR = 5$ since $OR = 1$ . Also $AT = 35$ so $RT = OQ = 30$\nWe can apply the same principle again to find that $PT = 27/2$ , and since $QT = 1$ , we find that $PQ = 27/2 - 1 = 25/2$ . The locus of all possible centers of the circle on this \"half\" of the rectangle is triangle $\\triangle OPQ$ . There exists another congruent triangle that is symmetric over $AC$ that has the same area as triangle $\\triangle OPQ$ $\\triangle APQ$ has area $\\frac {1}{2}\\cdot OP \\cdot PQ = \\frac {1}{2}\\cdot 30\\cdot \\frac {25}{2}$ , since $\\angle OQP$ is right. Thus the total area that works is $30\\cdot \\frac {25}{2} = 375$ , and the area of the locus of all centers of any circle with radius 1 is $34\\cdot 13 = 442$ . Hence, the desired probability is $\\frac {375}{442}$ , and our answer is $\\boxed{817}$", "2004 I AIME-10b.png\nAgain, the location of the center of the circle must be in the $34 \\times 13$ rectangle that is one unit away from the sides of rectangle $ABCD$ . We want to find the area of the right triangle with hypotenuse one unit away from $\\overline{AC}$\nLet $A$ be at the origin, $B (36,0)$ $C (36,15)$ $D (0,15)$ . The slope of $\\overline{AC}$ is $\\frac{15}{36} = \\frac{5}{12}$ . Let $\\triangle A'B'C'$ be the right triangle with sides one unit inside $\\triangle ABC$ . Since $\\overline{AC} || \\overline{A'C'}$ , they have the same slope, and the equation of $A'C'$ is $y = \\frac{5}{12}x + c$ . Manipulating, $5x - 12y + 12c = 0$ . We need to find the value of $c$ , which can be determined since $\\overline{AC}$ is one unit away from $\\overline{A'C'}$ . Since the diagonal contains the origin, we can use the distance from a point to the line formula at the origin:\n\\[\\left|\\frac{Ax + By + C}{\\sqrt{A^2+B^2}}\\right| = 1 \\Longrightarrow \\left|\\frac{(5)(0) + (-12)(0) + 12c}{\\sqrt{5^2 + (-12)^2}}\\right| = 1\\] \\[c = \\pm \\frac{13}{12}\\]\nThe two values of $c$ correspond to the triangle on top and below the diagonal. We are considering $A'B'C'$ which is below, so $c = -\\frac{13}{12}$ . Then the equation of $\\overline{A'C'}$ is $y = \\frac{5}{12}x - \\frac{13}{12}$ . Solving for its intersections with the lines $y = 1, x = 35$ (boundaries of the internal rectangle), we find the coordinates of $A'B'C'$ are at $A'\\ (5,1)\\ B'\\ (35,1)\\ C'\\ (35,\\frac{27}{2})$ . The area is $\\frac{1}{2}bh = \\frac{1}{2}(35-5)\\left(\\frac{27}{2} - 1\\right) = \\frac{375}{2}$\nFinally, the probability is $\\frac{2\\cdot \\mathrm{area\\ of\\ triangle}}{34 \\times 13} = \\frac{375}{442}$ , and $m + n = \\boxed{817}$" ]

[ "We begin as in the first solution. Once we see that $\\triangle EOF$ has side lengths $12$ $20$ , and $24$ , we can compute its area with Heron's formula:\n\\[K = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{28\\cdot 16\\cdot 8\\cdot 4} = 32\\sqrt{14}.\\]\nThus, the circumradius of triangle $\\triangle EOF$ is $R = \\frac{abc}{4K} = \\frac{45}{\\sqrt{14}}$ . Looking at $EPFO$ , we see that $\\angle OEP = \\angle OFP = 90^\\circ$ , which makes it a cyclic quadrilateral. This means $\\triangle EOF$ 's circumcircle and $EPFO$ 's inscribed circle are the same.\nSince $EPFO$ is cyclic with diameter $OP$ , we have $OP = 2R = \\frac{90}{\\sqrt{14}}$ , so $OP^2 = \\frac{4050}{7}$ and the answer is $\\boxed{057}$", "Let $OP=x$\nProceed as the first solution in finding that quadrilateral $EPFO$ has side lengths $OE=20$ $OF=24$ $EP=\\sqrt{x^2-20^2}$ , and $PF=\\sqrt{x^2-24^2}$ , and diagonals $OP=x$ and $EF=12$\nWe note that quadrilateral $EPFO$ is cyclic and use Ptolemy's theorem to solve for $x$\n\\[20\\cdot \\sqrt{x^2-24^2} + 12\\cdot x = 24\\cdot \\sqrt{x^2-20^2}\\]\nSolving, we have $x^2=\\frac{4050}{7}$ so the answer is $\\boxed{057}$", "Let $M$ be the midpoint of $AB$ and $N$ of $CD$ . As $\\angle OMP = \\angle ONP$ , quadrilateral $OMPN$ is cyclic with diameter $OP$ . By Cyclic quadrilaterals note that $\\angle MPO = \\angle MNO$\nThe area of $\\triangle MNP$ can be computed by Herons as \\[[MNO] = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{28\\cdot 16\\cdot 8\\cdot 4} = 32\\sqrt{14}.\\] The area is also $\\frac{1}{2}ON \\cdot MN \\sin{\\angle MNO}$ . Therefore, \\begin{align*} \\sin{\\angle MNO} &= \\frac{2[MNO]}{ON \\cdot MN} \\\\ &= \\frac{2}{9}\\sqrt{14} \\\\ \\sin{\\angle MNO} &= \\frac{OM}{OP} \\\\ &= \\frac{2}{9}\\sqrt{14} \\\\ OP &= \\frac{90\\sqrt{14}}{14} \\\\ OP^2 &= \\frac{4050}{7} \\implies \\boxed{057}", "Define $M$ and $N$ as the midpoints of $AB$ and $CD$ , respectively. Because $\\angle OMP = \\angle ONP = 90^{\\circ}$ , we have that $ONPM$ is a cyclic quadrilateral. Hence, $\\angle PNM = \\angle POM.$ Then, let these two angles be denoted as $\\alpha$ . \nNow, assume WLOG that $PD = x < 7$ and $PB = y < 15$ (We can do this because one of $PD$ or $PC$ must be less than 7, and similarly for $PB$ and $PA$ ). Then, by Power of a Point on P with respect to the circle with center $O$ , we have that \\[(14-x)x = (30-y)y\\] \\[(7-x)^{2}+176=(15-y)^{2}.\\] Then, let $z = (7-x)^{2}$ . From Law of Cosines on $\\triangle NMP$ , we have that \\[\\textrm{cos } \\angle MNP = \\frac{NP^{2}+MN^{2}-MP^{2}}{2 \\cdot NP \\cdot MN}\\] \\[\\textrm{cos } \\alpha = \\frac{(7-x)^{2} + 12^{2} - (14-x)^{2}}{24 \\cdot (7-x)}.\\] Plugging in $z$ in gives \\[\\textrm{cos } \\alpha = \\frac{-32}{24 \\cdot \\sqrt{z}}\\] \\[\\textrm{cos } \\alpha = \\frac{-4}{3\\sqrt{z}}\\] \\[\\textrm{cos }^{2} \\alpha = \\frac{16}{9z}.\\] Hence, \\[\\textrm{tan }^{2} \\alpha = \\frac{\\frac{9z-16}{9z}}{\\frac{16}{9z}} = \\frac{9z-16}{16}.\\] Then, we also know that \\[\\textrm{tan } \\alpha = \\textrm{tan } \\angle MOP = \\frac{MP}{OM} = \\frac{14-y}{20}.\\] Squaring this, we get \\[\\textrm{tan }^{2} \\alpha = \\frac{z+176}{400}.\\] Equating our expressions for $z$ , we get $\\frac{z+176}{400} = \\frac{9z-16}{16}.$ Solving gives us that $z = \\frac{18}{7}$ .\nSince $\\angle ONP = 90^{\\circ}$ , from the Pythagorean Theorem, $OP^{2} = ON^{2}+PN^{2} = 25^{2}-7^{2} + z = 576+z = \\frac{4050}{7}$ ,\nand thus the answer is $4050+7 = 4057$ , which when divided by a thousand leaves a remainder of $\\boxed{57}.$", "2011 AIMEII Problem 10 CASE 2.png\nLet $E$ and $F$ be the midpoints of $\\overline{AB}$ and $\\overline{CD}$ , respectively, such that $\\overline{BE}$ intersects $\\overline{CF}$\nSince $E$ and $F$ are midpoints, $BE = 15$ and $CF = 7$\n$B$ and $C$ are located on the circumference of the circle, so $OB = OC = 25$\nSince $\\overline{OE}\\perp \\overline{AB}$ and $\\overline{OF}\\perp \\overline{CD}$ $OE = \\sqrt{OB^2-BE^2}=20$ and $OF = \\sqrt{OC^2-OF^2}=24$\nWith law of cosines, $\\cos \\angle EOF = \\frac{OE^2+OF^2-EF^2}{2\\cdot OE\\cdot OF} = \\frac{13}{15}$\nSince $EF < OF$ $\\angle EOF$ is acute angle. $\\sin \\angle EOF = \\sqrt{1-\\cos^2 \\angle EOF} = \\frac{\\sqrt{56}}{15}$ and $\\tan \\angle EOF = \\frac{\\sqrt{56}}{13}$\nLet $\\overline{OF}$ line be $x$ axis.\nLine $\\overline{DC}$ equation is $x = OF$\nSince line $\\overline{AB}$ passes point $E$ and perpendicular to $\\overline{OD}$ , its equation is $y - E_y = -\\frac{1}{\\tan \\angle EOF} (x - E_x)$\nwhere $E_x = OE\\cos{\\angle EOF}$ $E_y = OE\\sin{\\angle EOF}$\nSince $P$ is the intersection of $\\overline{AB}$ and $\\overline{CD}$\n$P_x = OF = 24$\n$P_y = E_y -\\frac{1}{\\tan \\angle EOF} (OF - E_x) = - \\frac{3\\sqrt{14}}{7}$ (Negative means point $P$ is between point $F$ and $C$\n$OP^2 = P_x^2 + P_y^2 = \\frac{4050}{7}$ and the answer is $\\boxed{057}$" ]

[ "Every pair of vertices of the polyhedron determines either an edge , a face diagonal or a space diagonal. We have ${26 \\choose 2} = \\frac{26\\cdot25}2 = 325$ total line segments determined by the vertices. Of these, $60$ are edges. Each triangular face has $0$ face diagonals and each quadrilateral face has $2$ , so there are $2 \\cdot 12 = 24$ face diagonals. This leaves $325 - 60 - 24 = \\boxed{241}$ segments to be the space diagonals." ]

[ "The polyhedron described looks like this, a truncated cuboctahedron.\nThe number of segments joining the vertices of the polyhedron is ${48\\choose2} = 1128$ . We must now subtract out those segments that lie along an edge or a face.\nSince every vertex of the polyhedron lies on exactly one vertex of a square/hexagon/octagon, we have that $V = 12 \\cdot 4 = 8 \\cdot 6 = 6 \\cdot 8 = 48$\nEach vertex is formed by the intersection of 3 edges. Since every edge is counted twice, once at each of its endpoints, the number of edges $E$ is $\\frac{3}{2}V = 72$\nEach of the segments lying on a face of the polyhedron must be a diagonal of that face. Each square contributes $\\frac{n(n-3)}{2} = 2$ diagonals, each hexagon $9$ , and each octagon $20$ . The number of diagonals is thus $2 \\cdot 12 + 9 \\cdot 8 + 20 \\cdot 6 = 216$\nSubtracting, we get that the number of space diagonals is $1128 - 72 - 216 = \\boxed{840}$", "We first find the number of vertices on the polyhedron:\nThere are 4 corners per square, 6 corners per hexagon, and 8 corners per octagon. Each vertex is where 3 corners coincide, so we count the corners and divide by 3. $\\text{vertices} = \\frac{12 \\cdot 4 + 8 \\cdot 6 + 6 \\cdot 8}{3}=48$\nWe know that all vertices look the same (from the problem statement), so we should find the number of line segments originating from a vertex, and multiply that by the number of vertices, and divide by 2 (because each space diagonal is counted twice because it has two endpoints).\nCounting the vertices that are on the same face as an arbitrary vertex, we find that there are 13 vertices that aren't possible endpoints of a line originating from the vertex in the middle of the diagram. \nYou can draw a diagram to count this better: 1988AIME10.png Since 13 aren't possible endpoints, that means that there are 35 possible endpoints per vertex.\nThe total number of segments joining vertices that aren't on the same face is $48\\cdot 35\\cdot \\frac 12 = 24 \\cdot 35 = \\boxed{840}$", "Since at each vertex one square, one hexagon, and one octagon meet, then there are a total of $12 \\cdot 4 = 8 \\cdot 6 = 6 \\cdot 8 = 48$ vertices. This means that for each segment we have $48$ choices of vertices for the first endpoint of the segment.\nSince each vertex is the meeting point of a square, octagon, and hexagon, then there are $3$ other vertices of the square that are not the first one, and connecting the first point to any of these would result in a segment that lies on a face or edge.\nSimilarly, there are $5$ points on the adjacent hexagon and $7$ points on adjoining octagon that, when connected to the first point, would result in a diagonal or edge.\nHowever, the square and hexagon share a vertex, as do the square and octagon, and the hexagon and octagon.\nSubtracting these from the $47$ vertices we have left to choose from, and adding the $3$ that we counted twice, we get\n\\[48 \\cdot (47 - 3 - 5 - 7 + 3) = 48 \\cdot 35 = 1680\\]\nWe over-counted, however, as choosing vertex $A$ then $B$ is the same thing as choosing $B$ then $A$ , so we must divide $1680 / 2 = \\boxed{840}$", "In the same ways as above, we find that there are 48 vertices. Now, notice that there are $\\binom{48}{2}$ total possible ways to choose two vertices. However, we must remove the cases where the segments do not lie in the interior of the polyhedron. We get\n\\[\\binom{48}{2}-12\\binom{4}{2}-8\\binom{6}{2}-6\\binom{8}{2}=768\\]\nWe remover all the possible edges of the squares, hexagons, and octagons. However, we have undercounted! We must add back the number of edges because when we subtracted the three binomials from $\\binom{48}{2}$ we removed each edge twice (each edge is shared by two polygons). This means that we need to add back the number of edges, 72. Thus, we get $768+72=\\boxed{840}$" ]

[ "Clearly, at least $5$ tails must be flipped; any less, then by the Pigeonhole Principle there will be heads that appear on consecutive tosses.\nConsider the case when $5$ tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled $(H)$\nThere are six slots for the heads to be placed, but only $5$ heads remaining. Thus, using stars-and-bars there are ${6\\choose5}$ possible combinations of 6 heads. Continuing this pattern, we find that there are $\\sum_{i=6}^{11} {i\\choose{11-i}} = {6\\choose5} + {7\\choose4} + {8\\choose3} + {9\\choose2} + {{10}\\choose1} + {{11}\\choose0} = 144$ . There are a total of $2^{10}$ possible flips of $10$ coins, making the probability $\\frac{144}{1024} = \\frac{9}{64}$ . Thus, our solution is $9 + 64 = \\boxed{073}$", "Call the number of ways of flipping $n$ coins and not receiving any consecutive heads $S_n$ . Notice that tails must be received in at least one of the first two flips.\nIf the first coin flipped is a T, then the remaining $n-1$ flips must fall under one of the configurations of $S_{n-1}$\nIf the first coin flipped is a H, then the second coin must be a T. There are then $S_{n-2}$ configurations.\nThus, $S_n = S_{n-1} + S_{n-2}$ . By counting, we can establish that $S_1 = 2$ and $S_2 = 3$ . Therefore, $S_3 = 5,\\ S_4 = 8$ , forming the Fibonacci sequence . Listing them out, we get $2,3,5,8,13,21,34,55,89,144$ , and the 10th number is $144$ . Putting this over $2^{10}$ to find the probability, we get $\\frac{9}{64}$ . Our solution is $9+64=\\boxed{073}$", "We can also split the problem into casework.\nCase 1: 0 Heads\nThere is only one possibility.\nCase 2: 1 Head\nThere are 10 possibilities.\nCase 3: 2 Heads\nThere are 36 possibilities.\nCase 4: 3 Heads\nThere are 56 possibilities.\nCase 5: 4 Heads\nThere are 35 possibilities.\nCase 6: 5 Heads\nThere are 6 possibilities.\nWe have $1+10+36+56+35+6=144$ , and there are $1024$ possible outcomes, so the probability is $\\frac{144}{1024}=\\frac{9}{64}$ , and the answer is $\\boxed{073}$" ]

[ "If $f(2+x)=f(2-x)$ , then substituting $t=2+x$ gives $f(t)=f(4-t)$ . Similarly, $f(t)=f(14-t)$ . In particular, \\[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\\]\nSince $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\\pmod{10}$ are also roots. To see that these may be the only integer roots, observe that the function \\[f(x) = \\sin \\frac{\\pi x}{10}\\sin \\frac{\\pi (x-4)}{10}\\] satisfies the conditions and has no other roots.\nIn the interval $-1000\\leq x\\leq 1000$ , there are $201$ multiples of $10$ and $200$ numbers that are congruent to $4 \\pmod{10}$ , therefore the minimum number of roots is $\\boxed{401}$", "We notice that the function has reflectional symmetry across both $x=2$ and $x=7$ . We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\\pm 10$ as roots. Continuing this shows that the roots are $0 \\mod 10$ or $4 \\mod 10$ . There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of $\\boxed{401}$ $QED \\blacksquare$", "Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only counting the total number of x values so ignore the double counted values. This means that $x = 0, \\pm 5, \\pm 10, \\pm 15... \\pm 1000$ so the answer is 400 + 1 = $\\boxed{401}$", "Let $z$ be an arbitrary zero. If $z=2-x$ , then $x=2-z$ and $2+x=4-z$ . Repeat with other equation to find if $z$ is a zero then so are $4-z$ and $14-z$ . From $0$ , we get $4$ and $14$ . Now note that applying either of these twice will return $z$ , so we must apply them in an alternating fashion for distinct roots. Doing so to $4$ and $14$ returns $10$ and $-10$ , respectively. A pattern will emerge of each path hitting a multiple of $10$ after $2$ moves. Hence, we will reach $\\pm 1000$ after $200$ jumps in either direction. Including zero, there are $2\\cdot200+1=\\boxed{401}$" ]

[ "Let $x=AC=BF$ $y=AD=BE$ , and $z=AE=BD$\nPtolemy's Theorem on $ABCD$ gives $81y+31\\cdot 81=xz$ , and Ptolemy on $ACDF$ gives $x\\cdot z+81^2=y^2$ .\nSubtracting these equations give $y^2-81y-112\\cdot 81=0$ , and from this $y=144$ . Ptolemy on $ADEF$ gives $81y+81^2=z^2$ , and from this $z=135$ . Finally, plugging back into the first equation gives $x=105$ , so $x+y+z=105+144+135=\\boxed{384}$" ]

[ "Substitute the variables to determine the value of $n$ \\[n = 2 - (-2)^{2-(-2)}\\] \\[n = 2 - (-2)^4\\] \\[n = 2 - 16\\] \\[n = -14\\] The answer is $\\boxed{14}$" ]

[ "The slope of the line is $\\frac{1000-1}{100-1}=\\frac{111}{11},$ so all points on the line have the form $(1+11t, 1+111t)$ for some value of $t$ (the rise is 111 and the run is 11). Such a point has integer coordinates if and only if $t$ is an integer, and the point is strictly between $A$ and $B$ if and only if $0<t<9$ . Thus, there are $\\boxed{8}$ points with the required property.\n-Paixiao" ]

[ "We examine only the factors of $90, 91, 92, 93,$ and $94$ that are less than $13$ , because for a year to be lucky, it must have at least one factor between $1$ and $12$ to represent the month.\n$90$ has factors of $1, 2, 3, 5, 6, 9,$ and $10$ . Dividing $90$ by the last number $10$ , gives $9$ . Thus, $10/9/90$ is a valid date, and $1990$ is a lucky year.\n$91$ has factors of $1$ and $7$ . Dividing $91$ by $7$ gives $13$ , and $7/13/91$ is a valid date. Thus, $1991$ is a lucky year.\n$92$ has factors of $1, 2$ and $4$ . Dividing $92$ by $4$ gives $23$ , and $4/23/92$ is a valid date. Thus, $1992$ is a lucky year.\n$93$ has factors of $1$ and $3$ . Dividing $93$ by $3$ gives $31$ , and $3/31/93$ is a valid date. Thus, $1993$ is a lucky year.\n$94$ has factors of $1$ and $2$ . Dividing $94$ by $2$ gives $47$ , and no month has $47$ days. Thus, $1994$ is not a lucky year, and the answer is $\\boxed{1994}$" ]

[ "Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \\ldots, 99$ . Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as $110=77+22+11$ . Then, $N = 110$ , and the sum of the digits of $N$ is $1+1+0 = \\boxed{2}$", "We already know that two-digit palindromes can only be two-digit multiples of 11; which are: $11, 22, 33, 44, 55, 66, 77, 88,$ and $99$ . Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then, we start counting. $110 \\ldots$ Aha! This multiple of 11, 110, not only isn’t a palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of $N$ ’s digits is $1+1+0 = \\boxed{2}$", "As stated above, two-digit palindromes can only be two-digit multiples of 11. We can see that if we add anything that are multiples of 11 together, we will again get a multiple of 11. For instance, $11+22=33$ . Since we know this fact and we are finding the smallest value possible, we can start with the first three-digit multiple of 11 which is $110$ . Since this is not a palindrome and can be the sum of 3 two-digit palindromes (see above solutions for more details), $110$ fits the bill. We can see that the sum of $110$ 's digits is $1+1+0 = \\boxed{2}$" ]

[ "Substitute the points $(2,3)$ and $(4,3)$ into the given equation for $(x,y)$\nThen we get a system of two equations:\n$3=4+2b+c$\n$3=16+4b+c$\nSubtracting the first equation from the second we have:\n$0=12+2b$\n$b=-6$\nThen using $b=-6$ in the first equation:\n$0=1+-12+c$\n$c=\\boxed{11}$", "Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely $(3,2)$ . Thus, the form of the equation of the parabola is $y - 2 = (x - 3)^2$ . Expanding this out, we find that $c=\\boxed{11}$", "The points given have the same $y$ -value, so the vertex lies on the line $x=\\frac{2+4}{2}=3$\nThe $x$ -coordinate of the vertex is also equal to $\\frac{-b}{2a}$ , so set this equal to $3$ and solve for $b$ , given that $a=1$\n$x=\\frac{-b}{2a}$\n$3=\\frac{-b}{2}$\n$6=-b$\n$b=-6$\nNow the equation is of the form $y=x^2-6x+c$ . Now plug in the point $(2,3)$ and solve for $c$\n$y=x^2-6x+c$\n$3=2^2-6(2)+c$\n$3=4-12+c$\n$3=-8+c$\n$c=\\boxed{11}$", "Substituting y into the two equations, we get:\n$3=x^2+bx+c$\nWhich can be written as:\n$x^2+bx+c-3=0$\n$4$ and $2$ are the solutions to the quadratic. Thus:\n$c-3=4\\times2$\n$c-3=8$\n$c=\\boxed{11}$" ]

[ "Since the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follows that this triangle is a Pythagorean Triple . We know that $31$ and either $25,\\, 20$ must be the dimensions of the rectangle, since they are the largest lengths. With some trial-and-error, if we assign the shortest side, $13$ , to be the hypotenuse of the triangle, we see the $5-12-13$ triple. Indeed this works, by placing the $31$ side opposite from the $19$ side and the $25$ side opposite from the $20$ side, leaving the cutaway side to be, as before, $13$\nTo find the area of the pentagon, we subtract the area of the triangle from that of the big rectangle: $31\\cdot 25-\\frac{12\\cdot5}{2}=775-30=745\\Longrightarrow \\boxed{745}$" ]

[ "By the transversals that go through $P$ , all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = \\dfrac{ab\\sin C}{2}$ to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as $2x,\\ 3x,\\ 7x$ . Thus, the corresponding side on the large triangle is $12x$ , and the area of the triangle is $12^2 = \\boxed{144}$", "Alternatively, since the triangles are similar by $AA$ , then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume $\\dfrac{base}{height} = \\dfrac{2}{1}.$ That means that the base of $t_{1}$ is 4, the base of $t_{2}$ is 6, and the base of $t_{3}$ is 14. Since the quadrilaterals underneath $t_{1}$ and $t_{2}$ are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is $4 + 14 + 6 = 24$ . Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is $\\dfrac{1}{2} \\cdot 24 \\cdot 12 = \\boxed{144}$", "The base of $\\triangle{ABC}$ is $BC$ . Let the base of $t_1$ be $x$ , the base of $t_2$ be $y$ , and the base of $t_3$ be $z$ . Since $\\triangle{ABC}, t_1, t_2,$ and $t_3$ are all similar, the sections in $\\triangle{ABC}$ that aren't $t_1,t_2,$ or $t_3$ are all parallelograms. Hence, $BC=x+z+y$ . We can relate $t_1,t_2,$ and $t_3$ by the square root of the ratio of their areas. $\\sqrt{\\frac{4}{9}}=\\frac{2}{3}$ and $\\sqrt{\\frac{4}{49}}=\\frac{2}{7}$ so $y=\\frac{3x}{2}$ and $z=\\frac{7x}{2}$ $x+\\frac{7x}{2}+\\frac{3x}{2}=6x$ , so $\\triangle{ABC}$ has a base that is $6$ times $t_1$ $[\\triangle{ABC}]=36[t_1]=36 \\cdot 4=\\boxed{144}$", "Since the three lines through $P$ are parallel to the sides, $t_1$ $t_2$ $t_3$ , and $\\triangle{ABC}$ are similar by $AA$ similarity. Suppose the area of $\\triangle{ABC}$ is $x^2$ , so the ratio of the base of $t_1$ to the base of $t_2$ to the base of $t_3$ to the base of $\\triangle{ABC}$ is $2:3:7:x$ . Because the quadrilaterals below $t_1$ and $t_2$ are parallelograms, the base of $\\triangle{ABC}$ is equal to the sum of the bases of $t_1, t_2,$ and $t_3$ . Therefore, $x$ equals $2+3+7=12$ so the area of $\\triangle{ABC}$ equals $x^2=12^2=\\boxed{144}.$" ]

[ "Note that an ascending number is exactly determined by its digits : for any set of digits (not including 0, since the only position for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits.\nSo, there are nine digits that may be used: $1,2,3,4,5,6,7,8,9.$ Note that each digit may be present or may not be present. Hence, there are $2^9=512$ potential ascending numbers, one for each subset of $\\{1, 2, 3, 4, 5, 6, 7, 8, 9\\}$\nHowever, we've counted one-digit numbers and the empty set , so we must subtract them off to get our answer, $512-10=\\boxed{502}.$" ]

[ "We start by drawing a diagram;\n\nWe know that $\\sin \\alpha = \\frac{1}{5}$ . Since $\\sin \\alpha = \\cos (90- \\alpha)$ \\[\\cos (90- \\alpha) = \\frac{1}{5} \\implies \\cos (\\beta + \\gamma) = \\frac{1}{5}\\]\nUsing our angle sum identities, we expand this to $\\cos \\beta \\cdot \\cos \\gamma - \\sin \\beta \\cdot \\sin \\gamma = \\frac{1}{5}$ . \nWe can now use the right triangle definition of cosine and sine to rewrite this equation as;\n\\[\\frac{l}{40} \\cdot \\frac{w}{31} - \\frac{x}{40} \\cdot \\frac{y}{31} = \\frac{1}{5} \\implies lw- xy = 8 \\cdot 31 \\implies lw = xy + 31 \\cdot 8\\]\nHang on; $lw$ is the area we want to maximize! Therefore, to maximize this area we must maximize $xy = 40 \\sin \\beta \\cdot 31 \\sin \\gamma = 31 \\cdot 40 \\cdot \\frac{1}{2}( \\cos (\\beta - \\gamma) - \\cos ( \\beta + \\gamma)) = 31 \\cdot 20 \\cdot (\\cos(\\beta-\\gamma)-\\frac{1}{5})$ .\nSince $\\cos(\\beta-\\gamma)$ is the only variable component of this expression, to maximize the expression we must maximize $\\cos(\\beta-\\gamma)$ . The cosine function has a maximum value of 1, so our equation evaluates to $xy = 31 \\cdot 20 \\cdot (1-\\frac{1}{5}) = 31 \\cdot 20 \\cdot \\frac{4}{5} = 31 \\cdot 16$ (Note that at this max value, since $\\beta$ and $\\gamma$ are both acute, $\\beta-\\gamma=0 \\implies \\beta=\\gamma$ ).\nFinally, $lw = xy + 31 \\cdot 8 = 31 \\cdot 16 + 31\\cdot 8 = 31 \\cdot 24 = \\boxed{744}$", "As above, we note that angle $A$ must be acute. Therefore, let $A$ be the origin, and suppose that $Q$ is on the positive $x$ axis and $S$ is on the positive $y$ axis. We approach this using complex numbers. Let $w=\\text{cis} A$ , and let $z$ be a complex number with $|z|=1$ $\\text{Arg}(z)\\ge 0^\\circ$ and $\\text{Arg}(zw)\\le90^\\circ$ . Then we represent $B$ by $40z$ and $C$ by $31zw$ . The coordinates of $Q$ and $S$ depend on the real part of $40z$ and the imaginary part of $31zw$ . Thus \\[[AQRS]=\\Re(40z)\\cdot \\Im(31zw)=1240\\left(\\frac{z+\\overline{z}}{2}\\right)\\left(\\frac{zw-\\overline{zw}}{2i}\\right).\\] We can expand this, using the fact that $z\\overline{z}=|z|^2$ , finding \\[[AQRS]=620\\left(\\frac{z^2w-\\overline{z^2w}+w-\\overline{w}}{2i}\\right)=620(\\Im(z^2w)+\\Im(w)).\\] Now as $w=\\text{cis}A$ , we know that $\\Im(w)=\\frac15$ . Also, $|z^2w|=1$ , so the maximum possible imaginary part of $z^2w$ is $1$ . This is clearly achievable under our conditions on $z$ . Therefore, the maximum possible area of $AQRS$ is $620(1+\\tfrac15)=\\boxed{744}$", "Let $\\theta$ be the angle $\\angle BAQ$ . The height of the rectangle then can be expressed as $h = 31 \\sin (A+\\theta)$ , and the length of the rectangle can be expressed as $l = 40\\cos \\theta$ . The area of the rectangle can then be written as a function of $\\theta$ $[AQRS] = a(\\theta) = 31\\sin (A+\\theta)\\cdot 40 \\cos \\theta = 1240 \\sin (A+\\theta) \\cos \\theta$ . For now, we will ignore the $1240$ and focus on the function $f(\\theta) = \\sin (A+\\theta) \\cos \\theta = (\\sin A \\cos \\theta + \\cos A \\sin \\theta)(\\cos \\theta) = \\sin A \\cos^2 \\theta + \\cos A \\sin \\theta \\cos \\theta = \\sin A \\cos^2 \\theta + \\frac{1}{2} \\cos A \\sin 2\\theta$\nTaking the derivative, $f'(\\theta) = \\sin A \\cdot -2\\cos \\theta \\sin \\theta + \\cos A \\cos 2\\theta = \\cos A \\cos 2\\theta - \\sin A \\sin 2\\theta = \\cos(2\\theta + A)$ . Setting this equal to $0$ , we get $\\cos(2 \\theta + A) = 0 \\Rightarrow 2\\theta +A = 90, 270 ^\\circ$ . Since we know that $A+ \\theta < 90$ , the $270^\\circ$ solution is extraneous. Thus, we get that $\\theta = \\frac{90 - A}{2} = 45 - \\frac{A}{2}$\nPlugging this value into the original area equation, $a(45 - \\frac{A}{2}) = 1240 \\sin (45 - \\frac{A}{2} + A) \\cos (45 - \\frac{A}{2}) = 1240\\sin( 45+ \\frac{A}{2})\\cos(45 - \\frac{A}{2})$ . Using a product-to-sum formula, we get that: \\[1240\\sin( 45+ \\frac{A}{2})\\cos(45 - \\frac{A}{2}) =\\] \\[1240\\cdot \\frac{1}{2}\\cdot(\\sin((45 + \\frac{A}{2}) + (45 -\\frac{A}{2}))+\\sin((45 +\\frac{A}{2})-(45 - \\frac{A}{2})))=\\] \\[620 (\\sin 90^\\circ + \\sin A) = 620 \\cdot \\frac{6}{5} = \\boxed{744}\\]", "Let $\\alpha$ be the angle $\\angle CAS$ and $\\beta$ be the angle $\\angle BAQ$ . Then \\[\\alpha + \\beta + \\angle A = 90^\\circ \\Rightarrow \\alpha + \\beta = 90^\\circ - \\angle A\\] \\[\\cos(\\alpha + \\beta) = \\cos(90^\\circ - \\angle A)\\] \\[\\cos(\\alpha + \\beta) = \\sin(\\angle A) = \\frac{1}{5}\\] \\[\\cos\\alpha\\cos\\beta - \\sin\\alpha\\sin\\beta = \\frac{1}{5}\\] \\[\\cos\\alpha\\cos\\beta - \\sqrt{(1-\\cos^2\\alpha)(1-\\cos^2\\beta)} = \\frac{1}{5}\\] \\[\\cos\\alpha\\cos\\beta - \\sqrt{1-\\cos^2\\alpha-\\cos^2\\beta+\\cos^2\\alpha\\cos^2\\beta} = \\frac{1}{5}\\] However, by AM-GM: \\[\\cos^2\\alpha+\\cos^2\\beta \\ge 2\\cos\\alpha\\beta\\] Therefore, \\[1-\\cos^2\\alpha-\\cos^2\\beta+\\cos^2\\alpha\\cos^2\\beta \\le 1-2\\cos\\alpha\\beta+\\cos^2\\alpha\\cos^2\\beta = (1-\\cos\\alpha\\cos\\beta)^2\\] \\[\\sqrt{1-\\cos^2\\alpha-\\cos^2\\beta+\\cos^2\\alpha\\cos^2\\beta} \\le 1-\\cos\\alpha\\cos\\beta\\] So, \\[\\frac{1}{5} \\ge \\cos\\alpha\\cos\\beta - (1-\\cos\\alpha\\cos\\beta) = 2\\cos\\alpha\\cos\\beta-1\\] \\[\\frac{3}{5} \\ge \\cos\\alpha\\cos\\beta\\] .\nHowever, the area of the rectangle is just $AS \\cdot AQ = 31\\cos\\alpha \\cdot 40\\cos\\beta \\le 31 \\cdot 40 \\cdot \\frac{3}{5} = \\boxed{744}$" ]

[ "\nWe label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled $A$\nIf we define a \"move\" as each time Erin crawls along a single edge from one vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be one unit away from $A$ (since Erin cannot crawl back to $A$ ), this vertex must be $4$\nTherefore, we now just need to count the number of paths from $A$ to $4$ . To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to $4$ , and 2 choices for which vertex Erin was at 2 moves before $4$ . All of Erin's previous moves were forced, so the total number of legal paths from $A$ to $4$ is $3 \\cdot 2 = \\boxed{6}$", "Lets say that this cube is an unit cube and the given corner is $(0,0,0)$ . Because Erin cannot return back to her starting point, she cannot be on $(0,0,1)$ $(0,1,0)$ , or $(1,0,0)$ . She cannot be on $(1,1,0)$ $(1,0,1)$ , or $(0,1,1)$ , because after $7$ moves, the sum of all the coordinates has to be odd. Thus, Erin has to be at $(1,1,1)$ . Now, we draw a net and see that there are $3$ choices for the first move, $2$ for the second, and the rest are forced. Thus the answer is $3\\cdot2 = \\boxed{6}$", "Let's suppose the given corner on the cube is $(0,0,0)$ . Erin has 3 identical ways to proceed. Suppose she goes to $(0,1,0)$ . She now has two more identical ways to go. Let's say she goes to $(0,1,1)$ . She has to go to $(0,0,1)$ , otherwise, she will end up on that point after 7 moves. This is because once if she chooses the other path to $(1,1,1)$ , the endpoint is certain to be $(0,0,1)$ , which is directly connected to $(0,0,0)$ . After she goes to $(0,0,1)$ , the only option she has is to go to $(1,0,1)$ , then $(1,0,0)$ , after that $(1,1,0)$ , and finally $(1,1,1)$ . She is forced to go this way because she cannot end up on $(1,0,0)$ . At the start, she had 3 ways to choose, and after that 2 ways to choose, so $3\\cdot2=\\boxed{6}$" ]

[ "We can solve this problem by finding the cases where the number is divisible by $3$ or $4$ , then subtract from the cases where none of those cases divide $5$ . To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor function to every case to get $\\left\\lfloor \\frac{2001}{3} \\right\\rfloor$ $\\left\\lfloor \\frac{2001}{4} \\right\\rfloor$ , and $\\left\\lfloor \\frac{2001}{12} \\right\\rfloor$ . The first two floor functions were for calculating the number of individual cases for $3$ and $4$ . The third case was to find any overlapping numbers. The numbers were $667$ $500$ , and $166$ , respectively. We add the first two terms and subtract the third to get $1001$ . The first case is finished.\nThe second case is more or less the same, except we are applying $3$ and $4$ to $5$ . We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions $\\left\\lfloor \\frac{2001}{3\\cdot5} \\right\\rfloor$ $\\left\\lfloor \\frac{2001}{4\\cdot5} \\right\\rfloor$ , and $\\left\\lfloor \\frac{2001}{3\\cdot4\\cdot5} \\right\\rfloor$ yields the numbers $133$ $100$ , and $33$ . The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$ . The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$ . Subtracting this number from the original $1001$ numbers procures $\\boxed{801}$", "First find the number of such integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of $3$\nThere are $\\frac45\\cdot2000=1600$ numbers that are not multiples of $5$ $\\frac23\\cdot\\frac34\\cdot1600=800$ are not multiples of $3$ or $4$ , so $800$ numbers are. $800+1=\\boxed{801}$", "Take a good-sized sample of consecutive integers; for example, the first $25$ positive integers. Determine that the numbers $3, 4, 6, 8, 9, 12, 16, 18, 21,$ and $24$ exhibit the properties given in the question. $25$ is a divisor of $2000$ , so there are $\\frac{10}{25}\\cdot2000=800$ numbers satisfying the given conditions between $1$ and $2000$ . Since $2001$ is a multiple of $3$ , add $1$ to $800$ to get $800+1=\\boxed{801}$", "By PIE, there are $1001$ numbers that are multiples of $3$ or $4$ and less than or equal to $2001$ $80\\%$ of them will not be divisible by $5$ , and by far the closest number to $80\\%$ of $1001$ is $\\boxed{801}$", "Similar to some of the above solutions. \nWe can divide $2001$ by $3$ and $4$ to find the number of integers divisible by $3$ and $4$ . Hence, we find that there are $667$ numbers less than $2001$ that are divisible by $3$ , and $500$ numbers that are divisible by $4$ . However, we will need to subtract the number of multiples of $15$ from 667 and that of $20$ from $500$ , since they're also divisible by 5 which we don't want. There are $133$ $100$ $233$ such numbers. Note that during this process, we've subtracted the multiples of $60$ twice because they're divisible by both $15$ and $20$ , so we have to add $33$ back to the tally (there are $33$ multiples of $60$ that does not exceed $2001$ ). Lastly, we have to subtract multiples of both $3$ AND $4$ since we only want multiples of either $3$ or $4$ . This is tantamount to subtracting the number of multiples of $12$ . And there are $166$ such numbers. Let's now collect our numbers and compute the total: $667$ $500$ $133$ $100$ $33$ $166$ $\\boxed{801}$", "Similar to @above:\nLet the function $M_{2001}(n)$ return how many multiples of $n$ are there not exceeding $2001$ . Then we have that the desired number is: \\[M_{2001}(3)+M_{2001}(4)-M_{2001}(3\\cdot 4)-M_{2001}(3 \\cdot 5) - M_{2001}(4 \\cdot 5)+M_{2001}(3 \\cdot 4 \\cdot 5)\\]\nEvaluating each of these we get: \\[667+500-166-133-100+33 = 1100-299 = 801.\\]\nThus, the answer is $\\boxed{801}.$", "We can solve this problem by finding the cases where the number is divisible by $3$ or $4$ , then subtract from the cases where none of those cases divide $5$ . To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor function to every case to get $\\left\\lfloor \\frac{2001}{3} \\right\\rfloor$ $\\left\\lfloor \\frac{2001}{4} \\right\\rfloor$ , and $\\left\\lfloor \\frac{2001}{12} \\right\\rfloor$ . The first two floor functions were for calculating the number of individual cases for $3$ and $4$ . The third case was to find any overlapping numbers. The numbers were $667$ $500$ , and $166$ , respectively. We add the first two terms and subtract the third to get $1001$ . The first case is finished.\nThe second case is more or less the same, except we are applying $3$ and $4$ to $5$ . We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions $\\left\\lfloor \\frac{2001}{3\\cdot5} \\right\\rfloor$ $\\left\\lfloor \\frac{2001}{4\\cdot5} \\right\\rfloor$ , and $\\left\\lfloor \\frac{2001}{3\\cdot4\\cdot5} \\right\\rfloor$ yields the numbers $133$ $100$ , and $33$ . The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$ . The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$ . Subtracting this number from the original $1001$ numbers procures $\\boxed{801}$", "First find the number of such integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of $3$\nThere are $\\frac45\\cdot2000=1600$ numbers that are not multiples of $5$ $\\frac23\\cdot\\frac34\\cdot1600=800$ are not multiples of $3$ or $4$ , so $800$ numbers are. $800+1=\\boxed{801}$", "Take a good-sized sample of consecutive integers; for example, the first $25$ positive integers. Determine that the numbers $3, 4, 6, 8, 9, 12, 16, 18, 21,$ and $24$ exhibit the properties given in the question. $25$ is a divisor of $2000$ , so there are $\\frac{10}{25}\\cdot2000=800$ numbers satisfying the given conditions between $1$ and $2000$ . Since $2001$ is a multiple of $3$ , add $1$ to $800$ to get $800+1=\\boxed{801}$", "By PIE, there are $1001$ numbers that are multiples of $3$ or $4$ and less than or equal to $2001$ $80\\%$ of them will not be divisible by $5$ , and by far the closest number to $80\\%$ of $1001$ is $\\boxed{801}$", "Similar to some of the above solutions. \nWe can divide $2001$ by $3$ and $4$ to find the number of integers divisible by $3$ and $4$ . Hence, we find that there are $667$ numbers less than $2001$ that are divisible by $3$ , and $500$ numbers that are divisible by $4$ . However, we will need to subtract the number of multiples of $15$ from 667 and that of $20$ from $500$ , since they're also divisible by 5 which we don't want. There are $133$ $100$ $233$ such numbers. Note that during this process, we've subtracted the multiples of $60$ twice because they're divisible by both $15$ and $20$ , so we have to add $33$ back to the tally (there are $33$ multiples of $60$ that does not exceed $2001$ ). Lastly, we have to subtract multiples of both $3$ AND $4$ since we only want multiples of either $3$ or $4$ . This is tantamount to subtracting the number of multiples of $12$ . And there are $166$ such numbers. Let's now collect our numbers and compute the total: $667$ $500$ $133$ $100$ $33$ $166$ $\\boxed{801}$", "Similar to @above:\nLet the function $M_{2001}(n)$ return how many multiples of $n$ are there not exceeding $2001$ . Then we have that the desired number is: \\[M_{2001}(3)+M_{2001}(4)-M_{2001}(3\\cdot 4)-M_{2001}(3 \\cdot 5) - M_{2001}(4 \\cdot 5)+M_{2001}(3 \\cdot 4 \\cdot 5)\\]\nEvaluating each of these we get: \\[667+500-166-133-100+33 = 1100-299 = 801.\\]\nThus, the answer is $\\boxed{801}.$" ]

[ "Let $A$ $T$ be Kristin's annual income and the income tax total, respectively. Notice that \\begin{align*} T &= p\\%\\cdot28000 + (p + 2)\\%\\cdot(A - 28000) \\\\ &= [p\\%\\cdot28000 + p\\%\\cdot(A - 28000)] + 2\\%\\cdot(A - 28000) \\\\ &= p\\%\\cdot A + 2\\%\\cdot(A - 28000) \\end{align*} We are also given that \\[T = (p + 0.25)\\%\\cdot A = p\\%\\cdot A + 0.25\\%\\cdot A\\] Thus, \\[p\\%\\cdot A + 2\\%\\cdot(A - 28000) = p\\%\\cdot A + 0.25\\%\\cdot A\\] \\[2\\%\\cdot(A - 28000) = 0.25\\%\\cdot A\\] Solve for $A$ to obtain $A = \\boxed{32000}$", "Let $A$ $T$ be Kristin's annual income and the income tax total, respectively. Notice that \\begin{align*} T &= p\\%\\cdot28000 + (p + 2)\\%\\cdot(A - 28000) \\\\ &= [p\\%\\cdot28000 + p\\%\\cdot(A - 28000)] + 2\\%\\cdot(A - 28000) \\\\ &= p\\%\\cdot A + 2\\%\\cdot(A - 28000) \\end{align*} We are also given that \\[T = (p + 0.25)\\%\\cdot A = p\\%\\cdot A + 0.25\\%\\cdot A\\] Thus, \\[p\\%\\cdot A + 2\\%\\cdot(A - 28000) = p\\%\\cdot A + 0.25\\%\\cdot A\\] \\[2\\%\\cdot(A - 28000) = 0.25\\%\\cdot A\\] Solve for $A$ to obtain $A = \\boxed{32000}$" ]

[ "Let $a=1$ . Our list is $\\{1,2,3,4,5\\}$ with an average of $15\\div 5=3$ . Our next set starting with $3$ is $\\{3,4,5,6,7\\}$ . Our average is $25\\div 5=5$\nTherefore, we notice that $5=1+4$ which means that the answer is $\\boxed{4}$", "We are given that \\[b=\\frac{a+a+1+a+2+a+3+a+4}{5}\\] \\[\\implies b =a+2\\]\nWe are asked to find the average of the 5 consecutive integers starting from $b$ in terms of $a$ . By substitution, this is \\[\\frac{a+2+a+3+a+4+a+5+a+6}5=a+4\\]\nThus, the answer is $\\boxed{4}$", "The list of numbers is $\\left\\{a,\\ a+1,\\ b,\\ a+3,\\ a+4\\right\\}$ so $b=a+2$ . The new list is $\\left\\{a+2,\\ a+3,\\ a+4,\\ a+5,\\ a+6\\right\\}$ and the average is $a+4 \\Longrightarrow \\boxed{4}$", "Let $a=1$ . Our list is $\\{1,2,3,4,5\\}$ with an average of $15\\div 5=3$ . Our next set starting with $3$ is $\\{3,4,5,6,7\\}$ . Our average is $25\\div 5=5$\nTherefore, we notice that $5=1+4$ which means that the answer is $\\boxed{4}$", "We are given that \\[b=\\frac{a+a+1+a+2+a+3+a+4}{5}\\] \\[\\implies b =a+2\\]\nWe are asked to find the average of the 5 consecutive integers starting from $b$ in terms of $a$ . By substitution, this is \\[\\frac{a+2+a+3+a+4+a+5+a+6}5=a+4\\]\nThus, the answer is $\\boxed{4}$", "The list of numbers is $\\left\\{a,\\ a+1,\\ b,\\ a+3,\\ a+4\\right\\}$ so $b=a+2$ . The new list is $\\left\\{a+2,\\ a+3,\\ a+4,\\ a+5,\\ a+6\\right\\}$ and the average is $a+4 \\Longrightarrow \\boxed{4}$" ]

[ "We know from experience that the average of $5$ consecutive numbers is the $3^\\text{rd}$ one or the $1^\\text{st} + 2$ . With the logic, we find that $b=a+2$ $b+2=(a+2)+2=\\boxed{4}$", "We know from experience that the average of $5$ consecutive numbers is the $3^\\text{rd}$ one or the $1^\\text{st} + 2$ . With the logic, we find that $b=a+2$ $b+2=(a+2)+2=\\boxed{4}$" ]

[ "Without loss of generality, let there be $20$ students (the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points. Therefore, the mean\nis $87$ and the median is $90$\nThus, the solution is \\[90-87=3\\implies\\boxed{3}\\]", "The percentage who scored $100$ points is $100\\%-(10\\%+35\\%+30\\%)=100\\%-75\\%=25\\%$ . Now, we need to find the median, which is the score that splits the upper and lower $50\\%$ .The lower $10\\%+35\\%=45\\%$ scored $70$ or $80$ points, so the median is $90$ (since the upper $25\\%$ is $100$ points and the lower $45\\%$ is $70$ or $80$ ).The mean is $10\\%\\cdot70+35\\%\\cdot80+30\\%\\cdot90+25\\%\\cdot100=7+28+27+25=87$ .\nSo, our solution is $90-87=3\\Rightarrow\\boxed{3}$ ~sosiaops", "The $\\le 80$ -point scores make up $10\\%+35\\% = 45\\% < 50\\%$ of the scores, but the $\\le 90$ -point scores make up $45\\%+30\\% = 75\\% > 50\\%$ of the scores, so the median is $90$\n$10\\%$ of scores were $70-90 = -20$ more than the median, $35\\%$ were $-10$ more, $100\\%-75\\% = 25\\%$ were $10$ more, and the rest were equal. This means that the mean score is $10\\%\\cdot(-20)+35\\%\\cdot(-10)+25\\%\\cdot10 = -2 + (-3.5) + 2.5 = -3$ more than the median, so their difference is $\\left|-3\\right| = \\boxed{3}$ .\n~ emerald_block", "Without loss of generality, let there be $20$ students (the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points. Therefore, the mean\nis $87$ and the median is $90$\nThus, the solution is \\[90-87=3\\implies\\boxed{3}\\]", "The percentage who scored $100$ points is $100\\%-(10\\%+35\\%+30\\%)=100\\%-75\\%=25\\%$ . Now, we need to find the median, which is the score that splits the upper and lower $50\\%$ .The lower $10\\%+35\\%=45\\%$ scored $70$ or $80$ points, so the median is $90$ (since the upper $25\\%$ is $100$ points and the lower $45\\%$ is $70$ or $80$ ).The mean is $10\\%\\cdot70+35\\%\\cdot80+30\\%\\cdot90+25\\%\\cdot100=7+28+27+25=87$ .\nSo, our solution is $90-87=3\\Rightarrow\\boxed{3}$ ~sosiaops", "The $\\le 80$ -point scores make up $10\\%+35\\% = 45\\% < 50\\%$ of the scores, but the $\\le 90$ -point scores make up $45\\%+30\\% = 75\\% > 50\\%$ of the scores, so the median is $90$\n$10\\%$ of scores were $70-90 = -20$ more than the median, $35\\%$ were $-10$ more, $100\\%-75\\% = 25\\%$ were $10$ more, and the rest were equal. This means that the mean score is $10\\%\\cdot(-20)+35\\%\\cdot(-10)+25\\%\\cdot10 = -2 + (-3.5) + 2.5 = -3$ more than the median, so their difference is $\\left|-3\\right| = \\boxed{3}$ .\n~ emerald_block" ]

[ "Let's assume we don't stop picking until all of the chips are picked. To satisfy this condition, we have to arrange the letters: $W, W, R, R, R$ such that both $W$ 's appear in the first $4$ . We find the number of ways to arrange the white chips in the first $4$ and divide that by the total ways to choose all the chips. The probability of this occurring is $\\dfrac{\\dbinom{4}{2}}{\\dbinom{5}{2}} = \\boxed{35}$", "The amount of ways to end with a white chip is by having $RRWW, RWW,$ and $WW$ . The amount of arrangements for $RRWW$ with $W$ at the end is $3$ , the number of arrangements of $RWW$ with $W$ at the end is $2$ , and the number of arrangements with $WW$ is just $1$ . This gives us $6$ total ways to end with white. Next, the cases to end with a red are $RWRR$ , and $RRR$ $RWRR$ gives us $3$ ways and $RRR$ gives us $1$ way. So the number of ways to end with a red is $4$ . Thus, our answer is simply $\\frac{6}{4+6}$ $\\boxed{35}$", "Let's assume we don't stop picking until all of the chips are picked. To satisfy this condition, we have to arrange the letters: $W, W, R, R, R$ such that both $W$ 's appear in the first $4$ . We find the number of ways to arrange the white chips in the first $4$ and divide that by the total ways to choose all the chips. The probability of this occurring is $\\dfrac{\\dbinom{4}{2}}{\\dbinom{5}{2}} = \\boxed{35}$", "The amount of ways to end with a white chip is by having $RRWW, RWW,$ and $WW$ . The amount of arrangements for $RRWW$ with $W$ at the end is $3$ , the number of arrangements of $RWW$ with $W$ at the end is $2$ , and the number of arrangements with $WW$ is just $1$ . This gives us $6$ total ways to end with white. Next, the cases to end with a red are $RWRR$ , and $RRR$ $RWRR$ gives us $3$ ways and $RRR$ gives us $1$ way. So the number of ways to end with a red is $4$ . Thus, our answer is simply $\\frac{6}{4+6}$ $\\boxed{35}$" ]

[ "We can make three equations out of the information, and since the distances are the same, we can equate these equations.\n\\[\\frac{4t}{5}(x+\\frac{1}{2})=xt=(t+\\frac{5}{2})(x-\\frac{1}{2})\\] where $x$ is the man's rate and $t$ is the time it takes him.\nLooking at the first two parts of the equations,\n\\[\\frac{4t}{5}(x+\\frac{1}{2})=xt\\]\nwe note that we can solve for $x$ . Solving for $x$ , we get $x=2.$\nNow we look at the last two parts of the equation:\n\\[xt=(t+\\frac{5}{2})(x-\\frac{1}{2})\\]\nwe note that we can solve for $t$ and we get $t=\\frac{15}{2}.$ We want the find the distance, which is $xt= \\boxed{15}.$" ]

[ "We can list the first few numbers in the form $8 \\cdot (8....8)$\n(Hard problem to do without the multiplication, but you can see the pattern early on)\n$8 \\cdot 8 = 64$\n$8 \\cdot 88 = 704$\n$8 \\cdot 888 = 7104$\n$8 \\cdot 8888 = 71104$\n$8 \\cdot 88888 = 711104$\nBy now it's clear that the numbers will be in the form $7$ $k-2$ $1$ 's, and $04$ . We want to make the numbers sum to 1000, so $7+4+(k-2) = 1000$ . Solving, we get $k = 991$ , meaning the answer is $\\fbox{(D)}$\nAnother way to proceed is that we know the difference between the sum of the digits of each product and $k$ is always $9$ , so we just do $1000-9=\\boxed{991}$", "We can list the first few numbers in the form $8 \\cdot (8....8)$\n(Hard problem to do without the multiplication, but you can see the pattern early on)\n$8 \\cdot 8 = 64$\n$8 \\cdot 88 = 704$\n$8 \\cdot 888 = 7104$\n$8 \\cdot 8888 = 71104$\n$8 \\cdot 88888 = 711104$\nBy now it's clear that the numbers will be in the form $7$ $k-2$ $1$ 's, and $04$ . We want to make the numbers sum to 1000, so $7+4+(k-2) = 1000$ . Solving, we get $k = 991$ , meaning the answer is $\\fbox{(D)}$\nAnother way to proceed is that we know the difference between the sum of the digits of each product and $k$ is always $9$ , so we just do $1000-9=\\boxed{991}$" ]

[ "We start by noting that there are $10$ letters, meaning there are $10$ digits in total. Listing them all out, we have $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ . Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that.\nCase 1: $G$ $H$ $I$ , and $J$ are $7$ $5$ $3$ , and $1$ respectively.\nA cursory glance allows us to deduce that the smallest possible sum of $A + B + C$ is $11$ when $D$ $E$ , and $F$ are $8$ $6$ , and $4$ respectively, so this is out of the question.\nCase 2: $G$ $H$ $I$ , and $J$ are $3$ $5$ $7$ , and $9$ respectively.\nA cursory glance allows us to deduce the answer. Clearly, when $D$ $E$ , and $F$ are $6$ $4$ , and $2$ respectively, $A + B + C$ is $9$ when $A$ $B$ , and $C$ are $8$ $1$ , and $0$ respectively, giving us a final answer of $\\boxed{8}$", "We start by noting that there are $10$ letters, meaning there are $10$ digits in total. Listing them all out, we have $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ . Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that.\nCase 1: $G$ $H$ $I$ , and $J$ are $7$ $5$ $3$ , and $1$ respectively.\nA cursory glance allows us to deduce that the smallest possible sum of $A + B + C$ is $11$ when $D$ $E$ , and $F$ are $8$ $6$ , and $4$ respectively, so this is out of the question.\nCase 2: $G$ $H$ $I$ , and $J$ are $3$ $5$ $7$ , and $9$ respectively.\nA cursory glance allows us to deduce the answer. Clearly, when $D$ $E$ , and $F$ are $6$ $4$ , and $2$ respectively, $A + B + C$ is $9$ when $A$ $B$ , and $C$ are $8$ $1$ , and $0$ respectively, giving us a final answer of $\\boxed{8}$" ]

[ "We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$ , or $12$ cases, which is not difficult to list out. We systematically list out all $12$ cases.\nNow, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums $1, 2, 3, 4,$ and $5$ . By choosing the full circle, we can obtain $15$ . By choosing everything except for $1, 2, 3, 4,$ and $5$ , we can obtain subsets with sums of $10, 11, 12, 13,$ and $14$\nThis means that we now only need to check for $6, 7, 8,$ and $9$ . However, once we have found a set summing to $6$ , we can choose all remaining numbers and obtain a set summing to $15-6=9$ , and similarly for $7$ and $8$ . Thus, we only need to check each case for whether or not we can obtain $6$ or $7$\nWe can make $6$ by having $4, 2$ , or $3, 2, 1$ , or $5, 1$ . We can start with the group of three. To separate $3, 2, 1$ from each other, they must be grouped two together and one separate, like this.\n\nNow, we note that $x$ is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have $1$ , because it is part of the $5, 1$ pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same.\n\nWe have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$ , so we must place $4$ next to the $1$ and $5$ next to the $2$\n\nThis is the only solution to make $6$ \"bad.\"\nNext we move on to $7$ , which can be made by $3, 4$ , or $5, 2$ , or $4, 2, 1$ . We do this the same way as before. We start with the three group. Since we can't have $4$ or $2$ in the top slot, we must have one there, and $4$ and $2$ are next to each other on the bottom. When we have $3$ and $5$ left to insert, we place them such that we don't have the two pairs adjacent.\n\nThis is the only solution to make $7$ \"bad.\"\nWe've covered all needed cases, and the two examples we found are distinct, therefore the answer is $\\boxed{2}$", "We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$ , or $12$ cases, which is not difficult to list out. We systematically list out all $12$ cases.\nNow, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums $1, 2, 3, 4,$ and $5$ . By choosing the full circle, we can obtain $15$ . By choosing everything except for $1, 2, 3, 4,$ and $5$ , we can obtain subsets with sums of $10, 11, 12, 13,$ and $14$\nThis means that we now only need to check for $6, 7, 8,$ and $9$ . However, once we have found a set summing to $6$ , we can choose all remaining numbers and obtain a set summing to $15-6=9$ , and similarly for $7$ and $8$ . Thus, we only need to check each case for whether or not we can obtain $6$ or $7$\nWe can make $6$ by having $4, 2$ , or $3, 2, 1$ , or $5, 1$ . We can start with the group of three. To separate $3, 2, 1$ from each other, they must be grouped two together and one separate, like this.\n\nNow, we note that $x$ is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have $1$ , because it is part of the $5, 1$ pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same.\n\nWe have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$ , so we must place $4$ next to the $1$ and $5$ next to the $2$\n\nThis is the only solution to make $6$ \"bad.\"\nNext we move on to $7$ , which can be made by $3, 4$ , or $5, 2$ , or $4, 2, 1$ . We do this the same way as before. We start with the three group. Since we can't have $4$ or $2$ in the top slot, we must have one there, and $4$ and $2$ are next to each other on the bottom. When we have $3$ and $5$ left to insert, we place them such that we don't have the two pairs adjacent.\n\nThis is the only solution to make $7$ \"bad.\"\nWe've covered all needed cases, and the two examples we found are distinct, therefore the answer is $\\boxed{2}$" ]

[ "We first claim that $n$ must be divisible by $42$ . Since $\\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$ , we can first consider the special case where $a$ is prime and $a \\neq 0,1 \\pmod{43}$ . By Dirichlet's Theorem (Refer to the Remark section.), such $a$ always exists.\nThen $\\sigma(a^n)-1 = \\sum_{i=1}^n a^i = a\\left(\\frac{a^n - 1}{a-1}\\right)$ . In order for this expression to be divisible by $2021=43\\cdot 47$ , a necessary condition is $a^n - 1 \\equiv 0 \\pmod{43}$ . By Fermat's Little Theorem $a^{42} \\equiv 1 \\pmod{43}$ . Moreover, if $a$ is a primitive root modulo $43$ , then $\\text{ord}_{43}(a) = 42$ , so $n$ must be divisible by $42$\nBy similar reasoning, $n$ must be divisible by $46$ , by considering $a \\not\\equiv 0,1 \\pmod{47}$\nWe next claim that $n$ must be divisible by $43$ . By Dirichlet, let $a$ be a prime that is congruent to $1 \\pmod{43}$ . Then $\\sigma(a^n) \\equiv n+1 \\pmod{43}$ , so since $\\sigma(a^n)-1$ is divisible by $43$ $n$ is a multiple of $43$\nAlternatively, since $\\left(\\frac{a(a^n - 1^n)}{a-1}\\right)$ must be divisible by $43,$ by LTE, we have $v_{43}(a)+v_{43}{(a-1)}+v_{43}{(n)}-v_{43}{(a-1)} \\geq 1,$ which simplifies to $v_{43}(n) \\geq 1,$ which implies the desired result.\nSimilarly, $n$ is a multiple of $47$\nLastly, we claim that if $n = \\text{lcm}(42, 46, 43, 47)$ , then $\\sigma(a^n) - 1$ is divisible by $2021$ for all positive integers $a$ . The claim is trivially true for $a=1$ so suppose $a>1$ . Let $a = p_1^{e_1}\\ldots p_k^{e_k}$ be the prime factorization of $a$ . Since $\\sigma(n)$ is multiplicative , we have \\[\\sigma(a^n) - 1 = \\prod_{i=1}^k \\sigma(p_i^{e_in}) - 1.\\] We can show that $\\sigma(p_i^{e_in}) \\equiv 1 \\pmod{2021}$ for all primes $p_i$ and integers $e_i \\ge 1$ , so \\[\\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \\ldots + p_i^n) + (p_i^{n+1} + \\ldots + p_i^{2n}) + \\ldots + (p_i^{n(e_i-1)+1} + \\ldots + p_i^{e_in}),\\] where each expression in parentheses contains $n$ terms. It is easy to verify that if $p_i = 43$ or $p_i = 47$ then $\\sigma(p_i^{e_in}) \\equiv 1 \\pmod{2021}$ for this choice of $n$ , so suppose $p_i \\not\\equiv 0 \\pmod{43}$ and $p_i \\not\\equiv 0 \\pmod{47}$ . Each expression in parentheses equals $\\frac{p_i^n - 1}{p_i - 1}$ multiplied by some power of $p_i$ . If $p_i \\not\\equiv 1 \\pmod{43}$ , then FLT implies $p_i^n - 1 \\equiv 0 \\pmod{43}$ , and if $p_i \\equiv 1 \\pmod{43}$ , then $p_i + p_i^2 + \\ldots + p_i^n \\equiv 1 + 1 + \\ldots + 1 \\equiv 0 \\pmod{43}$ (since $n$ is also a multiple of $43$ , by definition). Similarly, we can show $\\sigma(p_i^{e_in}) \\equiv 1 \\pmod{47}$ , and a simple CRT argument shows $\\sigma(p_i^{e_in}) \\equiv 1 \\pmod{2021}$ . Then $\\sigma(a^n) \\equiv 1^k \\equiv 1 \\pmod{2021}$\nThen the prime factors of $n$ are $2,3,7,23,43,$ and $47,$ and the answer is $2+3+7+23+43+47 = \\boxed{125}$", "$n$ only needs to satisfy $\\sigma(a^n)\\equiv 1 \\pmod{43}$ and $\\sigma(a^n)\\equiv 1 \\pmod{47}$ for all $a$ . Let's work on the first requirement (mod 43) first. All $n$ works for $a=1$ . If $a>1$ , let $a$ 's prime factorization be $a=p_1^{e_1}p_2^{e_2}\\cdots p_k^{e_k}$ . The following three statements are the same:\nWe can show this by casework on $p$\nSimilar arguments for modulo $47$ lead to $46|n$ and $47|n$ . Therefore, we get $n=\\operatorname{lcm}[42,43,46,47]$ . Following the last paragraph of Solution 1 gives the answer $\\boxed{125}$", "We perform casework on $a:$\nFinally, the least such positive integer $n$ for all cases is \\begin{align*} n&=\\operatorname{lcm}(42,43,46,47) \\\\ &=\\operatorname{lcm}(2\\cdot3\\cdot7,43,2\\cdot23,47) \\\\ &=2\\cdot3\\cdot7\\cdot23\\cdot43\\cdot47, \\end{align*} so the sum of its prime factors is $2+3+7+23+43+47=\\boxed{125}.$", "Since the problem works for all positive integers $a$ , let's plug in $a=2$ and see what we get. Since $\\sigma({2^n}) = 2^{n+1}-1,$ we have $2^{n+1} \\equiv 2 \\pmod{2021}.$ Simplifying using CRT and Fermat's Little Theorem , we get that $n \\equiv 0 \\pmod{42}$ and $n \\equiv 0 \\pmod{46}.$ Then, we can look at $a$ being a $1\\pmod{43}$ prime and a $1\\pmod{47}$ prime, just like in Solution 1, to find that $43$ and $47$ also divide $n$ . There don't seem to be any other odd \"numbers\" to check, so we can hopefully assume that the answer is the sum of the prime factors of $\\text{lcm(42, 43, 46, 47)}.$ From here, follow solution 1 to get the final answer $\\boxed{125}$" ]

[ "The problem is asking for between how many consecutive powers of $5$ are there $3$ power of $2$\nThere can be either $2$ or $3$ powers of $2$ between any two consecutive powers of $5$ $5^n$ and $5^{n+1}$\nThe first power of $2$ is between $5^n$ and $2 \\cdot 5^n$\nThe second power of $2$ is between $2 \\cdot 5^n$ and $4 \\cdot 5^n$\nThe third power of $2$ is between $4 \\cdot 5^n$ and $8 \\cdot 5^n$ , meaning that it can be between $5^n$ and $5^{n+1}$ or not.\nIf there are only $2$ power of $2$ s between every consecutive powers of $5$ up to $5^{867}$ , there would be $867\\cdot 2 = 1734$ power of $2$ s. However, there are $2013$ powers of $2$ before $5^{867}$ , meaning the answer is $2013 - 1734 = \\boxed{279}$", "The problem is asking for between how many consecutive powers of $5$ are there $3$ power of $2$\nThere can be either $2$ or $3$ powers of $2$ between any two consecutive powers of $5$ $5^n$ and $5^{n+1}$\nThe first power of $2$ is between $5^n$ and $2 \\cdot 5^n$\nThe second power of $2$ is between $2 \\cdot 5^n$ and $4 \\cdot 5^n$\nThe third power of $2$ is between $4 \\cdot 5^n$ and $8 \\cdot 5^n$ , meaning that it can be between $5^n$ and $5^{n+1}$ or not.\nIf there are only $2$ power of $2$ s between every consecutive powers of $5$ up to $5^{867}$ , there would be $867\\cdot 2 = 1734$ power of $2$ s. However, there are $2013$ powers of $2$ before $5^{867}$ , meaning the answer is $2013 - 1734 = \\boxed{279}$" ]

[ "Let's use casework on the yellow house. The yellow house $(\\text{Y})$ is either the $3^\\text{rd}$ house or the last house.\nCase 1: $\\text{Y}$ is the $3^\\text{rd}$ house.\nThe only possible arrangement is $\\text{B}-\\text{O}-\\text{Y}-\\text{R}$\nCase 2: $\\text{Y}$ is the last house.\nThere are two possible arrangements:\n$\\text{B}-\\text{O}-\\text{R}-\\text{Y}$\n$\\text{O}-\\text{B}-\\text{R}-\\text{Y}$\nThe answer is $1+2=\\boxed{3}$", "There are $4!=24$ arrangements without restrictions. There are $3!\\cdot2!=12$ arrangements such that the blue house neighboring the yellow house (calculating the arrangments of [ $\\text{BY}$ ], $\\text{O}$ , and $\\text{R}$ ). Hence, there are $24-12=12$ arrangements with the blue and yellow houses non-adjacent.\nBy symmetry, exactly half of the $12$ arrangements have the blue house before the yellow house, and exactly half of those $6$ arrangements have the orange house before the red house, so our answer is $12\\cdot\\frac{1}{2}\\cdot\\frac{1}{2}= \\boxed{3}$", "To start with, the blue house is either the first or second house.\nIf the blue house is the first, then the orange must follow, leading to $2$ cases: $\\text{B-O-R-Y}$ and $\\text{B-O-Y-R}$\nIf the blue house is second, then the orange house must be first and the yellow house last, leading to $1$ case: $\\text{O-B-R-Y}$ .\nTherefore, our answer is $\\boxed{3}$", "Let's use casework on the yellow house. The yellow house $(\\text{Y})$ is either the $3^\\text{rd}$ house or the last house.\nCase 1: $\\text{Y}$ is the $3^\\text{rd}$ house.\nThe only possible arrangement is $\\text{B}-\\text{O}-\\text{Y}-\\text{R}$\nCase 2: $\\text{Y}$ is the last house.\nThere are two possible arrangements:\n$\\text{B}-\\text{O}-\\text{R}-\\text{Y}$\n$\\text{O}-\\text{B}-\\text{R}-\\text{Y}$\nThe answer is $1+2=\\boxed{3}$", "There are $4!=24$ arrangements without restrictions. There are $3!\\cdot2!=12$ arrangements such that the blue house neighboring the yellow house (calculating the arrangments of [ $\\text{BY}$ ], $\\text{O}$ , and $\\text{R}$ ). Hence, there are $24-12=12$ arrangements with the blue and yellow houses non-adjacent.\nBy symmetry, exactly half of the $12$ arrangements have the blue house before the yellow house, and exactly half of those $6$ arrangements have the orange house before the red house, so our answer is $12\\cdot\\frac{1}{2}\\cdot\\frac{1}{2}= \\boxed{3}$", "To start with, the blue house is either the first or second house.\nIf the blue house is the first, then the orange must follow, leading to $2$ cases: $\\text{B-O-R-Y}$ and $\\text{B-O-Y-R}$\nIf the blue house is second, then the orange house must be first and the yellow house last, leading to $1$ case: $\\text{O-B-R-Y}$ .\nTherefore, our answer is $\\boxed{3}$" ]

[ "Without loss of generality, assume that the set $\\{a\\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \\le n \\cdot a_1.$ Now set $b_i \\equiv \\frac{a_i}{a_1},$ and since a triangle with sidelengths from $\\{a\\}$ will be similar to the corresponding triangle from $\\{b\\},$ we simply have to show the existence of acute triangles in $\\{b\\}.$ Note that $b_1 = 1$ and for all $i$ $b_i \\le n.$\nNow three arbitrary sidelengths $x$ $y$ , and $z$ , with $x \\le y \\le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \\longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$ 's for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$ ).\nWe now make another substitution: $c_i \\equiv b_i ^2.$ So $c_1 = 1$ and for all $i$ $c_i \\le n^2.$ Now we examine the smallest possible sets $\\{c\\}$ for small $n$ for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If $n=3$ , then the smallest possible set, call it $\\{s_3\\},$ is trivially $\\{1,1,2\\}$ , since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\\{s_n\\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\\{s_n\\} = \\{F_0, F_1, ... F_n\\}$ , then $\\{s_{n+1}\\} = \\{F_0, F_1, ... F_n, c_{n+1}\\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\\{s_n\\}$ which are $F_{n-1}$ and $F_n$ . But these sum to $F_{n+1}$ so $\\{s_{n+1}\\} = \\{F_0, F_1, ... F_{n+1}\\}$ and our induction is complete.\nNow since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set $\\{c\\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\\{c\\}$ is bounded between $1$ and $n^2$ , then the conditions of the problem are met if and only if $F_{n-1} > n^2$ . The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\\boxed{13}$", "Outline:\n1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \\ge 3$\n2. If the chosen $n$ is such that $F_n \\le n^2$ , then choose the sequence $a_n$ such that $a_k = \\sqrt{F_k}$ for $1 \\le k \\le n$ . It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to $n$ times the smallest term. Also, because for any three terms $x = \\sqrt{F_a}, y = \\sqrt{F_b}, z = \\sqrt{F_c}$ with $a<b<c$ $x^2 + y^2 = F_a + F_b \\le F_{b-1} + F_b = F_{b+1} \\le F_c = z^2$ , x, y, z do not form an acute triangle. Thus, all $n$ such that $F_n \\le n^2$ do not work.\n3. It is easy to observe via a contradiction argument that all $n$ such that $F_n > n^2$ produce an acute triangle. (If, without loss of generality, $a_n$ is an increasing sequence, such that no three (in particular, consecutive) terms form an acute triangle, then $a_2^2 \\ge F_1a_1^2, a_3^2 \\ge a_2^2 + a_1^2 \\ge F_2a^2$ , and by induction $a_n^2 > F_na_1^2$ , a contradiction to the condition's inequality.)\n4. Note that $F_{12} = 144 = 12^2$ and $F_{13} = 233 > 169 = 13^2$ . It is easily to verify through strong induction that all $n$ greater than 12 make $F_n > n^2$ . Thus, $\\boxed{13}$ is the desired solution set.", "Without loss of generality, assume that the set $\\{a\\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \\le n \\cdot a_1.$ Now set $b_i \\equiv \\frac{a_i}{a_1},$ and since a triangle with sidelengths from $\\{a\\}$ will be similar to the corresponding triangle from $\\{b\\},$ we simply have to show the existence of acute triangles in $\\{b\\}.$ Note that $b_1 = 1$ and for all $i$ $b_i \\le n.$\nNow three arbitrary sidelengths $x$ $y$ , and $z$ , with $x \\le y \\le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \\longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$ 's for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$ ).\nWe now make another substitution: $c_i \\equiv b_i ^2.$ So $c_1 = 1$ and for all $i$ $c_i \\le n^2.$ Now we examine the smallest possible sets $\\{c\\}$ for small $n$ for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If $n=3$ , then the smallest possible set, call it $\\{s_3\\},$ is trivially $\\{1,1,2\\}$ , since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\\{s_n\\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\\{s_n\\} = \\{F_0, F_1, ... F_n\\}$ , then $\\{s_{n+1}\\} = \\{F_0, F_1, ... F_n, c_{n+1}\\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\\{s_n\\}$ which are $F_{n-1}$ and $F_n$ . But these sum to $F_{n+1}$ so $\\{s_{n+1}\\} = \\{F_0, F_1, ... F_{n+1}\\}$ and our induction is complete.\nNow since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set $\\{c\\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\\{c\\}$ is bounded between $1$ and $n^2$ , then the conditions of the problem are met if and only if $F_{n-1} > n^2$ . The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\\boxed{13}$", "Outline:\n1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \\ge 3$\n2. If the chosen $n$ is such that $F_n \\le n^2$ , then choose the sequence $a_n$ such that $a_k = \\sqrt{F_k}$ for $1 \\le k \\le n$ . It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to $n$ times the smallest term. Also, because for any three terms $x = \\sqrt{F_a}, y = \\sqrt{F_b}, z = \\sqrt{F_c}$ with $a<b<c$ $x^2 + y^2 = F_a + F_b \\le F_{b-1} + F_b = F_{b+1} \\le F_c = z^2$ , x, y, z do not form an acute triangle. Thus, all $n$ such that $F_n \\le n^2$ do not work.\n3. It is easy to observe via a contradiction argument that all $n$ such that $F_n > n^2$ produce an acute triangle. (If, without loss of generality, $a_n$ is an increasing sequence, such that no three (in particular, consecutive) terms form an acute triangle, then $a_2^2 \\ge F_1a_1^2, a_3^2 \\ge a_2^2 + a_1^2 \\ge F_2a^2$ , and by induction $a_n^2 > F_na_1^2$ , a contradiction to the condition's inequality.)\n4. Note that $F_{12} = 144 = 12^2$ and $F_{13} = 233 > 169 = 13^2$ . It is easily to verify through strong induction that all $n$ greater than 12 make $F_n > n^2$ . Thus, $\\boxed{13}$ is the desired solution set." ]

[ "Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$ . The middle of the three numbers is the median, $5$ . So $\\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$ , which can be solved to get $m=10$ .\nHence, the sum of the three numbers is $3\\cdot 10 = \\boxed{30}$", "Say the three numbers are $x$ $y$ and $z$ . When we arrange them in ascending order then we can assume $y$ is in the middle therefore $y = 5$\nWe can also assume that the smallest number is $x$ and the largest number of the three is $y$ .\nTherefore,\n\\[\\frac{x+y+z}{3} = x + 10 = z - 15\\] \\[\\frac{x+5+z}{3} = x + 10 = z - 15\\]\nTaking up the first equation $\\frac{x+5+z}{3} = x + 10$ and simplifying we obtain $z - 2x - 25 = 0$ doing so for the equation $\\frac{x+5+z}{3} = z - 15$ we obtain the equation $x - 2z + 50 = 0$\n\\[x - 2z + 50 = 2x - 4z + 100\\]\nwhen solve the above obtained equation and $z - 2x - 25 = 0$ we obtain the values of $z = 25$ and $x = 0$\nTherefore the sum of the three numbers is $25 + 5 + 0 = \\boxed{30}$", "Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$ . The middle of the three numbers is the median, $5$ . So $\\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$ , which can be solved to get $m=10$ .\nHence, the sum of the three numbers is $3\\cdot 10 = \\boxed{30}$", "Say the three numbers are $x$ $y$ and $z$ . When we arrange them in ascending order then we can assume $y$ is in the middle therefore $y = 5$\nWe can also assume that the smallest number is $x$ and the largest number of the three is $y$ .\nTherefore,\n\\[\\frac{x+y+z}{3} = x + 10 = z - 15\\] \\[\\frac{x+5+z}{3} = x + 10 = z - 15\\]\nTaking up the first equation $\\frac{x+5+z}{3} = x + 10$ and simplifying we obtain $z - 2x - 25 = 0$ doing so for the equation $\\frac{x+5+z}{3} = z - 15$ we obtain the equation $x - 2z + 50 = 0$\n\\[x - 2z + 50 = 2x - 4z + 100\\]\nwhen solve the above obtained equation and $z - 2x - 25 = 0$ we obtain the values of $z = 25$ and $x = 0$\nTherefore the sum of the three numbers is $25 + 5 + 0 = \\boxed{30}$" ]

[ "Let's multiply ticket costs by $2$ , then the half price becomes an integer, and the charity sold $140$ tickets worth a total of $4002$ dollars.\nLet $h$ be the number of half price tickets, we then have $140-h$ full price tickets. The cost of $140-h$ full price tickets is equal to the cost of $280-2h$ half price tickets.\nHence we know that $h+(280-2h) = 280-h$ half price tickets cost $4002$ dollars. Then a single full price ticket costs $\\frac{4002}{280-h}$ dollars, and this must be an integer. Thus $280-h$ must be a divisor of $4002$ . Keeping in mind that $0\\leq h\\leq 140$ , we are looking for a divisor between $140$ and $280$ , inclusive.\nThe prime factorization of $4002$ is $4002=2\\cdot 3\\cdot 23\\cdot 29$ . We can easily find out that the only divisor of $4002$ within the given range is $2\\cdot 3\\cdot 29 = 174$\nThis gives us $280-h=174$ , hence there were $h=106$ half price tickets and $140-h = 34$ full price tickets.\nIn our modified setting (with prices multiplied by $2$ ) the price of a half price ticket is $\\frac{4002}{174} = 23$ . In the original setting this is the price of a full price ticket. Hence $23\\cdot 34 = \\boxed{782}$ dollars are raised by the full price tickets.", "Let the cost of the full price ticket be $x$ , the number of full-price tickets be $A$ , and the number of half-price tickets be $B$\nLet's multiply both sides of the equation that naturally follows by 2. We have\n\\[2Ax+Bx=4002\\]\nAnd we have $A+B=140\\implies B=140-A$\nPlugging in, we get $\\implies 2Ax+(140-A)(x)=4002$\nSimplifying, we get $Ax+140x=4002$\nFactoring out the $x$ , we get $x(A+140)=4002\\implies x=\\frac{4002}{A+140}$\nWe see that the fraction has to simplify to an integer (the full price is a whole dollar amount)\nThus, $A+140$ must be a factor of 4002.\nConsider the prime factorization of $4002$ $2\\times3\\times23\\times29$\n$A$ must be a positive integer. So, we seek a factor of $4002$ to set equal to $A+140$ so that we get an integer solution for $A$ that is less than $140$ . By guess-and-check OR inspection, the appropriate factor is $174$ $2\\times3\\times29$ ), meaning that $A$ has a value of $34$ . Plug this into the above equation for $x$ to get $x = 23$\nTherefore, the price of full tickets out of $2001$ is $23\\times34=\\boxed{782}$", "Let $f$ equal the number of full-price tickets, and let $h$ equal the number of half-price tickets. Additionally, suppose that the price of $f$ is $p$ . We are trying to solve for $f \\cdot p$\nSince the total number of tickets sold is $140$ , we know that \\[f+h=140.\\] The sales from full-price tickets ( $f \\cdot p$ ) plus the sales from half-price tickets $\\Big(\\frac{h \\cdot p}{2}$ , because each hall-price ticket costs $\\frac{p}{2}$ dollars $\\Big)$ equals $2001.$ Then we can write \\[fx + \\frac{hx}{2}=2001.\\]\nSubstituting $h=140-f$ into the second equation, we get \\[f \\cdot p +\\frac{(140-f)p}{2}=f \\cdot p+\\frac{140p-f\\cdot p}{2}=\\frac{f\\cdot p+140p}{2}=2001.\\]\nMultiplying by $2$ and subtracting $140p$ gives us \\[f\\cdot p=4002-140p.\\]\nSince the problem states that $x$ is a whole number, $140p$ will be some integer multiple of $140$ that ends in a $0$ . Thus, $4002-140p$ will end in a $2$ . Looking at the answer choices, only $\\boxed{782}$ satisfies that condition.", "Let's multiply ticket costs by $2$ , then the half price becomes an integer, and the charity sold $140$ tickets worth a total of $4002$ dollars.\nLet $h$ be the number of half price tickets, we then have $140-h$ full price tickets. The cost of $140-h$ full price tickets is equal to the cost of $280-2h$ half price tickets.\nHence we know that $h+(280-2h) = 280-h$ half price tickets cost $4002$ dollars. Then a single full price ticket costs $\\frac{4002}{280-h}$ dollars, and this must be an integer. Thus $280-h$ must be a divisor of $4002$ . Keeping in mind that $0\\leq h\\leq 140$ , we are looking for a divisor between $140$ and $280$ , inclusive.\nThe prime factorization of $4002$ is $4002=2\\cdot 3\\cdot 23\\cdot 29$ . We can easily find out that the only divisor of $4002$ within the given range is $2\\cdot 3\\cdot 29 = 174$\nThis gives us $280-h=174$ , hence there were $h=106$ half price tickets and $140-h = 34$ full price tickets.\nIn our modified setting (with prices multiplied by $2$ ) the price of a half price ticket is $\\frac{4002}{174} = 23$ . In the original setting this is the price of a full price ticket. Hence $23\\cdot 34 = \\boxed{782}$ dollars are raised by the full price tickets.", "Let the cost of the full price ticket be $x$ , the number of full-price tickets be $A$ , and the number of half-price tickets be $B$\nLet's multiply both sides of the equation that naturally follows by 2. We have\n\\[2Ax+Bx=4002\\]\nAnd we have $A+B=140\\implies B=140-A$\nPlugging in, we get $\\implies 2Ax+(140-A)(x)=4002$\nSimplifying, we get $Ax+140x=4002$\nFactoring out the $x$ , we get $x(A+140)=4002\\implies x=\\frac{4002}{A+140}$\nWe see that the fraction has to simplify to an integer (the full price is a whole dollar amount)\nThus, $A+140$ must be a factor of 4002.\nConsider the prime factorization of $4002$ $2\\times3\\times23\\times29$\n$A$ must be a positive integer. So, we seek a factor of $4002$ to set equal to $A+140$ so that we get an integer solution for $A$ that is less than $140$ . By guess-and-check OR inspection, the appropriate factor is $174$ $2\\times3\\times29$ ), meaning that $A$ has a value of $34$ . Plug this into the above equation for $x$ to get $x = 23$\nTherefore, the price of full tickets out of $2001$ is $23\\times34=\\boxed{782}$", "Let $f$ equal the number of full-price tickets, and let $h$ equal the number of half-price tickets. Additionally, suppose that the price of $f$ is $p$ . We are trying to solve for $f \\cdot p$\nSince the total number of tickets sold is $140$ , we know that \\[f+h=140.\\] The sales from full-price tickets ( $f \\cdot p$ ) plus the sales from half-price tickets $\\Big(\\frac{h \\cdot p}{2}$ , because each hall-price ticket costs $\\frac{p}{2}$ dollars $\\Big)$ equals $2001.$ Then we can write \\[fx + \\frac{hx}{2}=2001.\\]\nSubstituting $h=140-f$ into the second equation, we get \\[f \\cdot p +\\frac{(140-f)p}{2}=f \\cdot p+\\frac{140p-f\\cdot p}{2}=\\frac{f\\cdot p+140p}{2}=2001.\\]\nMultiplying by $2$ and subtracting $140p$ gives us \\[f\\cdot p=4002-140p.\\]\nSince the problem states that $x$ is a whole number, $140p$ will be some integer multiple of $140$ that ends in a $0$ . Thus, $4002-140p$ will end in a $2$ . Looking at the answer choices, only $\\boxed{782}$ satisfies that condition." ]

[ "The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the GCD ( greatest common divisor ) of $a$ and $b$ . Therefore, the answer is $gcd(300,210)=\\boxed{30}.$", "Alternatively, note that $300p + 210g = 30(10p + 7g)$ is divisible by $30$ no matter what $p$ and $g$ are, so our answer must be divisible by $30$ . Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly. Since our supposition can be achieved, the answer is $\\boxed{30}$" ]

[ "Let the listed price be $x$ . Since all the answer choices are above $\\textdollar100$ , we can assume $x > 100$ . Thus the discounts after the coupons are used will be as follows:\nCoupon 1: $x\\times10\\%=.1x$\nCoupon 2: $20$\nCoupon 3: $18\\%\\times(x-100)=.18x-18$\nFor coupon $1$ to give a greater price reduction than the other coupons, we must have $.1x>20\\implies x>200$ and $.1x>.18x-18\\implies.08x<18\\implies x<225$\nThe only choice that satisfies such conditions is $\\boxed{219.95}$", "For coupon $1$ to be the most effective, we want 10% of the price to be greater than 20. This clearly occurs if the value is over 200. For coupon 1 to be more effective than coupon 3, we want to minimize the value over 200, so $\\boxed{219.95}$ is the smallest number over 200.", "Let the listed price be $x$ . Since all the answer choices are above $\\textdollar100$ , we can assume $x > 100$ . Thus the discounts after the coupons are used will be as follows:\nCoupon 1: $x\\times10\\%=.1x$\nCoupon 2: $20$\nCoupon 3: $18\\%\\times(x-100)=.18x-18$\nFor coupon $1$ to give a greater price reduction than the other coupons, we must have $.1x>20\\implies x>200$ and $.1x>.18x-18\\implies.08x<18\\implies x<225$\nThe only choice that satisfies such conditions is $\\boxed{219.95}$", "For coupon $1$ to be the most effective, we want 10% of the price to be greater than 20. This clearly occurs if the value is over 200. For coupon 1 to be more effective than coupon 3, we want to minimize the value over 200, so $\\boxed{219.95}$ is the smallest number over 200." ]

[ "Consider any single tile:\n\nIf the side of the small square is $a$ , then the area of the tile is $9a^2$ , with $4a^2$ covered by squares and $5a^2$ by pentagons.\nHence exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane. When expressed as a percentage, this is $55.\\overline{5}\\%$ , and the closest integer to this value is $\\boxed{56}$", "Consider any single tile:\n\nIf the side of the small square is $a$ , then the area of the tile is $9a^2$ , with $4a^2$ covered by squares and $5a^2$ by pentagons.\nHence exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane. When expressed as a percentage, this is $55.\\overline{5}\\%$ , and the closest integer to this value is $\\boxed{56}$" ]

[ "Rewrite the equation\\[ $\\frac{a}{b}+\\frac{14b}{9a}=k$ \\]in two different forms. First, multiply both sides by $b$ and subtract $a$ to obtain\\[ $\\frac{14b^2}{9a}=bk-a.$ \\]Because $a$ $b$ , and $k$ are integers, $14b^2$ must be a multiple of $a$ , and because $a$ and $b$ have no common factors greater than 1, it follows that 14 is divisible by $a$ . Next, multiply both sides of the original equation by $9a$ and subtract $14b$ to obtain\\[ $\\frac{9a^2}{b}=9ak-14b.$ \\]This shows that $9a^2$ is a multiple of $b$ , so 9 must be divisible by $b$ . Thus if $(a,b)$ is a solution, then $b=1$ $3$ , or $9$ , and $a=1$ , 2, 7, or 14. This gives a total of twelve possible solutions $(a,b)$ , each of which can be checked quickly. $\\[\n\\frac{a}{b}+\\frac{14b}{9a}\\]$ (Error compiling LaTeX. Unknown error_msg) will only be an integer when $(a,b) \\in \\{(1,3),(2,3), (7,3), (14,3)\\},$ for a total of $\\boxed{4}$ pairs." ]

[ "Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives.\nLet $d$ be the distance still needed to travel after $1$ hour. We have that $\\dfrac{d}{50}+1.5=\\dfrac{d}{35}$ , where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early.\nSimplifying gives $7d+525=10d$ , or $d=175$\nNow, we must add an extra $35$ miles traveled in the first hour, giving a total of $\\boxed{210}$ miles.", "Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices.\nQuickly checking, we know that neither choice $\\textbf{(A)}$ or choice $\\textbf{(B)}$ work, but $\\textbf{(C)}$ does. We can verify as follows. After $1$ hour at $35 \\text{ mph}$ , David has $175$ miles left. This then takes him $3.5$ hours at $50 \\text{ mph}$ . But $210/35 = 6 \\text{ hours}$ . Since $1+3.5 = 4.5 \\text{ hours}$ is $1.5$ hours less than $6$ , our answer is $\\boxed{210}$", "Let the total distance be $d$ . Then $d=35(t+1)$ . Since 1 hour has passed, and he increased his speed by $15$ miles per hour, and he had already traveled $35$ miles, the new equation is $d-35=50(t-1-\\frac{1}{2})=50(t-\\frac{3}{2})$ . Solving, $d=35t+35=50t-40$ $15t=75$ $t=5$ , and $d=35(5+1)=35\\cdot 6=210 \\Longrightarrow \\boxed{210}$", "Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives.\nLet $d$ be the distance still needed to travel after $1$ hour. We have that $\\dfrac{d}{50}+1.5=\\dfrac{d}{35}$ , where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early.\nSimplifying gives $7d+525=10d$ , or $d=175$\nNow, we must add an extra $35$ miles traveled in the first hour, giving a total of $\\boxed{210}$ miles.", "Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices.\nQuickly checking, we know that neither choice $\\textbf{(A)}$ or choice $\\textbf{(B)}$ work, but $\\textbf{(C)}$ does. We can verify as follows. After $1$ hour at $35 \\text{ mph}$ , David has $175$ miles left. This then takes him $3.5$ hours at $50 \\text{ mph}$ . But $210/35 = 6 \\text{ hours}$ . Since $1+3.5 = 4.5 \\text{ hours}$ is $1.5$ hours less than $6$ , our answer is $\\boxed{210}$", "Let the total distance be $d$ . Then $d=35(t+1)$ . Since 1 hour has passed, and he increased his speed by $15$ miles per hour, and he had already traveled $35$ miles, the new equation is $d-35=50(t-1-\\frac{1}{2})=50(t-\\frac{3}{2})$ . Solving, $d=35t+35=50t-40$ $15t=75$ $t=5$ , and $d=35(5+1)=35\\cdot 6=210 \\Longrightarrow \\boxed{210}$" ]

[ "We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\\boxed{6}$", "We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\\boxed{6}$" ]

[ "The numbers on one die total $1+2+3+4+5+6 = 21$ , so the numbers\non the three dice total $63$ . Numbers $1, 1, 2, 3, 4, 5, 6$ are visible, and these total $22$ .\nThis leaves $63 - 22 = \\boxed{41}$ not seen." ]

[ "The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the GCD ( greatest common divisor ) of $a$ and $b$ . Therefore, the answer is $gcd(300,210)=\\boxed{30}.$", "Alternatively, note that $300p + 210g = 30(10p + 7g)$ is divisible by $30$ no matter what $p$ and $g$ are, so our answer must be divisible by $30$ . Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly. Since our supposition can be achieved, the answer is $\\boxed{30}$" ]

[ "Denote the number of horizontal lines drawn as $x$ , and the number of vertical lines drawn as $y$ . The number of basic rectangles is $(x - 1)(y - 1)$ $5x + 4y = 2007 \\Longrightarrow y = \\frac{2007 - 5x}{4}$ . Substituting, we find that $(x - 1)\\left(-\\frac 54x + \\frac{2003}4\\right)$\nFOIL this to get a quadratic, $-\\frac 54x^2 + 502x - \\frac{2003}4$ . Use $\\frac{-b}{2a}$ to find the maximum possible value of the quadratic: $x = \\frac{-502}{-2 \\cdot \\frac 54} = \\frac{1004}5 \\approx 201$ . However, this gives a non-integral answer for $y$ . The closest two values that work are $(199,253)$ and $(203,248)$\nWe see that $252 \\cdot 198 = 49896 > 202 \\cdot 247 = 49894$ . The solution is $\\boxed{896}$" ]

[ "1990 AIME-12.png\nThe easiest way to do this seems to be to find the length of each of the sides and diagonals. To do such, draw the radii that meet the endpoints of the sides/diagonals; this will form isosceles triangles . Drawing the altitude of those triangles and then solving will yield the respective lengths.\nAdding all of these up, we get $12[6(\\sqrt{6} - \\sqrt{2}) + 12 + 12\\sqrt{2} + 12\\sqrt{3} + 6(\\sqrt{6}+\\sqrt{2})] + 6 \\cdot 24$\n$= 12(12 + 12\\sqrt{2} + 12\\sqrt{3} + 12\\sqrt{6}) + 144 = 288 + 144\\sqrt{2} + 144\\sqrt{3} + 144\\sqrt{6}$ . Thus, the answer is $144 \\cdot 5 = \\boxed{720}$", "A second method involves drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon. Since the distance from the center to a vertex of the 12-gon is $12$ , the Law of Cosines can be applied to this isosceles triangle, to give:\n$a^2 = 12^2 + 12^2 - 2\\cdot 12\\cdot 12\\cdot \\cos \\theta$\n$a^2 = 2\\cdot 12^2 - 2\\cdot 12^2 \\cos \\theta$\n$a^2 = 2\\cdot 12^2 (1 - \\cos \\theta)$\n$a = 12\\sqrt{2} \\cdot \\sqrt{1 - \\cos \\theta}$\nThere are six lengths of sides/diagonals, corresponding to $\\theta = {30^{\\circ}, 60^{\\circ}, 90^{\\circ}, 120^{\\circ}, 150^{\\circ}, 180^{\\circ}}$\nCall these lengths $a_1, a_2, a_3, a_4, a_5, a_6$ from shortest to longest. The total length $l$ that is asked for is\n$l = 12(a_1 + a_2 + a_3 + a_4 + a_5) + 6a_6$ , noting that $a_6$ as written gives the diameter of the circle, which is the longest diagonal.\n$l = 12[12\\sqrt{2} (\\sqrt{1 - \\cos 30^{\\circ}}+\\sqrt{1-\\cos 60^{\\circ}}+\\sqrt{1-\\cos 90^{\\circ}}+\\sqrt{1-\\cos 120^{\\circ}}+\\sqrt{1 - \\cos 150^{\\circ}})] + 6 \\cdot 24$\n$l = 144\\sqrt{2} \\left(\\sqrt{1 - \\frac{\\sqrt{3}}{2}} + \\sqrt{1 - \\frac{1}{2}} + \\sqrt{1-0} + \\sqrt{1 + \\frac{1}{2}} + \\sqrt{1 + \\frac{\\sqrt{3}}{2}}\\right) + 144$\n$l = 144\\sqrt{2} \\left(\\sqrt{1 - \\frac{\\sqrt{3}}{2}} + \\frac{\\sqrt{2}}{2} + 1 + \\frac{\\sqrt{6}}{2} + \\sqrt{1 + \\frac{\\sqrt{3}}{2}}\\right) + 144$\nTo simplify the two nested radicals, add them, and call the sum $x$\n$x = \\sqrt{1 - \\frac{\\sqrt{3}}{2}} + \\sqrt{1 + \\frac{\\sqrt{3}}{2}}$\nSquaring both sides, the F and L part of FOIL causes the radicals to cancel, leaving:\n$x^2 = 2 + 2\\sqrt{\\left(1 - \\frac{\\sqrt{3}}{2}\\right)\\left(1 + \\frac{\\sqrt{3}}{2}\\right)}$\n$x^2 = 2 + 2\\sqrt{1 - \\frac{3}{4}}$\n$x^2 = 2 + 2\\sqrt{\\frac{1}{4}}$\n$x^2 = 2 + 2\\cdot\\frac{1}{2}$\n$x = \\sqrt{3}.$\nPlugging that sum $x$ back into the equation for $l$ , we find\n$l = 144\\sqrt{2} \\left(\\frac{\\sqrt{2}}{2} + 1 + \\frac{\\sqrt{6}}{2} + \\sqrt{3}\\right) + 144$\n$l = 144 + 144\\sqrt{2} + 144\\sqrt{3} + 144\\sqrt{6} + 144$\nThus, the desired quantity is $144\\cdot 5 = \\boxed{720}$", "Begin as in solution 2, drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon. Apply law of cosines on $\\theta = {30^{\\circ}, 60^{\\circ}, 90^{\\circ}, 120^{\\circ}, 150^{\\circ}, 180^{\\circ}}$ to get $d^2 = 288 - 288 \\cos \\theta$ where $d$ is the diagonal or sidelength distance between two points on the 12-gon. Now, $d=\\sqrt{288-288 \\cos \\theta}$ . Instead of factoring out $288$ as in solution 2, factor out $\\sqrt{576}$ instead-the motivation for this is to make the expression look like the half angle identity, and the fact that $\\sqrt{576}$ is an integer doesn't hurt. Now, we have that $d=24 \\sin \\frac{\\theta}{2}$ , which simplifies things quite nicely. Continue as in solution 2, computing half-angle sines instead of nested radicals, to obtain $\\boxed{720}$" ]

[ "If a point $z = r\\text{cis}\\,\\theta$ is in $R$ , then the point $\\frac{1}{z} = \\frac{1}{r} \\text{cis}\\, \\left(-\\theta\\right)$ is in $S$ (where cis denotes $\\text{cis}\\, \\theta = \\cos \\theta + i \\sin \\theta$ ). Since $R$ is symmetric every $60^{\\circ}$ about the origin, it suffices to consider the area of the result of the transformation when $-30 \\le \\theta \\le 30$ , and then to multiply by $6$ to account for the entire area.\nWe note that if the region $S_2 = \\left\\lbrace\\frac{1}{z}|z \\in R_2\\right\\rbrace$ , where $R_2$ is the region (in green below) outside the circle of radius $1/\\sqrt{3}$ centered at the origin, then $S_2$ is simply the region inside a circle of radius $\\sqrt{3}$ centered at the origin. It now suffices to find what happens to the mapping of the region $R-R_2$ (in blue below).\nThe equation of the hexagon side in that region is $x = r \\cos \\theta = \\frac{1}{2}$ , which is transformed to $\\frac{1}{r} \\cos (-\\theta) = \\frac{1}{r} \\cos \\theta =$ 2 . Let $r\\text{cis}\\,\\theta = a+bi$ where $a,b \\in \\mathbb{R}$ ; then $r = \\sqrt{a^2 + b^2}, \\cos \\theta = \\frac{a}{\\sqrt{a^2 + b^2}}$ , so the equation becomes $a^2 - 2a + b^2 = 0 \\Longrightarrow (a-1)^2 + b^2 = 1$ . Hence the side is sent to an arc of the unit circle centered at $(1,0)$ , after considering the restriction that the side of the hexagon is a segment of length $1/\\sqrt{3}$\nIncluding $S_2$ , we find that $S$ is the union of six unit circles centered at $\\text{cis}\\, \\frac{k\\pi}{6}$ $k = 0,1,2,3,4,5$ , as shown below.\nThe area of the regular hexagon is $6 \\cdot \\left( \\frac{\\left(\\sqrt{3}\\right)^2 \\sqrt{3}}{4} \\right) = \\frac{9}{2}\\sqrt{3}$ . The total area of the six $120^{\\circ}$ sectors is $6\\left(\\frac{1}{3}\\pi - \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\sqrt{3}\\right) = 2\\pi - \\frac{3}{2}\\sqrt{3}$ . Their sum is $2\\pi + \\sqrt{27}$ , and $a+b = \\boxed{029}$ .\n- Th3Numb3rThr33", "We can describe the line parallel to the imaginary axis $x=\\frac{1}{2}$ using polar coordinates as $r(\\theta)=\\dfrac{1}{2\\cos{\\theta}},$\nwhich rearranges to $z=\\left(\\dfrac{1}{2\\cos{\\theta}}\\right)(cis{\\theta})\\implies \\frac{1}{z}=2\\cos{\\theta}cis(-\\theta).$\nDenote the area of $S$ as $[S].$ Now, dividing the hexagon to 12 equal parts, we have the following integral:\n$[S] = 12\\int_{0}^{\\frac{\\pi}{6}}\\frac{1}{2}r^2 d\\theta = 12\\int_{0}^{\\frac{\\pi}{6}}\\frac{1}{2}(2\\cos\\theta)^2 d\\theta.$\nThankfully, this is a routine computation:\n$[S] = 12\\int_{0}^{\\frac{\\pi}{6}}2(\\cos\\theta)^2 d\\theta = 12\\int_{0}^{\\frac{\\pi}{6}}(\\cos{2\\theta}+1)d\\theta$\n$[S] = 12\\int_{0}^{\\frac{\\pi}{6}}(\\cos{2\\theta}+1)d\\theta = 12\\left[\\frac{1}{2}\\sin{2\\theta}+\\theta\\right]_0^{\\frac{\\pi}{6}}=12\\left(\\frac{\\sqrt{3}}{4}+\\frac{\\pi}{6}\\right)=2\\pi+3\\sqrt{3}=2\\pi + \\sqrt{27}$\n$a+b = \\boxed{029}$" ]

[ "Think about each plane independently. From the top vertex, we can go down to any of 5 different points in the second plane. From that point, there are 9 ways to go to another point within the second plane: rotate as many as 4 points clockwise or as many 4 counterclockwise, or stay in place. Then, there are 2 paths from that point down to the third plane. Within the third plane, there are 9 paths as well (consider the logic from the second plane). Finally, there is only one way down to the bottom vertex. Therefore there are $5 \\cdot 9 \\cdot 2 \\cdot 9 \\cdot 1=\\boxed{810}$ paths.", "Assume an ant is on the top of this icosahedron. Note that the icosahedron has two pentagon planes and two points where the ant starts and ends. Also note that when the ant hits a vertex of the pentagon, there is only two ways to go down. When the ant ends up at the last vertex and is about to head down, there is only one way to go down.\nCase $1$ : The ant move down to the first pentagon, then the next pentagon, and finally to the last point in a total of four moves; there are a total of $5 \\cdot 2 \\cdot 1 = 10$ ways to achieve this.\nCase $2$ : The ant goes to the first pentagon and moves to another vertex in the same pentagon. The ant has a total of $8$ paths doing this since it may go through $1, 2, 3,$ or $4$ edges from either two directions when going across the edges of the pentagon (recall the ant may not repeat a move to another vertex) . Now the ant moves to the second pentagon below. The ant again has a total of $8$ moves if he wanders around the pentagon. Finally, the ant moves down after wandering to the last vertex. There is a total of $5 \\cdot 8 \\cdot 2 \\cdot 8 = 640$ ways.\nCase $3$ : Assume the ant goes to the first pentagon and wanders around. Then the ant goes the next pentagon and then heads directly to the last vertex. There are $5 \\cdot 8 \\cdot 2 = 80$ ways.\nCase $4$ : Now let the ant go to the first pentagon and then go directly down to the next pentagon. The ant wanders around the second pentagon before heading to the last vertex. Like case $3$ above, there are $5 \\cdot 2 \\cdot 8 = 80$ ways.\nAdding up all four cases, we get $10 + 640 + 80 + 80 =\\boxed{810}$ total paths, as desired. ~skyscraper", "Go to 2020 AMC 10A #19 , and connect all of the centers of the faces on the dodecahedron to get the icosahedron. The answer is $\\boxed{810}$" ]

[ "We color face $6$ red and face $5$ yellow. Note that from the octahedron, face $5$ and face $?$ do not share anything in common. From the net, face $5$ shares at least one vertex with all other faces except face $1,$ which is shown in green: Therefore, the answer is $\\boxed{1}.$", "We label the octohedron going triangle by triangle until we reach the $?$ triangle. The triangle to the left of the $Q$ should be labeled with a $6$ . Underneath triangle $6$ is triangle $5$ . The triangle to the right of triangle $5$ is triangle $4$ and further to the right is triangle $3$ . Finally, the side of triangle $3$ under triangle $Q$ is $2$ , so the triangle to the right of $Q$ is $\\boxed{1}$", "Notice that the triangles labeled $2, 3, 4,$ and $5$ make the bottom half of the octahedron, as shown below: Therefore, $\\textbf{(B)}, \\textbf{(C)}, \\textbf{(D)},$ and $\\textbf{(E)}$ are clearly not the correct answer. Thus, the only choice left is $\\boxed{1}$", "The first half of the octahedron will need $4$ triangles connected to one another to form it. We can choose the triangles $4$ $5$ $6$ , and $7$ and form the half around the vertex they all share. That leaves triangles $1$ $3$ $2$ , and $Q$ to form the second half. Triangle $3$ will definitely share its sides with triangles $1$ and $2$ , leaving them to share their second side with triangle $Q$ . Since triangle $Q$ will certainly share its left side with triangle $2$ , the only triangle left to share its right side is triangle $\\boxed{1}$" ]

[ "The easiest way is to unwrap the cone into a circular sector. Center the sector at the origin with one radius on the positive $x$ -axis and the angle $\\theta$ going counterclockwise. The circumference of the base is $C=1200\\pi$ . The sector's radius (cone's sweep) is $R=\\sqrt{r^2+h^2}=\\sqrt{600^2+(200\\sqrt{7})^2}=\\sqrt{360000+280000}=\\sqrt{640000}=800$ . Setting $\\theta R=C\\implies 800\\theta=1200\\pi\\implies\\theta=\\frac{3\\pi}{2}$\nIf the starting point $A$ is on the positive $x$ -axis at $(125,0)$ then we can take the end point $B$ on $\\theta$ 's bisector at $\\frac{3\\pi}{4}$ radians along the $y=-x$ line in the second quadrant. Using the distance from the vertex puts $B$ at $(-375,-375)$ . Thus the shortest distance for the fly to travel is along segment $AB$ in the sector, which gives a distance $\\sqrt{(-375-125)^2+(-375-0)^2}=125\\sqrt{4^2+3^2}=\\boxed{625}$" ]

[ "We can just look at a quarter circle inscribed in a $45-45-90$ right triangle. We can then extend a radius, $r$ to one of the sides creating an $r,r, r\\sqrt{2}$ right triangle. This means that we have $r + r\\sqrt{2} = 8\\sqrt{2}$ so $r = \\frac{8\\sqrt{2}}{1+\\sqrt{2}} = 16 - 8\\sqrt{2}$ . Then the diameter is $32 - \\sqrt{512}$ giving us $32 + 512 = \\boxed{544}$", "We proceed by finding the area of the square in 2 different ways. The square is obviously 8*8=64, but we can also find the area in terms of d. From the center of the circle, draw radii that hit the points where the square is tangent to the semicircle. Then the square's area is the area of the small square +2* the area of the trapezoids on the corners+ the area of an isoceles triangle. Adding these all up gives $64=\\frac{d^2}{4}+\\frac{d^2}{4}+(8-\\frac{d\\sqrt{2}}{2}+\\frac{d}{2})(8-\\frac{d}{2})$ Simplifying gives $d-16\\sqrt{2}+d\\sqrt{2}=0$ . Solving gives $d=32-16\\sqrt{2}=32-\\sqrt{512}$ so the answer is $32+512= \\boxed{544}$", "It is easy after getting the image, after drawing labeling the lengths of those segments, assume the radius is $x$ , we can see $x=\\sqrt{2}(8-x)$ and we get $2x=32-\\sqrt{512}$ and we have the answer $\\boxed{544}$ ~bluesoul" ]

[ "Define the sum as $s$ . Since $a_n\\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$ , the sum will be:\nThus $s = \\frac{a_{28} + a_{30}}{2}$ , and $a_{28},\\,a_{30}$ are both given; the last four digits of their sum is $3668$ , and half of that is $1834$ . Therefore, the answer is $\\boxed{834}$ .−", "Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:\n$a_{1}\\equiv 1 \\pmod {1000} \\\\ a_{2}\\equiv 1 \\pmod {1000} \\\\ a_{3}\\equiv 1 \\pmod {1000} \\\\ a_{4}\\equiv 3 \\pmod {1000} \\\\ a_{5}\\equiv 5 \\pmod {1000} \\\\ \\cdots \\\\ a_{25} \\equiv 793 \\pmod {1000} \\\\ a_{26} \\equiv 281 \\pmod {1000} \\\\ a_{27} \\equiv 233 \\pmod {1000} \\\\ a_{28} \\equiv 307 \\pmod {1000}$\nAdding all the residues shows the sum is congruent to $\\boxed{834}$ mod $1000$", "All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given $a_{28}, a_{29},$ and $a_{30}$ , so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some $p, q, r$ such that $\\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}$ . From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that $(p, q, r) = (\\frac{1}{2}, 0, \\frac{1}{2})$ , at least for the first few terms. From this, we have that $\\sum_{k=1}^{28}{a_k} = \\frac{a_{28}+a_{30}}{2} \\equiv{\\boxed{834}(\\mod 1000)$" ]

[ "Define a function $f(n)$ on the non-negative integers, as \\[f(n) = \\frac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \\frac{a_n}{a_{n+1}}+\\frac{a_{n+1}}{a_{n}}\\] We want $\\left\\lfloor f(2006) \\right\\rfloor$\nConsider the relation $a_{n+1}a_{n-1}=a_{n}^{2}+2007$ . Dividing through by $a_{n}a_{n-1}$ , we get \\[.\\phantom{------------} \\frac{a_{n+1}}{a_{n}} = \\frac{a_{n}}{a_{n-1}} + \\frac{2007}{a_{n}a_{n-1}} \\phantom{------------} (1)\\] and dividing through by $a_{n}a_{n+1}$ , we get \\[.\\phantom{------------} \\frac{a_{n-1}}{a_{n}} = \\frac{a_{n}}{a_{n+1}} + \\frac{2007}{a_{n}a_{n+1}} \\phantom{------------} (2)\\] Adding LHS of $(1)$ with RHS of $(2)$ (and vice-versa), we get \\[\\frac{a_{n+1}}{a_{n}} + \\frac{a_{n}}{a_{n+1}} + \\frac{2007}{a_{n}a_{n+1}} = \\frac{a_{n}}{a_{n-1}} + \\frac{a_{n-1}}{a_{n}} + \\frac{2007}{a_{n}a_{n-1}}\\] i.e. \\[f(n)+ \\frac{2007}{a_{n}a_{n+1}} = f(n-1) + \\frac{2007}{a_{n}a_{n-1}}\\] Summing over $n=1$ to $n=2006$ , we notice that most of the terms on each side cancel against the corresponding term on the other side. We are left with \\[f(2006) + \\frac{2007}{a_{2006}a_{2007}} = f(0) + \\frac{2007}{a_{1}a_{0}}\\] We have $f(0) = 2$ , and $2007/a_0a_1 = 2007/9 = 223$ . So \\[f(2006) = 2 + 223 - \\frac{2007}{a_{2006}a_{2007}} = 224 + \\left( 1 - \\frac{2007}{a_{2006}a_{2007}}\\right)\\] Since all the $a_i$ are positive, $(1)$ tells us that the ratio $a_{n+1}/a_n$ of successive terms is increasing. Since this ratio starts with $a_1/a_0 = 1$ , this means that the sequence $(a_n)$ is increasing. Since $a_3=672$ already, we must have $a_{2006}a_{2007} > 672^2 > 2007$ . It follows that $\\left\\lfloor f(2006) \\right\\rfloor = \\boxed{224}$", "We are given that\n$a_{n+1}a_{n-1}= a_{n}^{2}+2007$ $a_{n-1}^{2}+2007 = a_{n}a_{n-2}$\nAdd these two equations to get\nThis is an invariant . Defining $b_{i}= \\frac{a_{i}}{a_{i-1}}$ for each $i \\ge 2$ , the above equation means\n$b_{n+1}+\\frac{1}{b_{n}}= b_{n}+\\frac{1}{b_{n-1}}$\nWe can thus calculate that $b_{2007}+\\frac{1}{b_{2006}}= b_{3}+\\frac{1}{b_{2}}= 225$ . Using the equation $a_{2007}a_{2005}=a_{2006}^{2}+2007$ and dividing both sides by $a_{2006}a_{2005}$ , notice that $b_{2007}= \\frac{a_{2007}}{a_{2006}}= \\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \\frac{a_{2006}}{a_{2005}}= b_{2006}$ . This means that\n$b_{2007}+\\frac{1}{b_{2007}}< b_{2007}+\\frac{1}{b_{2006}}= 225$ . It is only a tiny bit less because all the $b_i$ are greater than $1$ , so we conclude that the floor of $\\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\\frac{1}{b_{2007}}$ is $\\boxed{224}$", "The equation $a_{n+1}a_{n-1}-a_n^2=2007$ looks like the determinant \\[\\left|\\begin{array}{cc}a_{n+1}&a_n\\\\a_n&a_{n-1}\\end{array}\\right|=2007.\\] Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence $b_n$ defined by $b_1=b_2=3$ and $b_{n+1}=\\alpha b_n+\\beta b_{n-1}$ for $n\\ge 2$ . We wish to find $\\alpha$ and $\\beta$ such that $a_n=b_n$ for all $n\\ge 1$ . To do this, we use the following matrix form of a linear recurrence relation\n\\[\\left(\\begin{array}{cc}b_{n+1}&b_n\\\\b_n&b_{n-1}\\end{array}\\right)=\\left(\\begin{array}{cc}\\alpha&\\beta\\\\1&0\\end{array}\\right)\\left(\\begin{array}{cc}b_{n}&b_{n-1}\\\\b_{n-1}&b_{n-2}\\end{array}\\right).\\]\nWhen we take determinants, this equation becomes\n\\[\\text{det}\\left(\\begin{array}{cc}b_{n+1}&b_n\\\\b_n&b_{n-1}\\end{array}\\right)=\\text{det}\\left(\\begin{array}{cc}\\alpha&\\beta\\\\1&0\\end{array}\\right)\\text{det}\\left(\\begin{array}{cc}b_{n}&b_{n-1}\\\\b_{n-1}&b_{n-2}\\end{array}\\right).\\]\nWe want \\[\\text{det}\\left(\\begin{array}{cc}b_{n+1}&b_n\\\\b_n&b_{n-1}\\end{array}\\right)=2007\\] for all $n$ . Therefore, we replace the two matrices by $2007$ to find that\n\\[2007=\\text{det}\\left(\\begin{array}{cc}\\alpha&\\beta\\\\1&0\\end{array}\\right)\\cdot 2007\\] \\[1=\\text{det}\\left(\\begin{array}{cc}\\alpha&\\beta\\\\1&0\\end{array}\\right)=-\\beta.\\] Therefore, $\\beta=-1$ . Computing that $a_3=672$ , and using the fact that $b_3=\\alpha b_2-b_1$ , we conclude that $\\alpha=225$ . Clearly, $a_1=b_1$ $a_2=b_2$ , and $a_3=b_3$ . We claim that $a_n=b_n$ for all $n\\ge 1$ . We proceed by induction . If $a_k=b_k$ for all $k\\le n$ , then clearly, \\[b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.\\] We also know by the definition of $b_{n+1}$ that\n\\[\\text{det}\\left(\\begin{array}{cc}b_{n+1}&b_n\\\\b_n&b_{n-1}\\end{array}\\right)=\\text{det}\\left(\\begin{array}{cc}\\alpha&\\beta\\\\1&0\\end{array}\\right)\\text{det}\\left(\\begin{array}{cc}b_{n}&b_{n-1}\\\\b_{n-1}&b_{n-2}\\end{array}\\right).\\]\nWe know that the RHS is $2007$ by previous work. Therefore, $b_{n+1}b_{n-1}-b_n^2=2007$ . After substuting in the values we know, this becomes $b_{n+1}a_{n-1}-a_n^2=2007$ . Thinking of this as a linear equation in the variable $b_{n+1}$ , we already know that this has the solution $b_{n+1}=a_{n+1}$ . Therefore, by induction, $a_n=b_n$ for all $n\\ge 1$ . We conclude that $a_n$ satisfies the linear recurrence $a_{n+1}=225a_n-a_{n-1}$\nIt's easy to prove that $a_n$ is a strictly increasing sequence of integers for $n\\ge 3$ . Now\n\\[\\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\\frac{a_{2007}}{a_{2006}}+\\frac{a_{2006}}{a_{2007}}=\\frac{225a_{2006}-a_{2005}}{a_{2006}}+\\frac{a_{2006}}{a_{2007}}.\\]\n\\[=225+\\frac{a_{2006}}{a_{2007}}-\\frac{a_{2005}}{a_{2006}}=225+\\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.\\]\n\\[=225-\\frac{2007}{a_{2005}a_{2006}}.\\]\nThe sequence certainly grows fast enough such that $\\frac{2007}{a_{2005}a_{2006}}<1$ . Therefore, the largest integer less than or equal to this value is $\\boxed{224}$", "We will try to manipulate $\\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\\frac{a_1^2+a_2^2}{a_1a_2}$\n$\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_0+\\frac{a_1^2}{a_0}}{a_1} = \\frac{a_0+\\frac{a_1^2+2007}{a_0}-\\frac{2007}{a_0}}{a_1}$ Using the recurrence relation, $= \\frac{a_0+a_2-\\frac{2007}{a_0}}{a_1} = \\frac{a_0a_2+a_2^2-\\frac{2007a_2}{a_0}}{a_1a_2}$ Applying the relation to $a_0a_2$ $= \\frac{a_1^2+2007+a_2^2-\\frac{2007a_2}{a_0}}{a_1a_2} = \\frac{a_1^2+a_2^2}{a_1a_2}+\\frac{2007}{a_1a_2}-\\frac{2007}{a_0a_1}$\nWe can keep on using this method to get that $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\\frac{2007}{a_{2006}a_{2007}}-\\frac{2007}{a_{2005}a_{2006}}+\\frac{2007}{a_{2005}a_{2006}}-\\ldots-\\frac{2007}{a_{0}a_{1}}$\nThis telescopes to $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\\frac{2007}{a_{2006}a_{2007}}-\\frac{2007}{a_{0}a_{1}}$\nor $\\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \\frac{a_0^2+a_1^2}{a_0a_1}+\\frac{2007}{a_{0}a_{1}}-\\frac{2007}{a_{2006}a_{2007}}$\nFinding the first few values, we notice that they increase rapidly, so $\\frac{2007}{a_{2006}a_{2007}} < 1$ . Calculating the other values, $\\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\\frac{2007}{a_{2006}a_{2007}}$\nThe greatest number that does not exceed this is $\\boxed{224}$", "Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that $a_0 = a_1 = 0$ , and solving for $a_2$ and $a_3$ using the given relation we get $a_2 = 672 = 3(224)$ and $a_3 = 3(224^{2} + 223)$ , respectively. It will be clear why I decided to factor these expressions as I did momentarily.\nNext, let's see what the expression $\\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}$ looks like for small values of $n$ . For $n = 1$ , we get $\\frac{1 + 224^2}{224}$ , the floor of which is clearly $224$ because the $1$ in the numerator is insignificant. Repeating the procedure for $n + 1$ is somewhat messier, but we end up getting $\\frac{224^4 + 224^2\\cdot223\\cdot2 + 224^2 + 223^2}{224^3 + 224\\cdot223}$ . It's not too hard to see that $224^4$ is much larger than the sum of the remaining terms in the numerator, and that $224^3$ is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than $224^3$ , while the second-largest term in the denominator is smaller than $224^2$ . Thus, the floor of this expression will come out to be $224$ as well.\nNow we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time $n$ increases by $1$ , the degrees of both the numerator and denominator increase by $2$ , because we are squaring the $n+1th$ term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation $a_{n+1}^2 = (\\frac{a_{n}^2 + 2007}{a_{n-1}})^2$ ). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the $\\approx224:1$ ratio between the two.\nFor the non-greatest terms in the expression to offset this ratio for values of $n$ in the ballpark of $2006$ , they would have to have massive coefficients, because or else they are dwarfed by the additional $224$ attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to $\\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }$ for all $k\\geq2$ , whose $\\lim_{k\\to\\infty}=$ $\\boxed{224}$" ]

[ "The problem gives us a sequence defined by a recursion , so let's calculate a few values to get a feel for how it acts. We aren't given initial values, so let $a_1 = a$ and $a_2 = b$ . Then $a_3 = b - a$ $a_4 = (b - a) - b = -a$ $a_5 = -a - (b - a) = -b$ $a_6 = -b - (-a) = a - b$ $a_7 = (a - b) - (-b) = a$ and $a_8 = a - (a - b) = b$ . Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat. Thus, in particular $a_{j + 6} = a_j$ for all $j$ , and so repeating this $n$ times, $a_{j + 6n} = a_j$ for all integers $n$ and $j$\nBecause of this, the sum of the first 1492 terms can be greatly simplified: $1488 = 6 \\cdot 248$ is the largest multiple of 6 less than 1492, so \\[\\sum_{i = 1}^{1492} a_i = (a_{1489} + a_{1490} + a_{1491} + a_{1492})+ \\sum_{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + \\sum_{n = 0}^{247}\\sum_{j = 1}^6 a_{6n + j}\\] \\[=(a + b + (b - a) + (-a)) + \\sum_{n = 0}^{247}\\sum_{j = 1}^6 a_j = 2b - a,\\] where we can make this last step because $\\sum_{j = 1}^6 a_j = 0$ and so the entire second term of our expression is zero\nSimilarly, since $1980 = 6 \\cdot 330$ $\\sum_{i = 1}^{1985} a_i = (a_1 + a_2 + a_3 + a_4 + a_5) + \\sum_{i = 1}^{1980}a_i = a + b + (b - a) + (-a) + (-b) = b - a$\nFinally, $\\sum_{i = 1}^{2001}a_i = a_1 + a_2 + a_3 + \\sum_{i = 1}^{1998} a_i = a + b + (b - a) = 2b$\nThen by the givens, $2b - a = 1985$ and $b - a = 1492$ so $b = 1985 - 1492 = 493$ and so the answer is $2\\cdot 493 = \\boxed{986}$ .\nMinor edit by: PlainOldNumberTheory" ]

[ "Let the sum of all of the terms in the sequence be $\\mathbb{S}$ . Then for each integer $k$ $x_k = \\mathbb{S}-x_k-k \\Longrightarrow \\mathbb{S} - 2x_k = k$ . Summing this up for all $k$ from $1, 2, \\ldots, 100$\n\\begin{align*}100\\mathbb{S}-2(x_1 + x_2 + \\cdots + x_{100}) &= 1 + 2 + \\cdots + 100\\\\ 100\\mathbb{S} - 2\\mathbb{S} &= \\frac{100 \\cdot 101}{2} = 5050\\\\ \\mathbb{S}&=\\frac{2525}{49}\\end{align*}\nNow, substituting for $x_{50}$ , we get $2x_{50}=\\frac{2525}{49}-50=\\frac{75}{49} \\Longrightarrow x_{50}=\\frac{75}{98}$ , and the answer is $75+98=\\boxed{173}$", "Consider $x_k$ and $x_{k+1}$ . Let $S$ be the sum of the rest 98 terms. Then $x_k+k=S+x_{k+1}$ and $x_{k+1}+(k+1)=S+x_k.$ Eliminating $S$ we have $x_{k+1}-x_k=-\\dfrac{1}{2}.$ So the sequence is arithmetic with common difference $-\\dfrac{1}{2}.$\nIn terms of $x_{50},$ the sequence is $x_{50}+\\dfrac{49}{2}, x_{50}+\\dfrac{48}{2},\\cdots,x_{50}+\\dfrac{1}{2}, x_{50}, x_{50}-\\dfrac{1}{2}, \\cdots, x_{50}-\\dfrac{49}{2}, x_{50}-\\dfrac{50}{2}.$ Therefore, $x_{50}+50=99x_{50}-\\dfrac{50}{2}$\nSolving, we get $x_{50}=\\dfrac{75}{98}.$ The answer is $75+98=\\boxed{173}.$" ]

[ "Let $x = a_2$ ; then solving for the next several terms, we find that $a_3 = x^2,\\ a_4 = x(2x-1),\\ a_5$ $= (2x-1)^2,\\ a_6$ $= (2x-1)(3x-2)$ , and in general, $a_{2n} = f(n-1)f(n)$ $a_{2n+1} = f(n)^2$ , where $f(n) = nx - (n-1)$\nFrom \\[a_9 + a_{10} = f(4)^2 + f(4)f(5) = (4x-3)(9x-7) = 646 = 2\\cdot 17 \\cdot 19\\] , we find that by either the quadratic formula or trial-and-error/modular arithmetic that $x=5$ . Thus $f(n) = 4n+1$ , and we need to find the largest $n$ such that either $f(n)^2\\, \\mathrm{or}\\, f(n)f(n-1) < 1000$ . This happens with $f(7)f(8) = 29 \\cdot 33 = 957$ , and this is the $2(8) = 16$ th term of the sequence.\nThe answer is $957 + 16 = \\boxed{973}$", "Let $x = a_2$ . It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what $x$ can be. Finding that $x = 5$ works, after bashing out the rest of the terms we find that $a_{16} = 957$ and $a_{17} = 1089$ , hence our answer is $957 + 16 = \\boxed{973}$", "We can find the value of $a_{9}$ by its bounds using three conditions:\nRearranging conditions 1 and 2, we get:\n\\[\\frac{646}{3} < a_{9} < \\frac{646}{2}\\]\ntrying all the terms from the third condition, it is clear that $a_9 = 289$ is the only solution. \nThen we can calculate the next few terms from there since we have $a_{10}$ as well, to find that $a_{16} = 957$ and $a_{17} = 1089$ , thus we have our answer of $957 + 16 = \\boxed{973}$" ]

[ "Let $d$ be the common difference. Then $9$ $9+d+2=11+d$ $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and the third term is $2(-14) + 29 = 1\\Rightarrow\\boxed{1}$", "Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ $11+d$ , and $29+2d$ . Thus, we get the following equations:\n$9r=11+d\\Rightarrow d=9r-11$\n$9r^2=29+2d$\nPlugging in the first equation into the second, our equation becomes $9r^2=29+18r-22\\Longrightarrow9r^2-18r-7=0$ . By the quadratic formula, $r$ can either be $-\\frac{1}{3}$ or $\\frac{7}{3}$ . If $r$ is $-\\frac{1}{3}$ , the third term (of the geometric sequence) would be $1$ , and if $r$ is $\\frac{7}{3}$ , the third term would be $49$ . Clearly the minimum possible value for the third term of the geometric sequence is $\\boxed{1}$", "Let the three numbers be, in increasing order, $9,y,z$\nHence, we have that $9-y=y-z\\implies 9+z=2y$\nAlso, from the second part of information given, we get that\n$\\frac{9}{y+2}=\\frac{y+2}{z+20}\\implies 9(z+20)=(y+2)^2\\implies y=3(\\sqrt{z+20})-2$\nPlugging back in...\n$9+z=6(\\sqrt{z+20})-4\\implies (9+z)^2=36(z+20)$\nSimplifying, we get that $z^2-10z-551=0$\nApplying the quadratic formula, we get that $z=\\frac{10\\pm \\sqrt{2304}}{2}\\implies \\frac{10\\pm48}{2}$\nObviously, in order to minimize the value of $z$ , we have to subtract. Hence, $z=-19$\nHowever, the problem asks for the minimum value of the third term in a geometric progression.\nHence, the answer is $-19+20=\\boxed{1}$", "Let the arithmetic sequence be $9,9+x,9+2x$ and let the geometric sequence be $9,11+x,29+2x$ . Now, we just try all the solutions. If the last term is $1$ , then $x=-14$ . This gives the geometric sequence $9,-3,1$ which indeed works. The answer is $\\boxed{1}$", "The terms of the arithmetic progression are 9, $9+d$ , and $9+2d$ for some real number $d$ . The terms of the geometric progression are 9, $11+d$ , and $29+2d$ . Therefore $(11+d)^{2} = 9(29+2d) \\quad\\text{so}\\quad d^{2}+4d-140 = 0.$ Thus $d=10$ or $d=-14$ . The corresponding geometric progressions are $9, 21, 49$ and $9, -3, 1,$ so the smallest possible value for the third term of the geometric progression is $\\boxed{1}$", "List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is $\\boxed{1}$" ]

[ "Let $d$ be the common difference. Then $9$ $9+d+2=11+d$ $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and the third term is $2(-14) + 29 = 1\\Rightarrow\\boxed{1}$", "Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ $11+d$ , and $29+2d$ . Thus, we get the following equations:\n$9r=11+d\\Rightarrow d=9r-11$\n$9r^2=29+2d$\nPlugging in the first equation into the second, our equation becomes $9r^2=29+18r-22\\Longrightarrow9r^2-18r-7=0$ . By the quadratic formula, $r$ can either be $-\\frac{1}{3}$ or $\\frac{7}{3}$ . If $r$ is $-\\frac{1}{3}$ , the third term (of the geometric sequence) would be $1$ , and if $r$ is $\\frac{7}{3}$ , the third term would be $49$ . Clearly the minimum possible value for the third term of the geometric sequence is $\\boxed{1}$", "Let the three numbers be, in increasing order, $9,y,z$\nHence, we have that $9-y=y-z\\implies 9+z=2y$\nAlso, from the second part of information given, we get that\n$\\frac{9}{y+2}=\\frac{y+2}{z+20}\\implies 9(z+20)=(y+2)^2\\implies y=3(\\sqrt{z+20})-2$\nPlugging back in...\n$9+z=6(\\sqrt{z+20})-4\\implies (9+z)^2=36(z+20)$\nSimplifying, we get that $z^2-10z-551=0$\nApplying the quadratic formula, we get that $z=\\frac{10\\pm \\sqrt{2304}}{2}\\implies \\frac{10\\pm48}{2}$\nObviously, in order to minimize the value of $z$ , we have to subtract. Hence, $z=-19$\nHowever, the problem asks for the minimum value of the third term in a geometric progression.\nHence, the answer is $-19+20=\\boxed{1}$", "Let the arithmetic sequence be $9,9+x,9+2x$ and let the geometric sequence be $9,11+x,29+2x$ . Now, we just try all the solutions. If the last term is $1$ , then $x=-14$ . This gives the geometric sequence $9,-3,1$ which indeed works. The answer is $\\boxed{1}$", "The terms of the arithmetic progression are 9, $9+d$ , and $9+2d$ for some real number $d$ . The terms of the geometric progression are 9, $11+d$ , and $29+2d$ . Therefore $(11+d)^{2} = 9(29+2d) \\quad\\text{so}\\quad d^{2}+4d-140 = 0.$ Thus $d=10$ or $d=-14$ . The corresponding geometric progressions are $9, 21, 49$ and $9, -3, 1,$ so the smallest possible value for the third term of the geometric progression is $\\boxed{1}$", "List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is $\\boxed{1}$" ]

[ "Our original solid has volume equal to $V = \\frac13 \\pi r^2 h = \\frac13 \\pi 3^2\\cdot 4 = 12 \\pi$ and has surface area $A = \\pi r^2 + \\pi r \\ell$ , where $\\ell$ is the slant height of the cone. Using the Pythagorean Theorem , we get $\\ell = 5$ and $A = 24\\pi$\nLet $x$ denote the radius of the small cone. Let $A_c$ and $A_f$ denote the area of the painted surface on cone $C$ and frustum $F$ , respectively, and let $V_c$ and $V_f$ denote the volume of cone $C$ and frustum $F$ , respectively. Because the plane cut is parallel to the base of our solid, $C$ is similar to the uncut solid and so the height and slant height of cone $C$ are $\\frac{4}{3}x$ and $\\frac{5}{3}x$ , respectively. Using the formula for lateral surface area of a cone, we find that $A_c=\\frac{1}{2}c\\cdot \\ell=\\frac{1}{2}(2\\pi x)\\left(\\frac{5}{3}x\\right)=\\frac{5}{3}\\pi x^2$ . By subtracting $A_c$ from the surface area of the original solid, we find that $A_f=24\\pi - \\frac{5}{3}\\pi x^2$\nNext, we can calculate $V_c=\\frac{1}{3}\\pi r^2h=\\frac{1}{3}\\pi x^2 \\left(\\frac{4}{3}x\\right)=\\frac{4}{9}\\pi x^3$ . Finally, we subtract $V_c$ from the volume of the original cone to find that $V_f=12\\pi - \\frac{4}{9}\\pi x^3$ . We know that $\\frac{A_c}{A_f}=\\frac{V_c}{V_f}=k.$ Plugging in our values for $A_c$ $A_f$ $V_c$ , and $V_f$ , we obtain the equation $\\frac{\\frac{5}{3}\\pi x^2}{24\\pi - \\frac{5}{3}\\pi x^2}=\\frac{\\frac{4}{9}\\pi x^3}{12\\pi - \\frac{4}{9}\\pi x^3}$ . We can take reciprocals of both sides to simplify this equation to $\\frac{72}{5x^2} - 1 = \\frac{27}{x^3} - 1$ and so $x = \\frac{15}{8}$ . Then $k = \\frac{\\frac{5}{3}\\pi x^2}{24\\pi - \\frac{5}{3}\\pi x^2}= \\frac{125}{387} = \\frac mn$ so the answer is $m+n=125+387=\\boxed{512}$", "Our original solid $V$ has surface area $A_v = \\pi r^2 + \\pi r \\ell$ , where $\\ell$ is the slant height of the cone. Using the Pythagorean Theorem or Pythagorean Triple knowledge, we obtain $\\ell = 5$ and lateral area $A_\\ell = 15\\pi$ . The area of the base is $A_B = 3^2\\pi = 9\\pi$\n$V$ and $C$ are similar cones, because the plane that cut out $C$ was parallel to the base of $V$ . Let $x$ be the scale factor between the original cone and the small cone $C$ in one dimension. Because the scale factor is uniform in all dimensions, $x^2$ relates corresponding areas of $C$ and $V$ , and $x^3$ relates corresponding volumes. Then, the ratio of the painted areas $\\frac{A_c}{A_f}$ is $\\frac{15\\pi x^2}{9\\pi + 15\\pi - 15\\pi x^2} = \\frac{5 x^2}{8 - 5 x^2} = k$ and the ratio of the volumes $\\frac{V_c}{V_f}$ is $\\frac{x^3}{1 - x^3} = k$ . Since both ratios are equal to $k$ , they are equal to each other. Therefore, $\\frac{5 x^2}{8 - 5 x^2} = \\frac{x^3}{1 - x^3}$\nNow we must merely solve for x and substitute back into either ratio. Cross multiplying gives $5 x^2(1 - x^3) = x^3(8 - 5 x^2)$ . Dividing both sides by $x^2$ and distributing the $x$ on the right, we have $5 - 5 x^3 = 8 x - 5 x^3$ , and so $8 x = 5$ and $x = \\frac{5}{8}$ . Substituting back into the easier ratio, we have $\\frac{(\\frac{5}{8})^3}{1 - (\\frac{5}{8})^3} = \\frac{\\frac{125}{512}}{\\frac{387}{512}} = \\frac{125}{387}$ . And so we have $m + n = 125 + 387 = \\boxed{512}$" ]

[ "We rotate the smaller cube around the sphere such that two opposite vertices of the cube are on opposite faces of the larger cube. Thus the main diagonal of the smaller cube is the side length of the outer square.\nLet $S$ be the surface area of the inner square. The ratio of the areas of two similar figures is equal to the square of the ratio of their sides. As the diagonal of a cube has length $s\\sqrt{3}$ where $s$ is a side of the cube, the ratio of a side of the inner square to a side of the outer square is $\\frac{1}{\\sqrt{3}}$ (since the side of the outer square = the diagonal of the inner square). So we have $\\frac{S}{24} = \\left(\\frac{1}{\\sqrt{3}}\\right)^2$ . Thus $S = 8\\Rightarrow \\mathrm{\\boxed{8}$", "First of all, it is pretty easy to see the length of each edge of the bigger cube is $2$ so the radius of the sphere is $1$ . We know that when a cube is inscribed in a sphere, assuming the edge length of the square is $x$ the radius can be presented as $\\frac {\\sqrt3x}{2} = 1$ , we can solve that $x=\\frac {2\\sqrt3}{3}$ and now we only need to apply the basic formula to find the surface and we got our final answer as $\\frac {2\\sqrt3}{3}*\\frac {2\\sqrt3}{3}*6=8$ and the final answer is $\\boxed{8}$" ]

[ "Since the surface area of the original cube is 24 square meters, each face of the cube has a surface area of $24/6 = 4$ square meters, and the side length of this cube is 2 meters. The sphere inscribed within the cube has diameter 2 meters, which is also the length of the diagonal of the cube inscribed in the sphere. Let $l$ represent the side length of the inscribed cube. Applying the Pythagorean Theorem twice gives \\[l^2 + l^2 + l^2 = 2^2 = 4.\\] Hence each face has surface area \\[l^2 = \\frac{4}{3} \\ \\text{square meters}.\\] So the surface area of the inscribed cube is $6\\cdot \\frac43 = \\boxed{8}$ square meters." ]

[ "\nThe center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius. $I$ must also be a distance $r$ from the plane $ABC$\nThe signed distance between a plane and a point $I$ can be calculated as $\\frac{(I-G) \\cdot P}{|P|}$ , where G is any point on the plane, and P is a vector perpendicular to ABC.\nA vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)\\times(B-C)=\\langle 8, 12, 24 \\rangle$\nThus $\\frac{(I-C) \\cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$\n\\begin{align*}\\frac{(I-C) \\cdot P}{|P|}&=-r\\\\ \\frac{(\\langle r, r, r \\rangle-\\langle 0, 0, 2 \\rangle) \\cdot P}{|P|}&=-r\\\\ \\frac{\\langle r, r, r-2 \\rangle \\cdot \\langle 8, 12, 24 \\rangle}{\\langle 8, 12, 24 \\rangle}&=-r\\\\ \\frac{44r -48}{28}&=-r\\\\ 44r-48&=-28r\\\\ 72r&=48\\\\ r&=\\frac{2}{3} \\end{align*}\nFinally $2+3=\\boxed{005}$", "Notice that we can split the tetrahedron into $4$ smaller tetrahedrons such that the height of each tetrahedron is $r$ and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the tetrahedrons are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be $V$ and surface area be $F$ , using the volume formula for each pyramid(base times height divided by 3) we have $\\dfrac{rF}{3}=V$ . The surface area of the pyramid is $\\dfrac{6\\cdot{4}+6\\cdot{2}+4\\cdot{2}}{2}+[ABC]=22+[ABC]$ . We know triangle ABC's side lengths, $\\sqrt{2^{2}+4^{2}}, \\sqrt{2^{2}+6^{2}},$ and $\\sqrt{4^{2}+6^{2}}$ , so using the expanded form of heron's formula, \\begin{align*}[ABC]&=\\sqrt{\\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}\\\\ &=\\sqrt{2(5\\cdot{13}+10\\cdot{5}+13\\cdot{10})-5^{2}-10^{2}-13^{2}}\\\\ &=\\sqrt{196}\\\\ &=14\\end{align*} Therefore, the surface area is $14+22=36$ , and the volume is $\\dfrac{[BCD]\\cdot{6}}{3}=\\dfrac{4\\cdot{2}\\cdot{6}}{3\\cdot{2}}=8$ , and using the formula above that $\\dfrac{rF}{3}=V$ , we have $12r=8$ and thus $r=\\dfrac{2}{3}$ , so the desired answer is $2+3=\\boxed{005}$", "First let us find the equation of the plane passing through $(6,0,0), (0,0,2), (0,4,0)$ . The \"point-slope form\" is $A(6-x1)+B(0-y1)+C(0-z1)=0.$ Plugging in $(0,0,2)$ gives $A(6)+B(0)+C(-2)=0.$ Plugging in $(0,4,0)$ gives $A(6)+B(-4)+C(0)=0.$ We can then use Cramer's rule/cross multiplication to get $A/(0-8)=-B/(0+12)=C/(-24)=k.$ Solve for A, B, C to get $2k, 3k, 6k$ respectively. We can then get $2k(x-x1)+3k(y-y1)+6k(z-z1)=0.$ Cancel out k on both sides. Next, let us substitute $(0,0,2)$ . We can then get $2x+3y+6z=12$ as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get $2x/7+2y/7+6z/7=12/7$ to be the normal form. Note that the point is going to be at $(r,r,r).$ We find the distance from $(r,r,r)$ to the plane as $2/7r+3/7r+6/7r-12/7/(\\sqrt{(4/49+9/49+36/49)})$ , which is $+/-(11r/7-12/7)$ . We take the negative value of this because if we plug in $(0,0,0)$ to the equation of the plane we get a negative value. We equate that value to r and we get the equation $-(11r/7-12/7)=r$ to solve $r={2/3}$ , so the answer is $\\boxed{005}$", "Clearly, if the radius of the sphere is $r$ , the center of the sphere lies on $(r, r, r)$\nWe find the equation of plane $ABC$ to be $\\frac16 x+\\frac14 y+\\frac12 z=1$ . From the definition of the insphere, it must be true that the distance from the center of the sphere to plane $ABC$ is equal to the length of the radius of the sphere. By point-to-plane, we have \\[r=\\frac{|\\frac16 r+\\frac14 r+\\frac12 r-1|}{\\sqrt{\\left(\\frac16\\right)^2+\\left(\\frac14\\right)^2+\\left(\\frac12\\right)^2}} \\implies r=\\frac23,\\] so the answer is $\\boxed{005}$" ]

[ "The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$ . The integers from $25$ to $100$ are left. They can be paired so the sum is $125$ $25+100$ $26+99$ $27+98$ $\\ldots$ $62+63$ . That is $38$ pairs, and at most one number from each pair can be included in the set. The total is $24 + 38 = \\boxed{62}$", "\"Cut\" $125$ into half. The maximum integer value in the smaller half is $62$ . Thus the answer is $\\boxed{62}$", "The maximum possible number of elements includes the smallest numbers. So, subset $B = \\{1,2,3....n-1,n\\}$ where n is the maximum number of elements in subset $B$ . So, we have to find two consecutive numbers, $n$ and $n+1$ , whose sum is $125$ . Setting up our equation, we have $n+(n+1) = 2n+1 = 125$ . When we solve for $n$ , we get $n =\\boxed{62}$", "We can put all odd numbers into subset B, or we can put all even numbers into subset B, so now there are 50 numbers in the set. I will use all even numbers in this solution. Now, we need to add other odd(or even!) numbers possible in this subset, which is all odd(or even) numbers that can be added so that the sum with 100(or 99) plus the biggest possible odd number(or even) to get 123. This will get us the numbers 1,3,5...21,23(or numbers 2,4,6...22,24), which gives us 12 more numbers, and adding that to the 50 original numbers, we get $B =\\boxed{62}$" ]

[ "Use the angle bisector theorem to find that the angle bisector of $\\angle P$ divides $QR$ into segments of length $\\frac{25}{x} = \\frac{15}{20 -x} \\Longrightarrow x = \\frac{25}{2},\\ \\frac{15}{2}$ . It follows that $\\frac{QP'}{RP'} = \\frac{5}{3}$ , and so $P' = \\left(\\frac{5x_R + 3x_Q}{8},\\frac{5y_R + 3y_Q}{8}\\right) = (-5,-23/2)$\nThe desired answer is the equation of the line $PP'$ $PP'$ has slope $\\frac{-11}{2}$ , from which we find the equation to be $11x + 2y + 78 = 0$ . Therefore, $a+c = \\boxed{089}$", "Transform triangle $PQR$ so that $P$ is at the origin. Note that the slopes do not change when we transform the triangle.\nWe know that the slope of $PQ$ is $\\frac{24}{7}$ and the slope of $PR$ is $-\\frac{4}{3}$ . Thus, in the complex plane, they are equivalent to $\\tan(\\alpha)=\\frac{24}{7}$ and $\\tan(\\beta)=-\\frac{4}{3}$ , respectively. Here $\\alpha$ is the angle formed by the $x$ -axis and $PQ$ , and $\\beta$ is the angle formed by the $x$ -axis and $PR$ . The equation of the angle bisector is $\\tan\\left(\\frac{\\alpha+\\beta}{2}\\right)$\nAs the tangents are in very neat Pythagorean triples , we can easily calculate $\\cos(\\alpha)$ and $\\cos(\\beta)$\nAngle $\\alpha$ is in the third quadrant, so $\\cos(\\alpha)$ is negative. Thus $\\cos(\\alpha)=-\\frac{7}{25}$\nAngle $\\beta$ is in the fourth quadrant, so $\\cos(\\beta)$ is positive. Thus $\\cos(\\beta)=\\frac{3}{5}$\nBy the Half-Angle Identities $\\tan\\left(\\frac{\\alpha}{2}\\right)=\\pm\\sqrt{\\frac{1-\\cos(\\alpha)}{1+\\cos(\\alpha)}}=\\pm\\sqrt{\\frac{\\frac{32}{25}}{\\frac{18}{25}}}=\\pm\\sqrt{\\frac{16}{9}}=\\pm\\frac{4}{3}$ and $\\tan\\left(\\frac{\\beta}{2}\\right)=\\pm\\sqrt{\\frac{1-\\cos(\\beta)}{1+\\cos(\\beta)}}=\\pm\\sqrt{\\frac{\\frac{2}{5}}{\\frac{8}{5}}}=\\pm\\sqrt{\\frac{1}{4}}=\\pm\\frac{1}{2}$\nSince $\\frac{\\alpha}{2}$ and $\\frac{\\beta}{2}$ must be in the second quadrant, their tangent values are both negative. Thus $\\tan\\left(\\frac{\\alpha}{2}\\right)=-\\frac{4}{3}$ and $\\tan\\left(\\frac{\\beta}{2}\\right)=-\\frac{1}{2}$\nBy the sum of tangents formula $\\tan\\left(\\frac{\\alpha}{2}+\\frac{\\beta}{2}\\right)=\\frac{\\tan\\left(\\frac{\\alpha}{2}\\right)+\\tan\\left(\\frac{\\beta}{2}\\right)}{1-\\tan\\left(\\frac{\\alpha}{2}\\right)\\tan\\left(\\frac{\\beta}{2}\\right)}=\\frac{-\\frac{11}{6}}{\\frac{1}{3}}=-\\frac{11}{2}$ , which is the slope of the angle bisector.\nFinally, the equation of the angle bisector is $y-5=-\\frac{11}{2}(x+8)$ or $y=-\\frac{11}{2}x-39$ . Rearranging, we get $11x+2y+78=0$ , so our sum is $a+c=11+78=\\boxed{089}$ . ~eevee9406" ]

[ "Let the $k$ th number in the $n$ th row be $a(n,k)$ . Writing out some numbers, we find that $a(n,k) = 2^{n-1}(n+2k-2)$\nWe wish to find all $(n,k)$ such that $67| a(n,k) = 2^{n-1} (n+2k-2)$ . Since $2^{n-1}$ and $67$ are relatively prime , it follows that $67|n+2k-2$ . Since every row has one less element than the previous row, $1 \\le k \\le 51-n$ (the first row has $50$ elements, the second $49$ , and so forth; so $k$ can range from $1$ to $50$ in the first row, and so forth). Hence\nit follows that $67| n - 2k + 2$ implies that $n-2k+2 = 67$ itself.\nNow, note that we need $n$ to be odd, and also that $n+2k-2 = 67 \\le 100-n \\Longrightarrow n \\le 33$\nWe can check that all rows with odd $n$ satisfying $1 \\le n \\le 33$ indeed contains one entry that is a multiple of $67$ , and so the answer is $\\frac{33+1}{2} = \\boxed{017}$", "The result above is fairly intuitive if we write out several rows and then divide all numbers in row $r$ by $2^{r-1}$ (we can do this because dividing by a power of 2 doesn't affect divisibility by $67$ ). The second row will be $2, 4, 6, \\cdots , 98$ , the third row will be $3, 5, \\cdots, 97$ , and so forth. Clearly, only the odd-numbered rows can have a term divisible by $67$ . However, with each row the row will have one less element, and the $99-67+1 = 33$ rd row is the last time $67$ will appear. Therefore the number of multiples of 67 in the entire array is $\\frac{33+1}{2} = \\boxed{017}$" ]

[ "Label each of the bottom squares as $x_0, x_1 \\ldots x_9, x_{10}$\nThrough induction , we can find that the top square is equal to ${10\\choose0}x_0 + {10\\choose1}x_1 + {10\\choose2}x_2 + \\ldots {10\\choose10}x_{10}$ . (This also makes sense based on a combinatorial argument: the number of ways a number can \"travel\" to the top position going only up is equal to the number of times it will be counted in the final sum.)\nExamine the equation $\\mod 3$ . All of the coefficients from $x_2 \\ldots x_8$ will be multiples of $3$ (since the numerator will have a $9$ ). Thus, the expression boils down to $x_0 + 10x_1 + 10x_9 + x_{10} \\equiv 0 \\mod 3$ . Reduce to find that $x_0 + x_1 + x_9 + x_{10} \\equiv 0 \\mod 3$ . Out of $x_0,\\ x_1,\\ x_9,\\ x_{10}$ , either all are equal to $0$ , or three of them are equal to $1$ . This gives ${4\\choose0} + {4\\choose3} = 1 + 4 = 5$ possible combinations of numbers that work.\nThe seven terms from $x_2 \\ldots x_8$ can assume either $0$ or $1$ , giving us $2^7$ possibilities. The answer is therefore $5 \\cdot 2^7 = \\boxed{640}$" ]