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[ "For $1\\leq k\\leq 12,$ suppose that cards $1, 2, \\ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\\ldots,13$ are picked up on the second pass.\nOnce we pick the spots for the cards on the first pass, there is only one way to arrange all $\\boldsymbol{13}$ cards.\nFor each value of $k,$ there are $\\binom{13}{k}-1$ ways to pick the $k$ spots for the cards on the first pass: We exclude the arrangement in which the first pass consists of all $13$ cards.\nTherefore, the answer is \\[\\sum_{k=1}^{12}\\left[\\binom{13}{k}-1\\right] = \\left[\\sum_{k=1}^{12}\\binom{13}{k}\\right]-12 = \\left[\\sum_{k=0}^{13}\\binom{13}{k}\\right]-14 = 2^{13} - 14 = \\boxed{8178}.\\]", "Since the $13$ cards are picked up in two passes, the first pass must pick up the first $n$ cards and the second pass must pick up the remaining cards $m$ through $13$ . \nAlso note that if $m$ , which is the card that is numbered one more than $n$ , is placed before $n$ , then $m$ will not be picked up on the first pass since cards are picked up in order. Therefore we desire $m$ to be placed before $n$ to create a second pass, and that after the first pass, the numbers $m$ through $13$ are lined up in order from least to greatest.\nTo construct this, $n$ cannot go in the $n$ th position because all cards $1$ to $n-1$ will have to precede it and there will be no room for $m$ . Therefore $n$ must be in slots $n+1$ to $13$ .\nLet's do casework on which slot $n$ goes into to get a general idea for how the problem works.\nCase 1: With $n$ in spot $n+1$ , there are $n$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve these slots for the $n-1$ cards is $\\binom{n}{n-1}$ . Then there is only $1$ way to order these cards (since we want them in increasing order). Then card $m$ goes into whatever slot is remaining, and the $13-m$ cards are ordered in increasing order after slot $n+1$ , giving only $1$ way. Therefore in this case there are $\\binom{n}{n-1}$ possibilities.\nCase 2: With $n$ in spot $n+2$ , there are $n+1$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve slots for these cards are $\\binom{n+1}{n-1}$ . Then there is one way to order these cards. Then cards $m$ and $m+1$ must go in the remaining two slots, and there is only one way to order them since they must be in increasing order. Finally, cards $m+2$ to $13$ will be ordered in increasing order after slot $n+1$ , which yields $1$ way. Therefore, this case has $\\binom{n+1}{n-1}$ possibilities.\nI think we can see a general pattern now. With $n$ in slot $x$ , there are $x-1$ slots to distribute to the previous $n-1$ cards, which can be done in $\\binom{x-1}{n-1}$ ways. Then the remaining cards fill in in just $1$ way. Since the cases of $n$ start in slot $n+1$ and end in slot $13$ , this sum amounts to: \\[\\binom{n}{n-1}+\\binom{n+1}{n-1}+\\binom{n+2}{n-1} + \\cdots + \\binom{12}{n-1}\\] for any $n$\nHmmm ... where have we seen this before?\nWe use wishful thinking to add a term of $\\binom{n-1}{n-1}$ \\[\\binom{n-1}{n-1}+\\binom{n}{n-1}+\\binom{n+1}{n-1}+\\binom{n+2}{n-1} + \\cdots + \\binom{12}{n-1}\\]\nThis is just the hockey stick identity! Applying it, this expression is equal to $\\binom{13}{n}$ . However, we added an extra term, so subtracting it off, the total number of ways to order the $13$ cards for any $n$ is \\[\\binom{13}{n}-1\\]\nFinally, to calculate the total for all $n$ , we sum from $n=0$ to $13$ . This yields us:\n\\[\\sum_{n=0}^{13} \\binom{13}{n}-1 \\implies \\sum_{n=0}^{13} \\binom{13}{n} - \\sum_{n=0}^{13} 1\\] \\[\\implies 2^{13} - 14 = 8192 - 14 = 8178 = \\boxed{8178}.\\]", "To solve this problem, we can use recursion on $n$ . Let $A_n$ be the number of arrangements for $n$ numbers. Now, let's look at how these arrangements are formed by case work on the first number $a_1$\nIf $a_1 = 1$ , the remaining $n-1$ numbers from $2$ to $n$ are arranged in the same way just like number 1 to $n-1$ in the case of $n-1$ numbers. So there are $A_{n-1}$ arrangements.\nIf $a_1 = 2$ , then we need to choose 1 position from position 2 to $n-1$ to put 1, and all remaining numbers must be arranged in increasing order, so there are $\\binom{n-1}{1}$ such arrangements.\nIf $a_1 = k$ , then we need to choose $k-1$ positions from position 2 to $n-1$ to put $1, 2,\\cdots k-1$ , and all remaining numbers must be arranged in increasing order, so there are $\\binom{n-1}{k-1}$ such arrangements.\nSo we can write \\[A_n = A_{n-1} + \\binom{n-1}{1} + \\binom{n-1}{2} + \\cdots + \\binom{n-1}{n-1}\\] which can be simplified to \\[A_n = A_{n-1} + 2^{n-1} - 1\\] We can solve this recursive sequence by summing up $n-1$ lines of the recursive formula \\[A_n - A_{n-1} = 2^{n-1} - 1\\] \\[A_{n-1} - A_{n-2} = 2^{n-2} - 1\\] \\[\\cdots\\] \\[A_2 - A_{1} = 2^{1} - 1\\] to get \\[A_n - A_1 = \\sum_{k=1}^{n-1} (2^k - 1) = 2^n - 2 - (n-1) = 2^n - n - 1\\] since $A_1=0$ , we have \\[A_n = 2^n - n - 1\\] and $A_{13} = 2^{13} - 14 = \\boxed{8178}$", "When we have $3$ cards arranged in a row, after listing out all possible arrangements, we see that we have $4$ ones: $(1, 3, 2), (2, 1, 3), (2, 3, 1),$ and $(3, 1, 2)$ . When we have $4$ cards, we find $11$ possible arrangements: $(1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (2, 1, 3, 4), (2, 3, 1, 4), (2, 3, 4, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 4, 1, 2),$ and $(4, 1, 2, 3).$ Hence, we recognize the pattern that for $n$ cards, we have $2^n - n - 1$ valid arrangements, so our answer is $2^{13} - 13 - 1 = \\boxed{8178}.$ ~eibc", "Notice that for each card \"position\", we can choose for it to be picked up on the first or second pass, for a total of $2^{13}$ options. However, if all of the cards selected to be picked up first are before all of the cards to be picked up second, then this means that the list is in consecutive ascending order (and thus all cards will be picked up on the first pass instead). This can happen in 14 ways, so our answer is $2^{13}-14=\\boxed{8178}$" ]

[ "We have $12^{mn} = (2\\cdot 2\\cdot 3)^{mn} = 2^{2mn} \\cdot 3^{mn} = (2^m)^{2n} \\cdot (3^n)^m = \\boxed{2}$" ]

[ "We have $12^{mn} = (2\\cdot 2\\cdot 3)^{mn} = 2^{2mn} \\cdot 3^{mn} = (2^m)^{2n} \\cdot (3^n)^m = \\boxed{2}$" ]

[ "The answer cannot be $0,$ as every nonconstant polynomial has at least $1$ distinct complex root (Fundamental Theorem of Algebra). Since $P(z) \\cdot Q(z)$ has degree $2 + 3 = 5,$ we conclude that $R(z) - P(z)\\cdot Q(z)$ has degree $6$ and is thus nonconstant.\nIt now suffices to illustrate an example for which $N = 1$ : Take \\begin{align*} P(z)&=z^2+1, \\\\ Q(z)&=z^3+2, \\\\ R(z)&=(z+1)^6 + P(z) \\cdot Q(z). \\end{align*} Note that $R(z)$ has degree $6$ and constant term $3,$ so it satisfies the conditions.\nWe need to find the solutions to \\begin{align*} P(z) \\cdot Q(z) &= (z+1)^6 + P(z) \\cdot Q(z) \\\\ 0 &= (z+1)^6. \\end{align*} Clearly, the only distinct complex root is $-1,$ so our answer is $N=\\boxed{1}.$" ]

[ "Let the lowest value be $L$ and the highest $G$ , and let the sum be $Z$ and the amount of numbers $n$ . We have $\\frac{Z-G}{n-1}=32$ $\\frac{Z-L-G}{n-2}=35$ $\\frac{Z-L}{n-1}=40$ , and $G=L+72$ . Clearing denominators gives $Z-G=32n-32$ $Z-L-G=35n-70$ , and $Z-L=40n-40$ . We use $G=L+72$ to turn the first equation into $Z-L=32n+40$ . Since $Z-L=40(n-1)$ we substitute it into the equation which gives $n=10$ . Turning the second into $Z-2L=35n+2$ using $G=L+72$ we see $L=8$ and $Z=368$ so the average is $\\frac{Z}{n}=\\boxed{36.8}$ ~aop2014", "Let $x$ be the greatest integer, $y$ be the smallest, $z$ be the sum of the numbers in S excluding $x$ and $y$ , and $k$ be the number of elements in S.\nThen, $S=x+y+z$\nFirst, when the greatest integer is removed, $\\frac{S-x}{k-1}=32$\nWhen the smallest integer is also removed, $\\frac{S-x-y}{k-2}=35$\nWhen the greatest integer is added back, $\\frac{S-y}{k-1}=40$\nWe are given that $x=y+72$\nAfter you substitute $x=y+72$ , you have 3 equations with 3 unknowns $S,$ $y$ and $k$\n$S-y-72=32k-32$\n$S-2y-72=35k-70$\n$S-y=40k-40$\nThis can be easily solved to yield $k=10$ $y=8$ $S=368$\n$\\therefore$ average value of all integers in the set $=S/k = 368/10 = \\boxed{36.8}$", "We should plug in $36.2$ and assume everything is true except the $35$ part. We then calculate that part and end up with $35.75$ . We also see with the formulas we used with the plug in that when you increase by $0.2$ the $35.75$ part decreases by $0.25$ . The answer is then $\\boxed{36.8}$ . You can work backwards because it is multiple choice and you don't have to do critical thinking. ~Lopkiloinm", "Let $S = \\{a_1, a_2, a_3, \\hdots, a_n\\}$ with $a_1 < a_2 < a_3 < \\hdots < a_n.$ We are given the following: \\[{\\begin{cases} \\sum_{i=1}^{n-1} a_i = 32(n-1) = 32n-32, \\\\ \\sum_{i=2}^n a_i = 40(n-1) = 40n-40, \\\\ \\sum_{i=2}^{n-1} a_i = 35(n-2) = 35n-70, \\\\ a_n-a_1 = 72 \\implies a_1 + 72 = a_n. \\end{cases}}\\] Subtracting the third equation from the sum of the first two, we find that \\[\\sum_{i=1}^n a_i = \\left(32n-32\\right) + \\left(40n-40\\right) - \\left(35n-70\\right) = 37n - 2.\\] Furthermore, from the fourth equation, we have \\[\\sum_{i=2}^{n} a_i - \\sum_{i=1}^{n-1} a_i = \\left[\\left(a_1 + 72\\right) + \\sum_{i=2}^{n-1} a_i\\right] - \\left[\\left(a_1\\right) + \\sum_{i=2}^{n-1} a_i\\right] = \\left(40n-40\\right)-\\left(32n-32\\right).\\] Combining like terms and simplifying, we have \\[72 = 8n-8 \\implies 8n = 80 \\implies n=10.\\] Thus, the sum of the elements in $S$ is $37 \\cdot 10 - 2 = 368,$ and since there are 10 elements in $S,$ the average of the elements in $S$ is $\\tfrac{368}{10}=\\boxed{36.8}$" ]

[ "Let the lowest value be $L$ and the highest $G$ , and let the sum be $Z$ and the amount of numbers $n$ . We have $\\frac{Z-G}{n-1}=32$ $\\frac{Z-L-G}{n-2}=35$ $\\frac{Z-L}{n-1}=40$ , and $G=L+72$ . Clearing denominators gives $Z-G=32n-32$ $Z-L-G=35n-70$ , and $Z-L=40n-40$ . We use $G=L+72$ to turn the first equation into $Z-L=32n+40$ . Since $Z-L=40(n-1)$ we substitute it into the equation which gives $n=10$ . Turning the second into $Z-2L=35n+2$ using $G=L+72$ we see $L=8$ and $Z=368$ so the average is $\\frac{Z}{n}=\\boxed{36.8}$ ~aop2014", "Let $x$ be the greatest integer, $y$ be the smallest, $z$ be the sum of the numbers in S excluding $x$ and $y$ , and $k$ be the number of elements in S.\nThen, $S=x+y+z$\nFirst, when the greatest integer is removed, $\\frac{S-x}{k-1}=32$\nWhen the smallest integer is also removed, $\\frac{S-x-y}{k-2}=35$\nWhen the greatest integer is added back, $\\frac{S-y}{k-1}=40$\nWe are given that $x=y+72$\nAfter you substitute $x=y+72$ , you have 3 equations with 3 unknowns $S,$ $y$ and $k$\n$S-y-72=32k-32$\n$S-2y-72=35k-70$\n$S-y=40k-40$\nThis can be easily solved to yield $k=10$ $y=8$ $S=368$\n$\\therefore$ average value of all integers in the set $=S/k = 368/10 = \\boxed{36.8}$", "We should plug in $36.2$ and assume everything is true except the $35$ part. We then calculate that part and end up with $35.75$ . We also see with the formulas we used with the plug in that when you increase by $0.2$ the $35.75$ part decreases by $0.25$ . The answer is then $\\boxed{36.8}$ . You can work backwards because it is multiple choice and you don't have to do critical thinking. ~Lopkiloinm", "Let $S = \\{a_1, a_2, a_3, \\hdots, a_n\\}$ with $a_1 < a_2 < a_3 < \\hdots < a_n.$ We are given the following: \\[{\\begin{cases} \\sum_{i=1}^{n-1} a_i = 32(n-1) = 32n-32, \\\\ \\sum_{i=2}^n a_i = 40(n-1) = 40n-40, \\\\ \\sum_{i=2}^{n-1} a_i = 35(n-2) = 35n-70, \\\\ a_n-a_1 = 72 \\implies a_1 + 72 = a_n. \\end{cases}}\\] Subtracting the third equation from the sum of the first two, we find that \\[\\sum_{i=1}^n a_i = \\left(32n-32\\right) + \\left(40n-40\\right) - \\left(35n-70\\right) = 37n - 2.\\] Furthermore, from the fourth equation, we have \\[\\sum_{i=2}^{n} a_i - \\sum_{i=1}^{n-1} a_i = \\left[\\left(a_1 + 72\\right) + \\sum_{i=2}^{n-1} a_i\\right] - \\left[\\left(a_1\\right) + \\sum_{i=2}^{n-1} a_i\\right] = \\left(40n-40\\right)-\\left(32n-32\\right).\\] Combining like terms and simplifying, we have \\[72 = 8n-8 \\implies 8n = 80 \\implies n=10.\\] Thus, the sum of the elements in $S$ is $37 \\cdot 10 - 2 = 368,$ and since there are 10 elements in $S,$ the average of the elements in $S$ is $\\tfrac{368}{10}=\\boxed{36.8}$" ]

[ "Let $M$ be the largest number in $S$ .\nWe categorize numbers $\\left\\{ 1, 2, \\ldots , M-1 \\right\\}$ (except $\\frac{M}{2}$ if $M$ is even) into $\\left\\lfloor \\frac{M-1}{2} \\right\\rfloor$ groups, such that the $i$ th group contains two numbers $i$ and $M-i$\nRecall that $M \\in S$ and the sum of two numbers in $S$ cannot be equal to $M$ , and the sum of numbers in each group above is equal to $S$ . Thus, each of the above $\\left\\lfloor \\frac{M-1}{2} \\right\\rfloor$ groups can have at most one number in $S$ .\nTherefore, \\begin{align*} |S| & \\leq 1 + \\left\\lfloor \\frac{M-1}{2} \\right\\rfloor \\\\ & \\leq 1 + \\left\\lfloor \\frac{25}{2} \\right\\rfloor \\\\ & = 13. \\end{align*}\nNext, we construct an instance of $S$ with $|S| = 13$ .\nLet $S = \\left\\{ 13, 14, \\ldots , 25 \\right\\}$ .\nThus, this set is feasible.\nTherefore, the most number of elements in $S$ is $\\boxed{13}$", "We know that two odd numbers sum to an even number, so we can easily say that odd numbers $1-25$ can be included in the list, making for $13$ elements. But, how do we know we can't include even numbers for a higher element value? Well, to get a higher element value than $13$ , odd numbers as well as even numbers would have to be included in the list (since there are only $12$ even numbers from $1-25$ , and many of those even numbers are the sum of even numbers). However, for every even value we add to our odd list, we have to take away an odd number because there are either two odd numbers that sum to that even value, or that even value and another odd number will sum to an odd number later in the list. So, $\\boxed{13}$ elements is the highest we can go.", "The smallest sum of a number $a + b$ where $b \\geq a$ is $a + a = 2a$ as we are using the smallest value of $b$ . Using this, we can say that if $12$ were an element of $S$ , then one of the sums (the smallest) would be $12 + 12 = 24 < 25$ . Thus $13$ must be the smallest element. So the largest amount of elements that could be in $S$ is the list of numbers from $13$ to $25$ as they all work. Because it is inclusive we have, $25 - 13 + 1 = \\boxed{13}$", "We construct a possible subset $S$ with $13$ elements by including all odd integers from $1$ to $25$ , inclusive. $S=\\left\\{ 1, 3, 5, \\cdots , 25 \\right\\}$ . The sum of any $2$ elements is even, and thus cannot be an element of $S$\nTo show that $S$ cannot have more than $13$ elements, assume for sake of contradiction that $|S| \\geq 14$ . Let $S=\\left\\{ x_1, x_2, \\cdots , x_n \\right\\}$ where $n \\geq 14$ and $x_1 < x_2 < \\cdots < x_n$ . Because the sums of any $2$ (not necessarily distinct) elements do not appear in $S$ $x_1+x_i$ is not an element of $S$ for all $1 \\leq i \\leq n$ . So, $x_1, x_2, \\cdots , x_n , x_1+x_1, x_1+x_2, \\cdots , x_1+x_n$ are all distinct integers. Let these integers be elements of the set $T$ $|T|=2n$ , and because $n \\geq 14$ $|T| \\geq 28$ . But all elements of $T$ must be $\\geq x_1$ and $\\leq x_1+x_n \\leq x_1+25$ , leaving only 26 possible values for the elements in $T$ . By the Pigeonhole Principle, the elements cannot be distinct, and we have a contradiction.\nThus, $\\boxed{13}$ is the maximum possible size of $S$", "We can start by building a list of the elements of $S$ ,and see if we can find a pattern. Let's start with $1$ . If we include 1, that means we cannot include $2$ (which is $1$ $1$ ), so we write the next valid number, $3$ . Similarly, we cannot include $4$ or $6$ , so we write $5$ . Proceeding, our list looks like this:\n$S$ = { $1$ $3$ $5$ $7$ ....\nWe've found a pattern! It's easy understand why this pattern of odd numbers will continue, seeing as all the even numbers can be written as sums between odds. Calculating how many odd numbers fit into the range $1$ $25$ , we conclude that the answer is $\\boxed{13}$" ]

[ "Plugging in some values, we see that the graph of the equation $|x+y|+|x-y| = 2$ is a square bounded by $x= \\pm 1$ and $y = \\pm 1$\nNotice that $x^2 - 6x + y^2 = (x-3)^2 + y^2 - 9$ means the square of the distance from a point $(x,y)$ to point $(3,0)$ minus 9. To maximize that value, we need to choose the point in the feasible region farthest from point $(3,0)$ , which is $(-1, \\pm 1)$ . Either one, when substituting into the function, yields $\\boxed{8}$", "Since the equation $|x+y|+|x-y| = 2$ is dealing with absolute values, the following could be deduced: $(x+y)+(x-y)=2$ $(x+y)-(x-y)=2$ $-(x+y)+(x-y)=2$ , and $-(x+y)-(x-y)=2$ . Simplifying would give $x=1$ $y=1$ $y=-1$ , and $x=-1$ . In $x^2-6x+y^2$ , we care most about $-6x,$ since both $x^2$ and $y^2$ are non-negative. To maximize $-6x$ , though, $x$ would have to be -1. Therefore, when $x=-1$ and $y=-1$ or $y=1$ , the equation evaluates to $\\boxed{8}$" ]

[ "Use the two trigonometric Pythagorean identities $1 + \\tan^2 x = \\sec^2 x$ and $1 + \\cot^2 x = \\csc^2 x$\nIf we square the given $\\sec x = \\frac{22}{7} - \\tan x$ , we find that\n\\begin{align*} \\sec^2 x &= \\left(\\frac{22}7\\right)^2 - 2\\left(\\frac{22}7\\right)\\tan x + \\tan^2 x \\\\ 1 &= \\left(\\frac{22}7\\right)^2 - \\frac{44}7 \\tan x \\end{align*}\nThis yields $\\tan x = \\frac{435}{308}$\nLet $y = \\frac mn$ . Then squaring,\n\\[\\csc^2 x = (y - \\cot x)^2 \\Longrightarrow 1 = y^2 - 2y\\cot x.\\]\nSubstituting $\\cot x = \\frac{1}{\\tan x} = \\frac{308}{435}$ yields a quadratic equation $0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)$ . It turns out that only the positive root will work, so the value of $y = \\frac{29}{15}$ and $m + n = \\boxed{044}$", "We are given that $\\frac{1+\\sin x}{\\cos x}=\\frac{22}7\\implies\\frac{1+\\sin x}{\\cos x}\\cdot\\frac{1-\\sin x}{1-\\sin x}=\\frac{1-\\sin^2x}{\\cos x(1-\\sin x)}=\\frac{\\cos^2x}{\\cos x(1-\\sin x)}$ $=\\frac{\\cos x}{1-\\sin x}$ , or equivalently, $\\cos x=\\frac{7+7\\sin x}{22}=\\frac{22-22\\sin x}7\\implies\\sin x=\\frac{22^2-7^2}{22^2+7^2}$ $\\implies\\cos x=\\frac{2\\cdot22\\cdot7}{22^2+7^2}$ . Note that what we want is just $\\frac{1+\\cos x}{\\sin x}=\\frac{1+\\frac{2\\cdot22\\cdot7}{22^2+7^2}}{\\frac{22^2-7^2}{22^2+7^2}}=\\frac{22^2+7^2+2\\cdot22\\cdot7}{22^2-7^2}=\\frac{(22+7)^2}{(22-7)(22+7)}=\\frac{22+7}{22-7}$ $=\\frac{29}{15}\\implies m+n=29+15=\\boxed{044}$", "Assign a right triangle with angle $x$ , hypotenuse $c$ , adjacent side $a$ , and opposite side $b$ .\nThen, through the given information above, we have that..\n$\\frac{c}{a}+\\frac{b}{a}=\\frac{22}{7}\\implies \\frac{c+b}{a}=\\frac{22}{7}$\n$\\frac{c}{b}+\\frac{a}{b}=\\frac{m}{n}\\implies \\frac{a+c}{b}=\\frac{m}{n}$\nHence, because similar right triangles can be scaled up by a factor, we can assume that this particular right triangle is indeed in simplest terms.\nHence, $a=7$ $b+c=22$\nFurthermore, by the Pythagorean Theorem, we have that\n$a^2+b^2=c^2\\implies 49+b^2=c^2$\nSolving for $c$ in the first equation and plugging in into the second equation...\n$49+b^2=(22-b)^2\\implies 49+b^2=484-44b+b^2\\implies 44b=435\\implies b=\\frac{435}{44}$\nHence, $c=22-\\frac{435}{44}=\\frac{533}{44}$\nNow, we want $\\frac{a+c}{b}$\nPlugging in, we find the answer is $\\frac{\\frac{7\\cdot{44}}{44}+\\frac{533}{44}}{\\frac{435}{44}}=\\frac{841}{435}=\\frac{29}{15}$\nHence, the answer is $29+15=\\boxed{044}$", "We know that $\\sec(x) = \\frac{h}{a}$ and that $\\tan(x) = \\frac{o}{a}$ where $h$ $a$ $o$ represent the hypotenuse, adjacent, and opposite (respectively) to angle $x$ in a right triangle. Thus we have that $\\sec(x) + \\tan(x) = \\frac{h+o}{a}$ . We also have that $\\csc(x) + \\cot(x) = \\frac{h}{o} + \\frac{a}{o} = \\frac{h+a}{o}$ . Set $\\sec(x) + \\tan(x) = \\alpha$ and csc(x)+cot(x) = $\\beta$ . Then, notice that $\\alpha + \\beta = \\frac{h+o}{a} + \\frac{h+a}{o} = \\frac{oh+ah+o^2 + a^2}{oa} = \\frac{h(o+a+h)}{oa}$ ( This is because of the Pythagorean Theorem, recall $o^2 +a^2 = h^2$ ). But then notice that $\\alpha \\cdot \\beta = \\frac{(o+h)(a+h)}{oa} = \\frac{oa +oh +ha +h^2}{oa} = 1+ \\frac{h(o+a+h)}{oa} = 1+ \\alpha + \\beta$ . From the information provided in the question, we can substitute $\\alpha$ for $\\frac{22}{7}$ . Thus, $\\frac{22 \\beta}{7}= \\beta + \\frac{29}{7} \\Longrightarrow 22 \\beta = 7 \\beta + 29 \\Longrightarrow 15 \\beta = 29 \\Longrightarrow \\beta = \\frac{29}{15}$ . Since, essentially we are asked to find the sum of the numerator and denominator of $\\beta$ , we have $29 + 15 = \\boxed{044}$", "Firstly, we write $\\sec x+\\tan x=a/b$ where $a=22$ and $b=7$ . This will allow us to spot factorable expressions later. Now, since $\\sec^2x-\\tan^2x=1$ , this gives us \\[\\sec x-\\tan x=\\frac{b}{a}\\] Adding this to our original expressions gives us \\[2\\sec x=\\frac{a^2+b^2}{ab}\\] or \\[\\cos x=\\frac{2ab}{a^2+b^2}\\] Now since $\\sin^2x+\\cos^2x=1$ $\\sin x=\\sqrt{1-\\cos^2x}$ So we can write \\[\\sin x=\\sqrt{1-\\frac{4a^2b^2}{(a^2+b^2)^2}}\\] Upon simplification, we get \\[\\sin x=\\frac{a^2-b^2}{a^2+b^2}\\] We are asked to find $1/\\sin x+\\cos x/\\sin x$ so we can write that as \\[\\csc x+\\cot x=\\frac{1}{\\sin x}+\\frac{\\cos x}{\\sin x}\\] \\[\\csc x+\\cot x=\\frac{a^2+b^2}{a^2-b^2}+\\frac{2ab}{a^2+b^2}\\frac{a^2+b^2}{a^2-b^2}\\] \\[\\csc x+\\cot x=\\frac{a^2+b^2+2ab}{a^2-b^2}\\] \\[\\csc x+\\cot x=\\frac{(a+b)^2}{(a-b)(a+b)}\\] \\[\\csc x+\\cot x=\\frac{a+b}{a-b}\\] Now using the fact that $a=22$ and $b=7$ yields, \\[\\csc x+\\cot x=\\frac{29}{15}=\\frac{p}{q}\\] so $p+q=15+29=\\boxed{44}$", "Rewriting $\\sec{x}$ and $\\tan{x}$ in terms of $\\sin{x}$ and $\\cos{x}$ , we know that $\\frac{1+\\sin{x}}{\\cos{x}}=\\frac{22}{7}.$\nClearing fractions, \\[22\\cos{x}=7+7\\sin{x}.\\]\nSquaring to get an expression in terms of $\\sin^2{x}$ and $\\cos^2{x}$ \\[484\\cos^2{x}=49+49\\sin^2{x}+98\\sin{x}.\\]\nSubstituting $\\cos^2{x}=1-\\sin^2{x},$\n\\[484(1-\\sin^2{x})=49+49\\sin^2{x}+98\\sin{x}.\\]\nExpanding then collecting terms yields a quadratic in $\\sin{x}:$\n\\[533\\sin^2{x}+98\\sin{x}-435=0.\\]\nTo make calculations easier, let $y=\\sin{x}.$\n\\[533y^2+98y-435=0.\\]\nUpon inspection, $y=-1$ is a root. Dividing by $y+1$\n\\[533y^2+98y-435=(533y-435)(y+1).\\]\nSubstituting $y=\\sin{x},$ we see that $\\sin{x}=-1$ doesn't work, as $\\cos{x}=0$ , leaving $\\tan{x}$ undefined.\nWe conclude that $\\sin{x}=\\frac{435}{533}.$\nSince $\\sin^2{x}+\\cos^2{x}=1,$\n\\[\\cos{x}=\\pm \\sqrt{\\frac{533^2-435^2}{533^2}}.\\] \\[=\\pm \\frac{308}{533}.\\]\nAfter checking via the given equation, we know that only the positive solution works.\nTherefore,\n\\[\\csc{x}+\\cot{x}=\\frac{1}{\\sin{x}}+\\frac{\\cos{x}}{\\sin{x}}\\] \\[=\\frac{533}{435}+\\frac{308}{435}\\] \\[=\\frac{29}{15}=\\frac{m}{n}.\\]\nAdding $m$ and $n$ , our answer is $\\boxed{044}.$" ]

[ "We begin by rotating $\\triangle{ APB}$ counterclockwise by $60^{\\circ}$ about $A$ , such that $P\\mapsto Q$ and $B\\mapsto C$ . We see that $\\triangle{ APQ}$ is equilateral with side length $1$ , meaning that $\\angle APQ = 60^{\\circ}$ . We also see that $\\triangle{CPQ}$ is a $30$ $60$ $90$ right triangle, meaning that $\\angle CPQ= 60^{\\circ}$ . Thus, by adding the two together, we see that $\\angle APC = 120^{\\circ}$ We can now use the law of cosines as following: \\begin{align*} s^2 &= (AP)^2 + (CP)^2 - 2\\cdot AP\\cdot CP\\cdot \\cos{\\angle{APC}} \\\\ &= 1 + 4 - 2\\cdot 1\\cdot 2\\cdot \\cos{120^{\\circ}} \\\\ &= 5 - 4\\left(-\\frac{1}{2}\\right) \\\\ &= 7, \\end{align*} giving us that $s = \\boxed{7}$", "Rotate $\\triangle CPA$ counterclockwise $60^\\circ$ around point $C$ to $\\triangle CQB$ . Then $CP=CQ, \\angle PCQ=60^\\circ$ , so $\\triangle CPQ$ is an equilateral triangle. Note that $\\triangle PQB$ is a $30^\\circ$ $60^\\circ$ $90^\\circ$ triangle, hence $\\angle BPQ=30^\\circ$ , and $\\angle BPC=90^\\circ$ , so \\[BC^2=PC^2+PB^2=2^2+3=7,\\] and the answer is $\\boxed{7}$", "Rotate $\\triangle BPC$ counterclockwise by $60^\\circ$ around point $B$ to $\\triangle BQA$ . Then $BP=BQ$ , and $\\angle PBQ=60^\\circ$ , so $\\triangle BPQ$ is an equilateral triangle. Note that $\\triangle QAP$ is a $30^\\circ$ $60^\\circ$ $90^\\circ$ triangle, hence $\\angle PQA=30^\\circ$ , and $\\angle BQA=90^\\circ$ , so \\[AB^2=PQ^2+AQ^2=2^2+3=7,\\] and the answer is $\\boxed{7}$", "\nSuppose that triangle $ABC$ had three segments of length $2$ , emanating from each of its vertices, making equal angles with each of its sides, and going into its interior. Suppose each of these segments intersected the segment clockwise to it precisely at its other endpoint and inside $ABC$ (as pictured in the diagram above). Clearly $s > 2$ and the triangle defined by these intersection points will be equilateral (pictured by the blue segments).\nTake this equilateral triangle to have side length $1$ . The portions of each segment outside this triangle (in red) have length $1$ . Take $P'$ to be the intersection of the segments emanating from $A$ and $C$ . By Law of Cosines, \\[BP' = \\sqrt{1 + 1 - 2\\cos{120^\\circ}} = \\sqrt{3}.\\] So, $P'$ actually satisfies the conditions of the problem, and we can obtain again by Law of Cosines \\[s = \\sqrt{4 + 1 - 4\\cos{120^\\circ}} = \\boxed{7}.\\]", "\nWe begin by dropping altitudes from point $P$ down to all three sides of the triangle as shown above. We can therefore make equations regarding the areas of triangles $\\triangle{APC}$ $\\triangle{APB}$ , and $\\triangle{BPC}$ . Let $s$ be the side of the equilateral triangle, we use the Heron's formula:\n\\[\\triangle{APC} = \\frac{s\\cdot PF}{2} = \\sqrt{\\frac{s+3}{2}\\left(\\frac{s+3}{2}-s\\right)\\left(\\frac{s+3}{2}-1\\right)\\left(\\frac{s+3}{2}-2\\right)}\\] \\[\\implies PF = \\frac{\\sqrt{10s^2-s^4-9}}{2s}\\]\nSimilarly, we obtain:\n\\[PE = \\frac{\\sqrt{8s^2-s^4-4}}{2s}\\] \\[PG = \\frac{\\sqrt{14s^2-s^4-1}}{2s}\\]\nBy Viviani's theorem, \\[\\frac{\\sqrt{10s^2-s^4-9}}{2s}+\\frac{\\sqrt{8s^2-s^4-4}}{2s}+\\frac{\\sqrt{14s^2-s^4-1}}{2s} = \\frac{\\sqrt{3}}{2}s\\] \\[\\sqrt{10s^2-s^4-9}+\\sqrt{8s^2-s^4-4}+\\sqrt{14s^2-s^4-1} = \\sqrt{3}s^2\\]\nNote that from now on, the algebra will get extremely ugly and almost impossible to do by hand within the time frame. However, we do see that it's extremely easy to check the answer choices with the equation in this form. Testing $s = \\sqrt{7}$ , We obtain $7\\sqrt{3}$ on both sides, revealing that our answer is in fact $\\boxed{7}$", "Instead of directly finding the side length of the equilateral triangle, we instead find the area and use it to find the side length.\nBegin by reflecting $P$ over each of the sides. Label these reflected points $P', P'', P'''$ . Connect these points to the vertices of the equilateral triangle, as well as to each other.\n\nObserve that the area of the equilateral triangle $ABC$ is half that of the hexagon $AP''CP'BP'''$\nNote that $AP=AP''=AP'''$ . The same goes for the other vertices. This means that $AP''P'''$ is isosceles. Using either the Law of Cosines or simply observing that $AP''P'''$ is comprised of two 30-60-90 triangles, we find that $P''P'''= \\sqrt{3}$ . Similarly (pun intended), $P'P'''=3$ and $P'P''=2\\sqrt{3}$ . Using the previous observation that $AP''P'''$ is two 30-60-90 triangles (as are the others) we find the areas of $AP''P''$ to be $\\frac{\\sqrt{3}}{4}$ . Again, using similarity we find the area of $BP'P'''$ to be $\\frac{3\\sqrt{3}}{4}$ and the area of $CP'P''$ to be $\\sqrt{3}$\nNext, observe that $P'P''P'''$ is a 30-60-90 right triangle. This right triangle therefore has an area of $\\frac{3\\sqrt{3}}{2}$\nAdding these areas together, we get the area of the hexagon as $\\frac{7\\sqrt{3}}{2}$ . This means that the area of $ABC$ is $\\frac{7\\sqrt{3}}{4}$\nThe formula for the area of an equilateral triangle with side length $s$ is $\\frac{s^2\\sqrt{3}}{4}$ (if you don't have this memorized it's not hard to derive). Comparing this formula to the area of $ABC$ , we can easily find that $s^2=7$ , which means that the side length of $ABC$ is $\\boxed{7}$", "Suppose $A(0,\\sqrt{3}a)$ $B(-a,0)$ $C(a,0)$ and $P(x,y)$ . So $s=2a$ . Since $BP = \\sqrt{3}$ and $CP = 2$ , we have \\[(x+a)^2+y^2=3\\] \\[(x-a)^2+y^2=4\\] Solving the equations, we have \\[x=-\\frac{1}{4a},~~y=\\sqrt{\\frac72-a^2-\\frac{1}{16a^2}}\\] From $AP=1$ (and a fair amount of algebra), we can have $a=\\sqrt{7}/2$ . The answer is $\\boxed{7}$", "Drawing out a rough sketch, it appears that $\\angle BPC = 90^{\\circ}$ . By Pythagorean, our answer is $\\sqrt{\\sqrt{3}^2 + 2^2} = \\boxed{7}$" ]

[ "We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like...\n\\[(a_{101} - a_1) + (a_{102} - a_2) +... + (a_{200} - a_{100}) = 200-100 = 100\\]\n...we get the value of the common difference of every hundred terms one hundred times. So we have to divide the answer by one hundred to get ...\n$\\frac{100}{100} = 1$\n...the common difference of every hundred terms. Then we have to simply divide the answer by hundred again to find the common difference between one term, therefore...\n$\\frac{1}{100} =\\boxed{0.01}$", "Adding the two given equations together gives\n$a_1+a_2+...+a_{200}=300$\nNow, let the common difference be $d$ . Notice that $a_2-a_1=d$ , so we merely need to find $d$ to get the answer. The formula for an arithmetic sum is\n$\\frac{n}{2}(2a_1+d(n-1))$\nwhere $a_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have $n=100$ . Therefore, we have\n$50(2a_1+99d)=100$\nor\n$2a_1+99d=2$ . * (1)\nFor the sum of the equations (shown at the beginning of the solution) we have $n=200$ , so\n$100(2a_1+199d)=300$\nor\n$2a_1+199d=3$ (2)\nNow we have a system of equations in terms of $a_1$ and $d$ .\nSubtracting (1) from (2) eliminates $a_1$ , yielding $100d=1$ , and $d=a_2-a_1=\\frac{1}{100}=\\boxed{0.01}$", "Subtracting the 2 given equations yields\n$(a_{101}-a_1)+(a_{102}-a_2)+(a_{103}-a_3)+...+(a_{200}-a_{100})=100$\nNow express each $a_n$ in terms of first term $a_1$ and common difference $x$ between consecutive terms\n$((a_1+100x)-(a_1))+((a_1+101x)-(a_1+x))+((a_1+102x)-(a_1+2x))+...+((a_1+199x)-(a_1+99x))=100$\nSimplifying and canceling $a_1$ and $x$ terms gives\n$100x+100x+100x+...+100x=100$\n$100x\\times100=100$\n$100x=1$\n$x=0.01=\\boxed{0.01}$" ]

[ "Let us plug in $(5 * x)=1$ into $3a-b$ . Thus it would be $3(5)-x$ . Now we have $2*(15-x)=1$ . Plugging $2*(15-x)$ into $3a-b$ , we have $6-15+x=1$ . Solving for $x$ we have \\[-9+x=1\\] \\[x=\\boxed{10}\\]", "Let us set a variable $y$ equal to $5 * x$ . Solving for y in the equation $3(2)-y=1$ , we see that y is equal to five. By substitution, we see that $5 * x$ = 5. Solving for x in the equation $5(3)-x = 5$ we get \\[x=\\boxed{10}\\]" ]

[ "Denote by $\\nu_p (x)$ the number of prime factor $p$ in number $x$\nWe index Equations given in this problem from (1) to (7).\nFirst, we compute $\\nu_2 (x)$ for $x \\in \\left\\{ a, b, c, d \\right\\}$\nEquation (5) implies $\\max \\left\\{ \\nu_2 (b), \\nu_2 (c) \\right\\} = 1$ .\nEquation (2) implies $\\max \\left\\{ \\nu_2 (a), \\nu_2 (b) \\right\\} = 3$ .\nEquation (6) implies $\\max \\left\\{ \\nu_2 (b), \\nu_2 (d) \\right\\} = 2$ .\nEquation (1) implies $\\nu_2 (a) + \\nu_2 (b) + \\nu_2 (c) + \\nu_2 (d) = 6$\nTherefore, all above jointly imply $\\nu_2 (a) = 3$ $\\nu_2 (d) = 2$ , and $\\left( \\nu_2 (b), \\nu_2 (c) \\right) = \\left( 0 , 1 \\right)$ or $\\left( 1, 0 \\right)$\nSecond, we compute $\\nu_3 (x)$ for $x \\in \\left\\{ a, b, c, d \\right\\}$\nEquation (2) implies $\\max \\left\\{ \\nu_3 (a), \\nu_3 (b) \\right\\} = 2$ .\nEquation (3) implies $\\max \\left\\{ \\nu_3 (a), \\nu_3 (c) \\right\\} = 3$ .\nEquation (4) implies $\\max \\left\\{ \\nu_3 (a), \\nu_3 (d) \\right\\} = 3$ .\nEquation (1) implies $\\nu_3 (a) + \\nu_3 (b) + \\nu_3 (c) + \\nu_3 (d) = 9$\nTherefore, all above jointly imply $\\nu_3 (c) = 3$ $\\nu_3 (d) = 3$ , and $\\left( \\nu_3 (a), \\nu_3 (b) \\right) = \\left( 1 , 2 \\right)$ or $\\left( 2, 1 \\right)$\nThird, we compute $\\nu_5 (x)$ for $x \\in \\left\\{ a, b, c, d \\right\\}$\nEquation (5) implies $\\max \\left\\{ \\nu_5 (b), \\nu_5 (c) \\right\\} = 2$ .\nEquation (2) implies $\\max \\left\\{ \\nu_5 (a), \\nu_5 (b) \\right\\} = 3$ .\nThus, $\\nu_5 (a) = 3$\nFrom Equations (5)-(7), we have either $\\nu_5 (b) \\leq 1$ and $\\nu_5 (c) = \\nu_5 (d) = 2$ , or $\\nu_5 (b) = 2$ and $\\max \\left\\{ \\nu_5 (c), \\nu_5 (d) \\right\\} = 2$\nEquation (1) implies $\\nu_5 (a) + \\nu_5 (b) + \\nu_5 (c) + \\nu_5 (d) = 7$ .\nThus, for $\\nu_5 (b)$ $\\nu_5 (c)$ $\\nu_5 (d)$ , there must be two 2s and one 0.\nTherefore, \\begin{align*} {\\rm gcd} (a,b,c,d) & = \\Pi_{p \\in \\{ 2, 3, 5\\}} p^{\\min\\{ \\nu_p (a), \\nu_p(b) , \\nu_p (c), \\nu_p(d) \\}} \\\\ & = 2^0 \\cdot 3^1 \\cdot 5^0 \\\\ & = \\boxed{3}" ]

[ "First, we have: \\[x^{(\\log_y z)^2} = x^{\\left( (\\log_y z)^2 \\right) } = x^{(\\log_y z) \\cdot (\\log_y z) } = \\left( x^{\\log_y z} \\right)^{\\log_y z}\\]\nNow, let $x=y^w$ , then we have: \\[x^{\\log_y z} = \\left( y^w \\right)^{\\log_y z} = y^{w\\log_y z} = y^{\\log_y (z^w)} = z^w\\]\nThis is all we need to evaluate the given formula. Note that in our case we have $27=3^3$ $49=7^2$ , and $\\sqrt{11}=11^{1/2}$ . We can now compute:\n\\[a^{(\\log_3 7)^2} = \\left( a^{\\log_3 7} \\right)^{\\log_3 7} = 27^{\\log_3 7} = (3^3)^{\\log_3 7} = 7^3 = 343\\]\nSimilarly, we get \\[b^{(\\log_7 11)^2} = (7^2)^{\\log_7 11} = 11^2 = 121\\]\nand \\[c^{(\\log_{11} 25)^2} = (11^{1/2})^{\\log_{11} 25} = 25^{1/2} = 5\\]\nand therefore the answer is $343+121+5 = \\boxed{469}$", "We know from the first three equations that $\\log_a27 = \\log_37$ $\\log_b49 = \\log_711$ , and $\\log_c\\sqrt{11} = \\log_{11}25$ . Substituting, we find\n\\[a^{(\\log_a27)(\\log_37)} + b^{(\\log_b49)(\\log_711)} + c^{(\\log_c\\sqrt {11})(\\log_{11}25)}.\\]\nWe know that $x^{\\log_xy} =y$ , so we find\n\\[27^{\\log_37} + 49^{\\log_711} + \\sqrt {11}^{\\log_{11}25}\\]\n\\[(3^{\\log_37})^3 + (7^{\\log_711})^2 + ({11^{\\log_{11}25}})^{1/2}.\\]\nThe $3$ and the $\\log_37$ cancel to make $7$ , and we can do this for the other two terms. Thus, our answer is\n\\[7^3 + 11^2 + 25^{1/2}\\] \\[= 343 + 121 + 5\\] \\[= \\boxed{469}.\\]" ]

[ "The repeating decimal $0.\\overline{ab}$ is equal to \\[\\frac{10a+b}{100} + \\frac{10a+b}{10000} + \\cdots = (10a+b)\\cdot\\left(\\frac 1{10^2} + \\frac 1{10^4} + \\cdots \\right) = (10a+b) \\cdot \\frac 1{99} = \\frac{10a+b}{99}\\]\nWhen expressed in the lowest terms, the denominator of this fraction will always be a divisor of the number $99 = 3\\cdot 3\\cdot 11$ . This gives us the possibilities $\\{1,3,9,11,33,99\\}$ . As $a$ and $b$ are not both nine and not both zero, the denominator $1$ can not be achieved, leaving us with $\\boxed{5}$ possible denominators.", "Another way to convert the decimal into a fraction (simplifying, I guess?). We have \\[100(0.\\overline{ab}) = ab.\\overline{ab}\\] \\[99(0.\\overline{ab}) = 100(0.\\overline{ab}) - 0.\\overline{ab} = ab.\\overline{ab} - 0.\\overline{ab} = ab\\] \\[0.\\overline{ab} = \\frac{ab}{99}\\] where $a, b$ are digits. Continuing in the same way by looking at the factors of 99, we have 5 different possibilities for the denominator. $\\boxed{5}$" ]

[ "Since $\\frac{1}{99}=0.\\overline{01}$ , we know that $0.\\overline{ab} = \\frac{ab}{99}$ . From here, we wish to find the number of factors of $99$ , which is $6$ . However, notice that $1$ is not a possible denominator, so our answer is $6-1=\\boxed{5}$ \\[\\] ~AopsUser101", "Since $0.\\overline{ab} = \\frac{ab}{99}$ , the denominator must be a factor of $99 = 3^2 \\cdot 11$ . The factors of $99$ are $1,$ $3,$ $9,$ $11,$ $33,$ and $99$ . Since $a$ and $b$ are not both nine, the denominator cannot be $1$ . By choosing $a$ and $b$ appropriately, we can make fractions with each of the other denominators.\nThus, the answer is $\\boxed{5}$" ]

[ "As $f(x)=ax^2 + bx + c$ , we have $f(1)=a\\cdot 1^2 + b\\cdot 1 + c = a+b+c$\nTo compute $f(1)$ , set $x=-2$ in the first formula. We get $f(1) = f(-2+3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = \\boxed{2}$", "Combining the two formulas, we know that $f(x+3) = a(x+3)^2 + b(x+3) + c$\nWe can rearrange the right hand side to $ax^2 + (6a+b)x + (9a+3b+c)$\nComparing coefficients we have $a=3$ $6a+b=7$ , and $9a+3b+c = 4$ . From the second equation we get $b=-11$ , and then from the third we get $c=10$ . Hence $a+b+c = 3-11+10 = \\boxed{2}$" ]

[ "$3 \\cdot 5^2m$ must be a perfect cube, so each power of a prime in the factorization for $3 \\cdot 5^2m$ must be divisible by $3$ . Thus the minimum value of $m$ is $3^2 \\cdot 5 = 45$ , which makes $n = \\sqrt[3]{3^3 \\cdot 5^3} = 15$ . The minimum possible value for the sum of $m$ and $n$ is $\\boxed{60}.$", "First, we need to prime factorize $75$ $75$ $5^2 \\cdot 3$ . We need $75m$ to be in the form $x^3y^3$ . Therefore, the smallest $m$ is $5 \\cdot 3^2$ $m$ = 45, and since $5^3 \\cdot 3^3 = 15^3$ , our answer is $45 + 15$ $\\boxed{60}$" ]

[ "Whenever $n$ is the product of three consecutive integers, $n$ is divisible by $3!$ , meaning it is divisible by $6$\nIt also mentions that it is divisible by $7$ , so the number is definitely divisible by all the factors of $42$\nIn our answer choices, the one that is not a factor of $42$ is $\\boxed{28}$", "We can look for counterexamples. For example, letting $n = 13 \\cdot 14 \\cdot 15$ , we see that $n$ is not divisible by 28, so $\\boxed{28}$ is our answer." ]

[ "The First (pun intended) thing to notice is that $xy$ and $zx$ have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is $e^{\\frac{3\\pi i}{2}}$ . Anyway this results in getting that $\\left(\\frac{-3i}{10}\\right)y=z$ . Then substitute this into $yz$ to get, after some calculation, that $y=10e^{\\frac{5\\pi i}{4}}\\sqrt{2}$ and $z=-3e^{\\frac{7\\pi i}{4}}\\sqrt{2}$ . Then plug $z$ into $zx$ , you could do the same thing with $xy$ but $zx$ looks like it's easier due to it being smaller. Anyway you get $x=20+12i$ . Then add all three up, it turns out easier than it seems because for $z$ and $y$ the $\\sqrt{2}$ disappears after you expand the root of unity (e raised to a specific power). Long story short, you get $x=20+12i, y=-3+3i, z=-10-10i \\implies x+y+z=7+5i \\implies a^2+b^2=\\boxed{074}$", "First we evaluate the magnitudes. $|xy|=80\\sqrt{17}$ $|yz|=60$ , and $|zx|=24\\sqrt{17}$ . Therefore, $|x^2y^2z^2|=17\\cdot80\\cdot60\\cdot24$ , or $|xyz|=240\\sqrt{34}$ . Divide to find that $|z|=3\\sqrt{2}$ $|x|=4\\sqrt{34}$ , and $|y|=10\\sqrt{2}$ This allows us to see that the argument of $y$ is $\\frac{\\pi}{4}$ , and the argument of $z$ is $-\\frac{\\pi}{4}$ . We need to convert the polar form to a standard form. Simple trig identities show $y=10+10i$ and $z=3-3i$ . More division is needed to find what $x$ is. \\[x=-20-12i\\] \\[x+y+z=-7-5i\\] \\[(-7)^2+(-5)^2=\\boxed{74}\\] \\[QED\\blacksquare\\] Written by a1b2", "Solve this system the way you would if the RHS of all equations were real. Multiply the first and 3rd equations out and then factor out $60$ to find $x^2$ , then use standard techniques that are used to evaluate square roots of irrationals. let \\[x = c+di\\] , then you get \\[c^2 - d^2 = 256\\] and \\[2cd = 480\\] Solve to get $x$ as $20+12i$ and $-20-12i$ . Both will give us the same answer, so use the positive one. Divide $-80-320i$ by $x$ , and you get $10+10i$ as $y$ . This means that $z$ is a multiple of $1-i$ to get a real product, so you find $z$ is $3-3i$ . Now, add the real and imaginary parts separately to get $-7-5i$ , and calculate $a^2 + b^2$ to get $\\boxed{74}$ . \n~minor latex improvements done by jske25 and jdong2006", "Dividing the first equation by the second equation given, we find that $\\frac{xy}{yz}=\\frac{x}{z}=\\frac{-80-320i}{60}=-\\frac{4}{3}-\\frac{16}{3}i \\implies x=z\\left(-\\frac{4}{3}-\\frac{16}{3}i\\right)$ . Substituting this into the third equation, we get $z^2=\\frac{-96+24i}{-\\frac{4}{3}-\\frac{16}{3}i}=3\\cdot \\frac{-24+6i}{-1-4i}=3\\cdot \\frac{(-24+6i)(-1+4i)}{1+16}=3\\cdot \\frac{-102i}{17}=-18i$ . Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of $y$ is the negative of that of $z$ , and their magnitudes multiply to $60$ . Thus, we have $z=\\sqrt{-18i}=3-3i$ and $3\\sqrt{2}\\cdot |y|=60 \\implies |y|=10\\sqrt{2} \\implies y=10+10i$ . To find $x$ , we can use the previous substitution we made to find that $x=z\\left(-\\frac{4}{3}-\\frac{16}{3}i\\right)=-\\frac{4}{3}\\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i$ .\nTherefore, $x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \\implies a^2+b^2=(-7)^2+(-5)^2=49+25=\\boxed{74}$", "We observe that by multiplying $xy,$ $yz,$ and $zx,$ we get $(xyz)^2=(-80-320i)(60)(-96+24i).$ Next, we divide $(xyz)^2$ by $(yz)^2$ to \nget $x^2.$ We have $x^2=\\frac{(-80-320i)(60)(-96+24i)}{3600}=256+480i.$ We can write $x$ in the form of $a+bi,$ so we get $(a+bi)^2=256+480i.$ Then, $a^2-b^2+2abi=256+480i,$ $a^2-b^2=256,$ and $2ab=480.$ Solving this system of equations is relatively \nsimple. We have two cases, $a=20, b=12,$ and $a=-20, b=-12.$ Case 1: $a=20, b=12,$ so $x=20+12i.$ We solve for $y$ and $z$ by plugging in $x$ to the two equations. We see $y=\\frac{-80-320i}{20+12i}=-10-10i$ and $z=\\frac{-96+24i}{20+12i}=-3+3i.$ $x+y+z=7+5i,$ so $a=7$ and $b=5.$ Solving, we end up with $7^2+5^2=\\boxed{074}$ as our answer. \nCase 2: $a=-20, b=-12,$ so $x=-20-12i.$ Again, we solve for $y$ and $z.$ We find $y=\\frac{-80-320i}{-20-12i}=10+10i,$ $z=\\frac{-96+24i}{-20-12i}=3-3i,$ so $x+y+z=-7-5i.$ We again have $(-7)^2+(-5)^2=\\boxed{074}.$ Solution by Airplane50", "According to the Euler's Theory, we can rewrite $x$ $y$ and $z$ as \\[x=r_{1}e^{i{\\theta}_1}\\] \\[y=r_{2}e^{i{\\theta}_2}\\] \\[x=r_{3}e^{i{\\theta}_3}\\] As a result, \\[|xy|=r_{1}r_{2}=\\sqrt{80^2+320^2}=80\\sqrt{17}\\] \\[|yz|=r_{2}r_{3}=60\\] \\[|xz|=r_{1}r_{3}=\\sqrt{96^2+24^2}=24\\sqrt{17}\\] Also, it is clear that \\[yz=r_{2}e^{i{\\theta}_2}r_{3}e^{i{\\theta}_3}=|yz|e^{i({\\theta}_2+{\\theta}_3)}=|yz|=60\\] So ${\\theta}_2+{\\theta}_3=0$ , or \\[{\\theta}_2=-{\\theta}_3\\] Also, we have \\[xy=-80\\sqrt{17}e^{i\\arctan{4}}\\] \\[yz=60\\] \\[xz=-24\\sqrt{17}e^{i\\arctan{-\\frac{1}{4}}}\\] So now we have $r_{1}r_{2}=80\\sqrt{17}$ $r_{2}r_{3}=60$ $r_{1}r_{3}=24\\sqrt{17}$ ${\\theta}_1+{\\theta}_2=\\arctan{4}$ and ${\\theta}_1-{\\theta}_2=\\arctan {-\\frac{1}{4}}$ . Solve these above, we get \\[r_{1}=4\\sqrt{34}\\] \\[r_{2}=10\\sqrt{2}\\] \\[r_{3}=3\\sqrt{2}\\] \\[{\\theta}_2=\\frac{\\arctan{4}-\\arctan{-\\frac{1}{4}}}{2}=\\frac{\\frac{\\pi}{2}}{2}=\\frac{\\pi}{4}\\] So we can get \\[y=r_{2}e^{i{\\theta}_2}=10\\sqrt{2}e^{i\\frac{\\pi}{4}}=10+10i\\] \\[z=r_{3}e^{i{\\theta}_3}=r_{3}e^{-i{\\theta}_2}=3\\sqrt{2}e^{-i\\frac{\\pi}{4}}=3-3i\\] Use $xy=-80-320i$ we can find that \\[x=-20-12i\\] So \\[x+y+z=-20-12i+10+10i+3-3i=-7-5i\\] So we have $a=-7$ and $b=-5$\nAs a result, we finally get \\[a^2+b^2=(-7)^2+(-5)^2=\\boxed{074}\\]", "We can turn the expression $x+y+z$ into $\\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}$ , and this would allow us to plug in the values after some computations.\nBased off of the given products, we have\n$xy^2z=60(-80-320i)$\n$xyz^2=60(-96+24i)$\n$x^2yz=(-96+24i)(-80-320i)$\nDividing by the given products, we have\n$y^2=\\frac{60(-80-320i)}{-96+24i}$\n$z^2=\\frac{60(-96+24i)}{-80-320i}$\n$x^2=\\frac{(-96+24i)(-80-320i)}{60}$\nSimplifying, we get that this expression becomes $\\sqrt{24+70i}$ . This equals $\\pm{(7+5i)}$ , so the answer is $7^2+5^2=\\boxed{74}$", "Multiplying $xy \\cdot yz \\cdot zx = (xyz)^2$ we obtain $60 \\cdot 960(16+30i)$ (too lazy to do $60 \\cdot 960$ , you don't need to). Taking the square root, we get $240\\sqrt{16+30i}$ . Letting $(a+bi)^2=16+30i,$ we have $a^2+2abi-b^2=16+30i.$ Thus, $(a+b)(a-b)=16,$ and $2ab=30.$ Guessing and checking, we get $a+bi=5+3i$ . Therefore, $xyz=240(5+3i).$ Dividing this by each of the equations provided in the original problem, we get $x=20+12i,y=-10-10i,$ and $z=-3+3i$ $20+12i-10-10i-3+3i=7+5i$ . Finally, $7^2+5^2=\\boxed{074}.$" ]

[ "We can rewrite $xyz=1$ as $\\frac{1}{z}=xy$\nSubstituting into one of the given equations, we have \\[x+xy=5\\] \\[x(1+y)=5\\] \\[\\frac{1}{x}=\\frac{1+y}{5}.\\]\nWe can substitute back into $y+\\frac{1}{x}=29$ to obtain \\[y+\\frac{1+y}{5}=29\\] \\[5y+1+y=145\\] \\[y=24.\\]\nWe can then substitute once again to get \\[x=\\frac15\\] \\[z=\\frac{5}{24}.\\] Thus, $z+\\frac1y=\\frac{5}{24}+\\frac{1}{24}=\\frac{1}{4}$ , so $m+n=\\boxed{005}$", "Let $r = \\frac{m}{n} = z + \\frac {1}{y}$\n\\begin{align*} (5)(29)(r)&=\\left(x + \\frac {1}{z}\\right)\\left(y + \\frac {1}{x}\\right)\\left(z + \\frac {1}{y}\\right)\\\\ &=xyz + \\frac{xy}{y} + \\frac{xz}{x} + \\frac{yz}{z} + \\frac{x}{xy} + \\frac{y}{yz} + \\frac{z}{xz} + \\frac{1}{xyz}\\\\ &=1 + x + z + y + \\frac{1}{y} + \\frac{1}{z} + \\frac{1}{x} + \\frac{1}{1}\\\\ &=2 + \\left(x + \\frac {1}{z}\\right) + \\left(y + \\frac {1}{x}\\right) + \\left(z + \\frac {1}{y}\\right)\\\\ &=2 + 5 + 29 + r\\\\ &=36 + r \\end{align*}\nThus $145r = 36+r \\Rightarrow 144r = 36 \\Rightarrow r = \\frac{36}{144} = \\frac{1}{4}$ . So $m + n = 1 + 4 = \\boxed{5}$", "(Hybrid between 1/2)\nBecause $xyz = 1, \\hspace{0.15cm} \\frac{1}{x} = yz, \\hspace{0.15cm} \\frac{1}{y} = xz,$ and $\\hspace{0.05cm}\\frac{1}{z} = xy$ . Substituting and factoring, we get $x(y+1) = 5$ $\\hspace{0.15cm}y(z+1) = 29$ , and $\\hspace{0.05cm}z(x+1) = k$ . Multiplying them all together, we get, $xyz(x+1)(y+1)(z+1) = 145k$ , but $xyz$ is $1$ , and by the Identity property of multiplication, we can take it out. So, in the end, we get $(x+1)(y+1)(z+1) = 145k$ . And, we can expand this to get $xyz+xy+yz+xz+x+y+z+1 = 145k$ , and if we make a substitution for $xyz$ , and rearrange the terms, we get $xy+yz+xz+x+y+z = 145k-2$ This will be important.\nNow, lets add the 3 equations $x(y+1) = 5, \\hspace{0.15cm}y(z+1) = 29$ , and $\\hspace{0.05cm}z(x+1) = k$ . We use the expand the Left hand sides, then, we add the equations to get $xy+yz+xz+x+y+z = k+34$ Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus $145k-2 = k+34$ We move all constant terms to the right, and all linear terms to the left, to get $144k = 36$ , so $k = \\frac{1}{4}$ which gives an answer of $1+4 = \\boxed{005}$", "Get rid of the denominators in the second and third equations to get $xz-5z=-1$ and $xy-29x=-1$ . Then, since $xyz=1$ , we have $\\tfrac 1y-5z=-1$ and $\\tfrac 1z-29x=-1$ . Then, since we know that $\\tfrac 1z+x=5$ , we can subtract these two equations to get that $30x=6\\implies x=5$ . The result follows that $z=\\tfrac 5{24}$ and $y=24$ , so $z+\\tfrac 1y=\\tfrac 1{24}+\\tfrac 5{24}=\\tfrac 14$ , and the requested answer is $1+4=\\boxed{005}.$", "Rewrite the equations in terms of x.\n$x+\\frac{1}{z}=5$ becomes $z=\\frac{1}{x+5}$\n$y+\\frac{1}{x}=29$ becomes $y=29-\\frac{1}{x}$\nNow express $xyz=1$ in terms of x.\n$\\frac{1}{5-x}\\cdot(29-\\frac{1}{x})\\cdot x=1$\nThis evaluates to $29x-1=5-x$ , giving us $x=\\frac{1}{5}$ . We can now plug x into the other equations to get $y=24$ and $z=\\frac{5}{24}$\nTherefore, $z+\\frac{1}{y}=\\frac{5}{24}+\\frac{1}{24}=\\frac{6}{24}=\\frac{1}{4}$\n$1+4=\\boxed{5}$ , and we are done.\n~MC413551" ]

[ "Substitute $y = \\frac34x$ into $x^y = y^x$ and solve. \\[x^{\\frac34x} = \\left(\\frac34x\\right)^x\\] \\[x^{\\frac34x} = \\left(\\frac34\\right)^x \\cdot x^x\\] \\[x^{-\\frac14x} = \\left(\\frac34\\right)^x\\] \\[x^{-\\frac14} = \\frac34\\] \\[x = \\frac{256}{81}\\] \\[y = \\frac34x = \\frac{192}{81}\\] \\[x + y = \\frac{448}{81}\\] \\[448 + 81 = \\boxed{529}\\]", "We solve in general using $c$ instead of $3/4$ . Substituting $y = cx$ , we have:\nDividing by $x^x$ , we get $(x^x)^{c - 1} = c^x$\nTaking the $x$ th root, $x^{c - 1} = c$ , or $x = c^{1/(c - 1)}$\nIn the case $c = \\frac34$ $x = \\Bigg(\\frac34\\Bigg)^{ - 4} = \\Bigg(\\frac43\\Bigg)^4 = \\frac {256}{81}$ $y = \\frac {64}{27}$ $x + y = \\frac {256 + 192}{81} = \\frac {448}{81}$ , yielding an answer of $448 + 81 = \\boxed{529}$", "Taking the logarithm base $x$ of both sides, we arrive with: \\[y = \\log_x y^x \\Longrightarrow \\frac{y}{x} = \\log_{x} y = \\log_x \\left(\\frac{3}{4}x\\right) = \\frac{3}{4}\\] Now we proceed by the logarithm rule $\\log(ab)=\\log a + \\log b$ . The equation becomes: \\[\\log_x \\frac{3}{4} + \\log_x x = \\frac{3}{4}\\] \\[\\Longleftrightarrow \\log_x \\frac{3}{4} + 1 = \\frac{3}{4}\\] \\[\\Longleftrightarrow \\log_x \\frac{3}{4} = -\\frac{1}{4}\\] \\[\\Longleftrightarrow x^{-\\frac{1}{4}} = \\frac{3}{4}\\] \\[\\Longleftrightarrow \\frac{1}{x^{\\frac{1}{4}}} = \\frac{3}{4}\\] \\[\\Longleftrightarrow x^{\\frac{1}{4}} = \\frac{4}{3}\\] \\[\\Longleftrightarrow \\sqrt[4]{x} = \\frac{4}{3}\\] \\[\\Longleftrightarrow x = \\left(\\frac{4}{3}\\right)^4=\\frac{256}{81}\\] Then find $y$ as in solution 3, and we get $\\boxed{529}$" ]

[ "Since $|x_i| < 1$ then\n\\[|x_1| + |x_2| + \\dots + |x_n| = 19 + |x_1 + x_2 + \\dots + x_n| < n.\\]\nSo $n \\ge 20$ . We now just need to find an example where $n = 20$ : suppose $x_{2k-1} = \\frac{19}{20}$ and $x_{2k} = -\\frac{19}{20}$ ; then on the left hand side we have $\\left|\\frac{19}{20}\\right| + \\left|-\\frac{19}{20}\\right| + \\dots + \\left|-\\frac{19}{20}\\right| = 20\\left(\\frac{19}{20}\\right) = 19$ . On the right hand side, we have $19 + \\left|\\frac{19}{20} - \\frac{19}{20} + \\dots - \\frac{19}{20}\\right| = 19 + 0 = 19$ , and so the equation can hold for $n = \\boxed{020}$", "Straight off, we notice that the RHS must be greater than or equal to 19, because an absolute value only gives nonnegative values. It then becomes clear that $|x_1|+|x_2|+...+|x_n|\\ge19$ . If each $x_n$ were equal to 1, then $n=19$ . However, $x_n<1$ , so there must be at least one extra term to satisfy the inequality. Therefore, $n=\\boxed{020}$" ]

[ "We approach this problem using Linearity of Expectation. Consider a pair of two people standing next to each other. Ignoring all other people, the probability that a boy is standing on the left position and a girl is standing on the right position is $\\frac7{20}\\cdot\\frac{13}{19}$ . Similarly, if a girl is standing on the left position and a boy is standing on the right position the probability is also $\\frac{7\\cdot 13}{20\\cdot 19}$ . Thus, the total probability of the two people being one boy and one girl is $\\frac{91}{190}$\nThere are a total of 19 different adjacent pairs, so by Linearity of Expectation, we have that the expected value of $S$ is $\\frac{91}{10} \\to \\boxed{9}$", "Suppose that the class tried every configuration. Boy $i$ and girl $j$ would stand next to each other in $2$ different orders, in $19$ different positions, $18!$ times each. Summing over all $i,j$ gives $7\\cdot13\\cdot2\\cdot19\\cdot18!=\\tfrac{91}{10}\\cdot20!$ , so the average value of $S$ is $\\boxed{9}$" ]

[ "The first triangle represents $1+3-4$ The 2nd triangle represents $2+5-6$\nSolving the first triangle, we get $0$ Solving the 2nd triangle, we get $1$\nSince we have to add the 2 triangles the final answer is $1$ , which is $\\boxed{1}$" ]

[ "If the vertex is at $\\left(\\frac{1}{4}, -\\frac{9}{8}\\right)$ , the equation of the parabola can be expressed in the form \\[y=a\\left(x-\\frac{1}{4}\\right)^2-\\frac{9}{8}.\\] Expanding, we find that \\[y=a\\left(x^2-\\frac{x}{2}+\\frac{1}{16}\\right)-\\frac{9}{8},\\] and \\[y=ax^2-\\frac{ax}{2}+\\frac{a}{16}-\\frac{9}{8}.\\] From the problem, we know that the parabola can be expressed in the form $y=ax^2+bx+c$ , where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$ $-\\frac{a}{2}=b$ , and $\\frac{a}{16}-\\frac{9}{8}=c$ . Adding up all of these gives us \\[\\frac{9a-18}{16}=a+b+c.\\] We know that $a+b+c$ is an integer, so $9a-18$ must be divisible by $16$ . Let $9a=z$ . If ${z-18}\\equiv {0} \\pmod{16}$ , then ${z}\\equiv {2} \\pmod{16}$ . Therefore, if $9a=2$ $a=\\frac{2}{9}$ . Adding up gives us $2+9=\\boxed{011}$", "Complete the square. Since $a>0$ , the parabola must be facing upwards. $a+b+c=\\text{integer}$ means that $f(1)$ must be an integer. The function can be recasted into $a\\left(x-\\frac{1}{4}\\right)^2-\\frac{9}{8}$ because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than $-\\frac{9}{8}$ is $-1$ . So the $y$ -coordinate must change by $\\frac{1}{8}$ and the $x$ -coordinate must change by $1-\\frac{1}{4}=\\frac{3}{4}$ . Thus, $a\\left(\\frac{3}{4}\\right)^2=\\frac{1}{8}\\implies \\frac{9a}{16}=\\frac{1}{8}\\implies a=\\frac{2}{9}$ . So $2+9=\\boxed{011}$", "To do this, we can use the formula for the minimum (or maximum) value of the $x$ coordinate at a vertex of a parabola, $-\\frac{b}{2a}$ and equate this to $\\frac{1}{4}$ . Solving, we get $-\\frac{a}{2}=b$ . Enter $x=\\frac{1}{4}$ to get $-\\frac{9}{8}=\\frac{a}{16}+\\frac{b}{4}+c=-\\frac{a}{16}+c$ so $c=\\frac{a-18}{16}$ . This means that $\\frac{9a-18}{16}\\in \\mathbb{Z}$ so the minimum of $a>0$ is when the fraction equals -1, so $a=\\frac{2}{9}$ . Therefore, $p+q=2+9=\\boxed{011}$ .\n-Gideontz" ]

[ "WLOG, let C be the largest angle in the triangle.\nAs above, we can see that $\\cos3A+\\cos3B-\\cos(3A+3B)=1$\nExpanding, we get\n$\\cos3A+\\cos3B-\\cos3A\\cos3B+\\sin3A\\sin3B=1$\n$\\cos3A\\cos3B-\\cos3A-\\cos3B+1=\\sin3A\\sin3B$\n$(\\cos3A-1)(\\cos3B-1)=\\sin3A\\sin3B$\nCASE 1: If $\\sin 3A = 0$ or $\\sin 3B = 0$\nThis implies one or both of A or B are 60 or 120.\nIf one of A or B is 120, we have a contradiction, since C must be the largest angle.\nOtherwise, if one of A or B is 60, WLOG, assume A = 60, we would have $\\cos(3B) + \\cos(3C) = 2$ , and thus, cos(3B) and cos(3C) both equal 1, implying $B = C = 120$ , a contradiction to the fact that the sum of the angles of a triangle must be 180 degrees.\nCASE 2: If $\\sin 3A \\neq 0$ and $\\sin 3B \\neq 0$\n$\\frac{\\cos3A-1}{\\sin3A}\\cdot\\frac{\\cos3B-1}{\\sin3B}=1$\n$\\tan{\\frac{3A}{2}}\\tan{\\frac{3B}{2}}=1$\nNote that $\\tan{x}=\\frac{1}{\\tan(90-x)}$ , or $\\tan{x}\\tan(90-x)=1$\nThus $\\frac{3A}{2}+\\frac{3B}{2}=90$ , or $A+B=60$\nNow we know that $C=120$ , so we can just use the Law of Cosines to get $\\boxed{399}$", "\\[\\cos3A+\\cos3B=1-\\cos(3C)=1+\\cos(3A+3B)\\] \\[2\\cos\\frac{3}{2}(A+B)\\cos\\frac{3}{2}(A-B)=2\\cos^2\\frac{3}{2}(A+B)\\] If $\\cos\\frac{3}{2}(A+B) = 0$ , then $\\frac{3}{2}(A+B)=90$ $A+B=60$ , so $C=120$ ; otherwise, \\[2\\cos\\frac{3}{2}(A-B)=2cos\\frac{3}{2}(A+B)\\] \\[\\sin\\frac{3}{2}A\\sin\\frac{3}{2}B=0\\] so either $\\sin\\frac{3}{2}A=0$ or $\\sin\\frac{3}{2}B=0$ , i.e., either $A=120$ or $B=120$ . In all cases, one of the angles must be 120, which opposes the longest side. Final result follows. $\\boxed{399}$", "Let $BC$ be the unknown side length. By Law of Cosines we have that $BC = \\sqrt{269-260\\cos{A}}$ . We notice that $\\cos{A}$ should be negative to optimize $BC$ so $A$ is between $90$ and $180$ degrees. We also know that the value inside the square root is an integer $m$ , so $269-260\\cos{A}$ should be an integer. We can then assume that $A$ is $120$ degrees so $\\cos{A} = \\frac{-1}{2}$ . We do this because $120$ degrees is a \"common\" value and it makes the value inside the square root an integer. Plugging this into $269-260\\cos{A}$ for $m$ we get that it is $\\boxed{399}$" ]

[ "The ratio $\\frac{400 \\text{ euros}}{500 \\text{ dollars}}$ can be simplified using conversion factors: \\[\\frac{400 \\text{ euros}}{500 \\text{ dollars}} \\cdot \\frac{1.3 \\text{ dollars}}{1 \\text{ euro}} = \\frac{520}{500} = 1.04\\] which means the money is greater by $\\boxed{4}$ percent.", "If we divide each of Etienne's and Diana's values by $100$ , the problem stays the same. Then, Etienne has $1.3$ times the amount of money Diana has, so Etienne has $5.2$ dollars. Since $\\dfrac{5.2}{5} = 1.04$ , Etienne has $\\boxed{4}$ percent more money than Diana. ~Extremelysupercooldude" ]

[ "There are $24 \\cdot 60=1440$ normal minutes in a day , and $10 \\cdot 100=1000$ metric minutes in a day. The ratio of normal to metric minutes in a day is $\\frac{1440}{1000}$ , which simplifies to $\\frac{36}{25}$ . This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to $\\text{6:36}$ AM, $6 \\cdot 60+36=396$ normal minutes pass. This can be viewed as $\\frac{396}{36}=11$ cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding $25 \\cdot 11=275$ to $\\text{0:00}$ gives $\\text{2:75}$ , so the answer is $\\boxed{275}$", "First we want to find out what fraction of a day has passed at 6:36 AM. One hour is $\\frac{1}{24}$ of a day, and 36 minutes is $\\frac{36}{60}=\\frac{3}{5}$ of an hour, so at 6:36 AM, $6 \\cdot \\frac{1}{24} + \\frac{1}{24} \\cdot \\frac{3}{5}=\\frac{1}{4}+\\frac{1}{40}=\\frac{11}{40}$ of a day has passed. Now the metric timing equivalent of $\\frac{11}{40}$ of a day is $\\frac{11}{40}\\cdot 1000=275$ metric minutes, which is equivalent to 2 metric hours and 75 metric minutes, so our answer is $\\boxed{275}$ - mathleticguyyy" ]

[ "Note that for all real numbers $k,$ we have $a^{2k} + a^{-2k} + 2 = (a^{k} + a^{-k})^2,$ from which \\[a^{2k} + a^{-2k} = (a^{k} + a^{-k})^2-2.\\] We apply this result twice to get the answer: \\begin{align*} a^4 + a^{-4} &= (a^2 + a^{-2})^2 - 2 \\\\ &= [(a + a^{-1})^2 - 2]^2 - 2 \\\\ &= \\boxed{194} ~Azjps (Fundamental Logic)", "Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$\nSquaring both sides of $a^2+a^{-2}=14$ gives $a^4+a^{-4}+2=196,$ from which $a^4+a^{-4}=\\boxed{194}.$", "The detailed explanation of Solution 2 is as follows: \\begin{alignat*}{8} a+a^{-1}&=4 \\\\ (a+a^{-1})^2&=4^2 \\\\ a^2+2aa^{-1}+a^{-2}&=16 \\\\ a^2+a^{-2}&=16-2&&=14 \\\\ (a^2+a^{-2})^2&=14^2 \\\\ a^4+2a^2a^{-2}+a^{-4}&=196 \\\\ a^4+a^{-4}&=196-2&&=\\boxed{194} ~MathFun1000 (Solution)", "Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$\nApplying the Binomial Theorem, we raise both sides of $a+a^{-1}=4$ to the fourth power: \\begin{align*} \\binom40a^4a^0+\\binom41a^3a^{-1}+\\binom42a^2a^{-2}+\\binom43a^1a^{-3}+\\binom44a^0a^{-4}&=256 \\\\ a^4+4a^2+6+4a^{-2}+a^{-4}&=256 \\\\ \\left(a^4+a^{-4}\\right)+4\\left(a^2+a^{-2}\\right)&=250 \\\\ \\left(a^4+a^{-4}\\right)+4(14)&=250 \\\\ a^4+a^{-4}&=\\boxed{194} ~MRENTHUSIASM", "We multiply both sides of $4=a+a^{-1}$ by $a,$ then rearrange: \\[a^2-4a+1=0.\\]\nWe apply the Quadratic Formula to get $a=2\\pm\\sqrt3.$\nBy Vieta's Formulas, note that the roots are reciprocals of each other. Therefore, both values of $a$ produce the same value of $a^4+a^{-4}:$ \\begin{align*} a^4+a^{-4}&=\\left(2+\\sqrt{3}\\right)^4 + \\left(2+\\sqrt{3}\\right)^{-4} \\\\ &=\\left(2+\\sqrt{3}\\right)^4+\\left(2-\\sqrt{3}\\right)^4 &&(*) \\\\ &=\\boxed{194}$", "From the first sentence of Solution 4, we conclude that $a$ and $a^{-1}$ are the roots of $x^2-4x+1=0.$ Let \\begin{align*} P_1&=a+a^{-1}, \\\\ P_2&=a^2+a^{-2}, \\\\ P_3&=a^3+a^{-3}, \\\\ P_4&=a^4+a^{-4}. \\end{align*} By Newton's Sums, we have \\begin{alignat*}{12} &1\\cdot P_1-4\\cdot 1&&=0 &&\\qquad\\implies\\qquad P_1&&=4, \\\\ &1\\cdot P_2-4\\cdot P_1+1\\cdot2 &&=0 &&\\qquad\\implies\\qquad P_2&&=14, \\\\ &1\\cdot P_3-4\\cdot P_2+1\\cdot P_1&&=0 &&\\qquad\\implies\\qquad P_3&&=52, \\\\ &1\\cdot P_4-4\\cdot P_3+1\\cdot P_2&&=0 &&\\qquad\\implies\\qquad P_4&&=\\boxed{194} ~Albert1993 (Fundamental Logic)", "Note that \\[a^{4} + a^{-4} = (a^{2} + a^{-2})^{2} - 2.\\] We guess that $a^{2} + a^{-2}$ is an integer, so the answer must be $2$ less than a perfect square. The only possibility is $\\boxed{194}.$" ]

[ "One way to solve this problem is by substitution . We have\n$x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10$\nHence observe that we can write $w=x+y$ and $z=xy$\nThis reduces the equations to $w^2-2z=7$ and $w(7-z)=10$\nBecause we want the largest possible $w$ , let's find an expression for $z$ in terms of $w$\n$w^2-7=2z \\implies z=\\frac{w^2-7}{2}$\nSubstituting, $w^3-21w+20=0$ , which factorizes as $(w-1)(w+5)(w-4)=0$ (the Rational Root Theorem may be used here, along with synthetic division).\nThe largest possible solution is therefore $x+y=w=\\boxed{004}$", "An alternate way to solve this is to let $x=a+bi$ and $y=c+di$\nBecause we are looking for a value of $x+y$ that is real, we know that $d=-b$ , and thus $y=c-bi$\nExpanding $x^2+y^2=7+0i$ will give two equations, since the real and imaginary parts must match up.\n$(a+bi)^2+(c-bi)^2=7+0i$\n$(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i$\nLooking at the imaginary part of that equation, $2ab-2cb=0$ , so $a=c$ , and $x$ and $y$ are actually complex conjugates.\nLooking at the real part of the equation and plugging in $a=c$ $2a^2-2b^2=7$ , or $2b^2=2a^2-7$\nNow, evaluating the real part of $(a+bi)^3+(a-bi)^3$ , which equals $10$ (ignoring the odd powers of $i$ , since they would not result in something in the form of $10+0i$ ):\n$a^3+3a(bi)^2+a^3+3a(-bi)^2=10$\n$2a^3-6ab^2=10$\nSince we know that $2b^2=2a^2-7$ , it can be plugged in for $b^2$ in the above equation to yield:\n$2a^3-3a(2a^2-7)=10$\n$-4a^3+21a=10$\n$4a^3-21a+10=0$\nSince the problem is looking for $x+y=2a$ to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, $a=10, a=5, a=\\frac{5}{2}$ all fail, but $a=2$ does work. Thus, the real part of both numbers is $2$ , and their sum is $\\boxed{004}$", "Begin by assuming that $x$ and $y$ are roots of some polynomial of the form $w^2+bw+c$ , such that by Vieta's Formulæ and some algebra (left as an exercise to the reader), $b^2-2c=7$ and $3bc-b^3=10$ .\nSubstituting $c=\\frac{b^2-7}{2}$ , we deduce that $b^3-21b-20=0$ , whose roots are $-4$ $-1$ , and $5$ .\nSince $-b$ is the sum of the roots and is maximized when $b=-4$ , the answer is $-(-4)=\\boxed{004}$", "$x^3 + y^3 = 10 = (x+y)(x^2-xy+y^2) = (x+y)(7-xy) \\implies xy = 7 - \\frac{10}{x+y}.$ Also, $(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 = 10 + 3xy(x+y).$ Substituting our above into this, we get $10 + 3(7-\\frac{10}{x+y})(x+y) = 21x+21y-20 = (x+y)^3$ . Letting $p = x+y$ , we have that $p^3 - 21p + 20 = 0$ . Testing $p = 1$ , we find that this is a root, to get $(p-1)(p^2+p-20) = 0 \\implies p = -5, 1, 4 \\implies \\boxed{4}$" ]

[ "We want the cost per person, which is \\begin{align*} \\frac{20\\text{ billion}}{250\\text{ million}} &= \\frac{20000\\text{ million}}{250\\text{ million}} \\\\ &= 80 \\rightarrow \\boxed{80}" ]

[ "We look for a pattern, hoping this sequence either settles down to one number, or that it forms a cycle that repeats.\nAfter $1$ press, the calculator displays $\\frac{1}{1 - 5} = -\\frac{1}{4}$\nAfter $2$ presses, the calculator displays $\\frac{1}{1 - (-\\frac{1}{4})} = \\frac{1}{\\frac{5}{4}} = \\frac{4}{5}$\nAfter $3$ presses, the calculator displays $\\frac{1}{1 - \\frac{4}{5}} = \\frac{1}{\\frac{1}{5}} = 5$\nThus, every three presses, the display will be $5$ . On press $3\\cdot 33 = 99$ , the display will be $5$ . One more press will give $-\\frac{1}{4}$ , which is answer $\\boxed{0.25}$" ]

[ "Rearranging, we find $3x+y=-2x+6y$ , or $5x=5y\\implies x=y$ .\nSubstituting, we can convert the second equation into $\\frac{x+3x}{3x-x}=\\frac{4x}{2x}=\\boxed{2}$", "Substituting each $x$ and $y$ with $1$ , we see that the given equation holds true, as $\\frac{3(1)+1}{1-3(1)} = -2$ . Thus, $\\frac{x+3y}{3x-y}=\\boxed{2}$", "Let $y=ax$ . The first equation converts into $\\frac{(3+a)x}{(1-3a)x}=-2$ , which simplifies to $3+a=-2(1-3a)$ . After a bit of algebra we found out $a=1$ , which means that $x=y$ . Substituting $y=x$ into the second equation it becomes $\\frac{4x}{2x}=\\boxed{2}$ - mathleticguyyy", "Let $x=1$ . Then $y=1$ . So the desired result is $2$ . Select $\\boxed{2}$" ]

[ "Rearranging, we find $3x+y=-2x+6y$ , or $5x=5y\\implies x=y$ .\nSubstituting, we can convert the second equation into $\\frac{x+3x}{3x-x}=\\frac{4x}{2x}=\\boxed{2}$", "Substituting each $x$ and $y$ with $1$ , we see that the given equation holds true, as $\\frac{3(1)+1}{1-3(1)} = -2$ . Thus, $\\frac{x+3y}{3x-y}=\\boxed{2}$", "Let $y=ax$ . The first equation converts into $\\frac{(3+a)x}{(1-3a)x}=-2$ , which simplifies to $3+a=-2(1-3a)$ . After a bit of algebra we found out $a=1$ , which means that $x=y$ . Substituting $y=x$ into the second equation it becomes $\\frac{4x}{2x}=\\boxed{2}$ - mathleticguyyy", "Let $x=1$ . Then $y=1$ . So the desired result is $2$ . Select $\\boxed{2}$" ]

[ "If Susan spent 12 dollars, then twice that much on rides, then she spent $12+12 \\times 2=36$ dollars in total. We subtract $36$ from $50$ to get $\\boxed{14}$" ]

[ "Denote by $x$ $y$ $z$ the amount of $20 bills, $50 bills and $100 bills, respectively.\nThus, we need to find the number of tuples $\\left( x , y, z \\right)$ with $x, y, z \\in \\Bbb N$ that satisfy \\[ 20 x + 50 y + 100 z = 800. \\]\nFirst, this equation can be simplified as \\[ 2 x + 5 y + 10 z = 80. \\]\nSecond, we must have $5 |x$ . Denote $x = 5 x'$ .\nThe above equation can be converted to \\[ 2 x' + y + 2 z = 16 . \\]\nThird, we must have $2 | y$ . Denote $y = 2 y'$ .\nThe above equation can be converted to \\[ x' + y' + z = 8 . \\]\nDenote $x'' = x' - 1$ $y'' = y' - 1$ and $z'' = z - 1$ .\nThus, the above equation can be written as \\[ x'' + y'' + z'' = 5 . \\]\nTherefore, the number of non-negative integer solutions $\\left( x'', y'', z'' \\right)$ is $\\binom{5 + 3 - 1}{3 - 1} = \\boxed{21}$", "We notice that each $100 can be split 3 ways: 5 $20 dollar bills, 2 $50 dollar bills, or 1 $100 dollar bill.\nThere are 8 of these $100 chunks in total--take away 3 as each split must be used at least once.\nNow there are five left--so we use stars and bars.\n5 chunks, 3 categories or 2 bars. This gives us $\\binom{5+2}{2}=\\boxed{21}$" ]

[ "Finding the area of the shaded walkway can be achieved by computing the total area of Tamara's garden and then subtracting the combined area of her six flower beds.\nSince the width of Tamara's garden contains three margins, the total width is $2\\cdot 6+3\\cdot 1 = 15$ feet.\nSimilarly, the height of Tamara's garden is $3\\cdot 2+4\\cdot 1 = 10$ feet.\nTherefore, the total area of the garden is $15\\cdot 10 =150$ square feet.\nFinally, since the six flower beds each have an area of $2\\cdot 6 = 12$ square feet, the area we seek is $150 - 6\\cdot 12$ , and our answer is $\\boxed{78}$", "The long horizontal walkways have an area of $15 ft^{2}$ each, and there are $4$ of them $\\longrightarrow 60 ft^{2}$\nThe short vertical walkways have an area of $2 ft^{2}$ each, and there are $9$ of them $\\longrightarrow 18 ft^{2}$\nAdding the areas together, we have $60 ft^{2} + 18 ft^{2} = 78 ft^{2} \\Longrightarrow \\boxed{78}$ .\n~JH. L" ]

[ "Suppose team $A$ has played $g$ games in total so that it has won $\\frac23g$ games.\nIt follows that team $B$ has played $g+14$ games in total so that it has won $\\frac23g+7$ games.\nWe set up and solve an equation for team $B$ 's win ratio: \\begin{align*} \\frac{\\frac23g+7}{g+14}&=\\frac58 \\\\ \\frac{16}{3}g+56&=5g+70 \\\\ \\frac13g&=14 \\\\ g&=\\boxed{42} ~MRENTHUSIASM", "If we consider the number of games team $B$ has played as $x$ and the number of games that team $A$ has played as $y$ , then we can set up the following system of equations: \\begin{align*} \\frac{5}{8}x &= \\frac{2}{3}y+7, \\\\ \\frac{3}{8}x &= \\frac{1}{3}y+7. \\end{align*} The first system equated the number of wins of each team, while the second system equates the number of losses by each team. By multiplying the second equation by $2$ and solving the system, we get $y = 42$ or answer choice $\\boxed{42}.$", "First, let us assign some variables. Let \\[A_w=2x, \\ A_l=x, \\ A_g=3x,\\] \\[B_w=5y, \\ B_l=3y, \\ B_g=8y,\\] where $X_w$ denotes number of games won, $X_l$ denotes number of games lost, and $X_g$ denotes total games played for $X\\in \\{A, B\\}$ . Using the given information, we can set up the following two equations: \\begin{align*} B_w=A_w+7&\\implies 5y=2x+7, \\\\ B_l=A_l+7&\\implies 3y=x+7. \\end{align*} We can solve through substitution, as the second equation can be written as $x=3y-7$ , and plugging this into the first equation gives $5y=6y-7\\implies y=7$ , which means $x=3(7)-7=14$ . Finally, we want the total number of games team $A$ has played, which is $A_g=3(14)=\\boxed{42}$", "Using the information from the problem, we can note that team $A$ has lost $\\frac{1}{3}$ of their matches. Using the answer choices, we can construct the following list of possible win-lose scenarios for $A,$ represented in the form $(w, l)$ for convenience: \\begin{align*} \\textbf{(A)} &\\implies (14, 7) \\\\ \\textbf{(B)} &\\implies (18, 9) \\\\ \\textbf{(C)} &\\implies (28, 14) \\\\ \\textbf{(D)} &\\implies (32, 16) \\\\ \\textbf{(E)} &\\implies (42, 21) \\end{align*} Thus, we have $5$ matching $B$ scenarios, simply adding $7$ to $w$ and $l.$ We can then test each of the five $B$ scenarios for $\\frac{w}{w+l} = \\frac{5}{8}$ and find that $(35, 21)$ fits this description. Then working backwards and subtracting $7$ from $w$ and $l$ gives us the point $(28, 14),$ making the answer $\\boxed{42}.$", "Let's say that team $A$ plays $n$ games in total. Therefore, team $B$ must play $n + 14$ games in total (7 wins, 7 losses) Since the ratio of $A$ is \\[\\frac{2}{3} \\implies n \\equiv 0 \\pmod{3}\\] Similarly, since the ratio of $B$ is \\[\\frac{5}{8} \\implies n + 14 \\equiv 0 \\pmod{8}\\] Now, we can go through the answer choices and see which ones work: \\begin{align*} \\textbf{(A) } 21 &\\implies 21 + 14 = 35 \\not \\equiv 0\\pmod{8} \\\\ \\textbf{(B) } 27 &\\implies 27 + 14 = 41 \\not \\equiv 0\\pmod{8} \\\\ \\textbf{(C) } 42 &\\implies 42 + 14 = 56 \\equiv 0\\pmod{8} \\\\ \\textbf{(D) } 48 &\\implies 48 + 14 = 62 \\not \\equiv 0\\pmod{8} \\\\ \\textbf{(E) } 63 &\\implies 63 + 14 = 77 \\not \\equiv 0\\pmod{8} \\\\ \\end{align*} So we can see $\\boxed{42}$ is the only valid answer." ]

[ "We are given that \\[2^m\\cdot\\sqrt{\\frac{1}{4096}} = 2\\cdot\\sqrt[m]{\\frac{1}{4096}}.\\] Converting everything into powers of $2,$ we have \\begin{align*} 2^m\\cdot(2^{-12})^{\\frac12} &= 2\\cdot (2^{-12})^{\\frac1m} \\\\ 2^{m-6} &= 2^{1-\\frac{12}{m}} \\\\ m-6 &= 1-\\frac{12}{m}. \\end{align*} We multiply both sides by $m$ , then rearrange as \\[m^2-7m+12=0.\\] By Vieta's Formulas, the sum of such values of $m$ is $\\boxed{7}.$", "We can rewrite the equation using fractional exponents and take logarithms of both sides: \\[\\log_2{(2^{m}\\cdot4096^{-1/2}}) = \\log_2{(2\\cdot4096^{-1/m})}.\\] We can then use the additive properties of logarithms to split them up: \\[\\log_2{(2^{m})} + \\log_2{(4096^{-1/2})} = \\log_2{2} + \\log_2{(4096^{-1/m})}.\\] Using the power rule, the fact that $4096 = 2^{12},$ and bringing the exponents down, we get \\begin{align*} m - 6 &= 1 - \\frac{12}{m} \\\\ m + \\frac{12}{m} &= 7 \\\\ m^{2} + 12 &= 7m \\\\ m^{2} - 7m + 12 &= 0 \\\\ (m-3)(m-4) &= 0, \\end{align*} from which $m = 3$ or $m = 4$ . Therefore, the answer is $3+4 = \\boxed{7}.$", "Since surd roots are conventionally positive integers, assume $m$ is an integer, so $m$ can only be $1$ $2$ $3$ $4$ $6$ , and $12$ $\\sqrt{\\frac{1}{4096}}=\\frac{1}{64}$ . Testing out $m$ , we see that only $3$ and $4$ work. Hence, $3+4=\\boxed{7}$" ]

[ "We use that fact that $d=rt$ . Let d= distance, r= rate or speed, and t=time. In this case, let $x$ represent the time.\nOn Monday, he was at a rate of $5 \\text{ m.p.h}$ . So, $5x = 2 \\text{ miles}\\implies x = \\frac{2}{5} \\text { hours}$\nFor Wednesday, he walked at a rate of $3 \\text{ m.p.h}$ . Therefore, $3x = 2 \\text{ miles}\\implies x = \\frac{2}{3} \\text { hours}$\nOn Friday, he walked at a rate of $4 \\text{ m.p.h}$ . So, $4x = 2 \\text{ miles}\\implies x=\\frac{2}{4}=\\frac{1}{2} \\text {hours}$\nAdding up the hours yields $\\frac{2}{5} \\text { hours}$ $\\frac{2}{3} \\text { hours}$ $\\frac{1}{2} \\text { hours}$ $\\frac{47}{30} \\text { hours}$\nWe now find the amount of time Grandfather would have taken if he walked at $4 \\text{ m.p.h}$ per day. Set up the equation, $4x = 2 \\text{ miles} \\times 3 \\text{ days}\\implies x = \\frac{3}{2} \\text { hours}$\nTo find the amount of time saved, subtract the two amounts: $\\frac{47}{30} \\text { hours}$ $\\frac{3}{2} \\text { hours}$ $\\frac{1}{15} \\text { hours}$ . To convert this to minutes, we multiply by $60$\nThus, the solution to this problem is $\\dfrac{1}{15}\\times 60=\\boxed{4}$" ]

[ "Label the left shoes be $L_1,\\dots, L_{10}$ and the right shoes $R_1,\\dots, R_{10}$ . Notice that there are $10!$ possible pairings.\nLet a pairing be \"bad\" if it violates the stated condition. We would like a better condition to determine if a given pairing is bad.\nNote that, in order to have a bad pairing, there must exist a collection of $k<5$ pairs that includes both the left and the right shoes of $k$ adults; in other words, it is bad if it is possible to pick $k$ pairs and properly redistribute all of its shoes to exactly $k$ people.\nThus, if a left shoe is a part of a bad collection, its corresponding right shoe must also be in the bad collection (and vice versa). To search for bad collections, we can start at an arbitrary right shoe (say $R_1$ ), check the left shoe it is paired with (say $L_i$ ), and from the previous observation, we know that $R_i$ must also be in the bad collection. Then we may check the left shoe paired with $R_i$ , find its counterpart, check its left pair, find its counterpart, etc. until we have found $L_1$ . We can imagine each right shoe \"sending\" us to another right shoe (via its paired left shoe) until we reach the starting right shoe, at which point we know that we have found a bad collection if we have done this less than $5$ times.\nEffectively we have just traversed a cycle. (Note: This is the cycle notation of permutations.) The only condition for a bad pairing is that there is a cycle with length less than $5$ ; thus, we need to count pairings where every cycle has length at least $5$ . This is only possible if there is a single cycle of length $10$ or two cycles of length $5$\nThe first case yields $9!$ working pairings. The second case yields $\\frac{{10\\choose 5}}{2}\\cdot{4!}^2=\\frac{10!}{2 \\cdot {5!}^2} \\cdot {4!}^2$ pairings. Therefore, taking these cases out of a total of $10!$ , the probability is $\\frac{1}{10}+\\frac{1}{50} = \\frac{3}{25}$ , for an answer of $\\boxed{028}$" ]

[ "We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs. There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent. There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10 \\cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have all $5$ adjacent, $10 \\cdot 4$ or $40$ have $4$ adjacent, and $10 \\cdot {5\\choose 2}$ or $100$ have $3$ adjacent. With $6$ chairs in the subset, $10$ have all $6$ adjacent, $10(3)$ or $30$ have $5$ adjacent, $10 \\cdot {4\\choose2}$ or $60$ have $4$ adjacent, $\\frac{10 \\cdot 3}{2}$ or $15$ have $2$ groups of $3$ adjacent chairs, and $10 \\cdot \\left({5\\choose2} - 3\\right)$ or $70$ have $1$ group of $3$ adjacent chairs. All possible subsets with more than $6$ chairs have at least $1$ group of $3$ adjacent chairs, so we add ${10\\choose7}$ or $120$ ${10\\choose8}$ or $45$ ${10\\choose9}$ or $10$ , and ${10\\choose10}$ or $1.$ Adding, we get $10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = \\boxed{581}.$", "Starting with small cases, we see that four chairs give $4 + 1 = 5$ , five chairs give $5 + 5 + 1 = 11$ , and six chairs give $6 + 6 + 6 + 6 + 1 = 25.$ Thus, n chairs should give $n 2^{n-4} + 1$ , as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are $n 2^{n-3}$ ways to arrange $3$ adjacent chairs, but then we subtract $n 2^{n-4}$ ways to arrange $4.$ Finally, we add $1$ to account for the full subset of chairs. Thus, for $n = 10$ we get a first count of $641.$\nHowever, we overcount cases in which there are two distinct groups of three or more chairs. We have $5$ cases for two groups of $3$ directly opposite each other, $5$ for two groups of four, $20$ for two groups of $3$ not symmetrically opposite, $20$ for a group of $3$ and a group of $4$ , and $10$ for a group of $3$ and a group of $5.$ Thus, we have $641 - 60 = \\boxed{581}$", "It is possible to use recursion to count the complement. Number the chairs $1, 2, 3, ..., 10.$ If chair $1$ is not occupied, then we have a line of $9$ chairs such that there is no consecutive group of three. If chair $1$ is occupied, then we split into more cases. If chairs $2$ and $10$ are empty, then we have a line of $7.$ If chair $2$ is empty but chair $10$ is occupied, then we have a line of $6$ chairs (because chair $9$ cannot be occupied); this is similar to when chair $2$ is occupied and chair $10$ is empty. Finally, chairs $2$ and $10$ cannot be simultaneously occupied. Thus, we have reduced the problem down to computing $T_9 + T_7 + 2T_6$ , where $T_n$ counts the ways to select a subset of chairs $\\textit{in a line}$ from a group of n chairs such that there is no group of $3$ chairs in a row.\nNow, we notice that $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ (representing the cases when the first, second, and/or third chair is unoccupied). Also, $T_0 = 1, T_1 = 2, T_2 = 4, T_3 = 7$ , and hence $T_4 = 13, T_5 = 24, T_6 = 44, T_7 = 81, T_8 = 149, T_9 = 274$ . Now we know the complement is $274 + 81 + 88 = 443$ , and subtracting from $2^{10} = 1024$ gives $1024 - 443 = \\boxed{581}$" ]

[ "For the first man, there are $10$ possible seats. For each subsequent man, there are $4$ $3$ $2$ , or $1$ possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is $10\\cdot 4\\cdot 3\\cdot 2\\cdot 1\\cdot 2 = \\boxed{480}$", "Label the seats ABCDEFGHIJ, where A is the top seat. The first man has $10$ possible seats. WLOG, assume he is in seat A in the diagram. Then, his wife can be in one of two seats, namely D or H. WLOG, assume she is in seat D. Now, in each structurally distinct solution we find, we know that there are $4! = 24$ ways to arrange the 4 other couples. Let there be x structurally distinct solutions under these conditions. We know the answer must be $10\\cdot 2\\cdot 24\\cdot x = 480x$ possible seating arrangements, and x is a nonnegative integer. There is only one answer that is a multiple of $480$ . So, our answer is $\\boxed{480}$" ]

[ "Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:\n\\begin{align*}3a + 4b + 6c &= 41\\\\ a + b + c &= 10\\end{align*}\nSubtracting 3 times the second from the first gives $b + 3c = 11$ , or $(b,c) = (2,3),(5,2),(8,1),(11,0)$ . The last doesn't work, obviously. This gives the three solutions $(a,b,c) = (5,2,3),(3,5,2),(1,8,1)$ . In terms of choosing which goes where, the first two solutions are analogous.\nFor $(5,2,3),(3,5,2)$ , we see that there are $2\\cdot\\dfrac{10!}{5!2!3!} = 10\\cdot9\\cdot8\\cdot7$ ways to stack the crates. For $(1,8,1)$ , there are $\\dfrac{10!}{8!1!1!} = 90$ . Also, there are $3^{10}$ total ways to stack the crates to any height.\nThus, our probability is $\\dfrac{10\\cdot9\\cdot8\\cdot7 + 90}{3^{10}} = \\dfrac{10\\cdot8\\cdot7 + 10}{3^{8}} = \\dfrac{570}{3^8} = \\dfrac{190}{3^{7}}$ . Our answer is the numerator, $\\boxed{190}$", "It would be helpful for this solution to be reformatted. To start with, let us observe the three numbers. Note that $3$ and $6$ are both divisible by $3$ , so the number of $4$ -crates must be congruent to $41\\bmod{3}$ , which is also congruent to $2\\bmod{3}$ . Our solutions for the number of $4$ -crates will repeat mod $3$ , so if $x$ is a solution, so is $x+3$ . By inspection, we have that $2$ is solution, and so are $5$ and $8$ . Each solution splits into its own case.We must solve the equation $41-4z=6x+3y$ , simultaneously with $x+y=10-z$ . Note that we already know the possible values of $z$ . Solving these (it's AIME $9$ , you should be able to do this and if anyone feels like they want to write a rundown of this please go ahead), we get the solution sets $\\{8,1,1\\},\\{5,2,3\\},$ and $\\{2,3,5\\}$ . We can count the number of possible arrangements for each solution by taking $\\dbinom{10}{z}$ and then multiplying by $\\dbinom{10-z}{x}$ (the solution sets, for the sake of consistency, are in the form $z,x,y$ ). Summing the results for all the solutions gives us $5130$ . Finally, to calculate the probability we must determine our denominator. Since we have $3$ ways to arrange each block, our denominator is $3^{10}$ $\\frac{5130}{3^{10}}=\\frac{190}{3^7}$ . The answer is $m=\\boxed{190}$", "Let's make two observations. We are trying to find the number of ways we can add $3\\text{s}, 4\\text{s}$ , and $6\\text{s}$ to get $41$ , and the total number of (non-distinct) sums possible is $3^{10}$ .\nThen we just use casework to easily and directly solve for the number of ways to get $41$ . To begin, the minimum sum is produced with $10$ threes, so WLOG we can solve for the number of ways to get $11$ with $0\\text{s}, 1\\text{s}$ , and $3\\text{s}$\nCase I: $0$ zeroes, $0$ threes, $11$ ones\nImpossible, because there are only ten available spots.\nCase II: $1$ zero, $1$ three, $8$ ones\nThis is just $\\frac{10!}{8!}$ , so there are $90$ possibilities.\nCase III: $3$ zeroes, $2$ threes, $5$ ones\nThis is just $\\frac{10!}{3!2!5!}$ . This gives $2520$ possibilities.\nCase IV: 5 zeroes, 3 threes, and 2 ones.\nThis is the same as case $3$ , so also $2520$ possibilities.\n$90+2520+2520=5130$\n$5130$ has three powers of $3$ , so $5130$ divided by $27$ is $\\boxed{190}$", "Note we are placing 10 crates where each \"height\" is 3, 4, 6 and we want all the heights to sum to 41. \nWe can model this as the generating function \\[\\left(x^3+x^4+x^6\\right)^{10}\\] where we want the coefficient of $x^{41}.$ First off, factor this to get \\[{x^3}^{10}\\left(1+x+x^3\\right)^{10}\\] and then see that we want the coefficient of $x^{11}$ in $\\left(1+x+x^3\\right)^{10}.$ From multinomial theorem, this expansion is \\[\\sum_{a+b+c=10}\\binom{10}{a,b,c}1^ax^bx^{3c}\\] If we want the coefficient of $x^{11}$ then we need $b + 3c = 11.$ with $b + c \\le 10$ (from the multinomial expansion).\nThis has the solutions $(b, c) = \\{8, 1\\}, \\{5, 2\\}, \\{2, 3\\}.$ Note that the denominator of the answer is just $3^{10}$ since there are 3 ways to orientate every crate and there are 10 creates. Thus, our answer is \\[\\frac{\\binom{10}{8,1,1} + \\binom{10}{5,2,3} + \\binom{10}{2,3,5}}{3^{10}} = \\frac{190}{3^7} \\rightarrow \\boxed{190}\\]" ]

[ "For $i\\in\\{1,2,3,\\ldots,10\\},$ suppose Person $i$ picks the number $a_i$ and announces the number $i.$ We wish to find $a_6.$\nTaking the indices modulo $10,$ we are given that $\\frac{a_{i-1}+a_{i+1}}{2}=i,$ from which $a_{i-1}+a_{i+1}=2i.$\nWe have ten equations: five with odd-numbered indices and five with even-numbered indices. Note that these two sets of equations are independent. The set that involves $a_6$ is \\begin{align*} a_2 + a_4 & = 6, &&(1) \\\\ a_4 + a_6 & = 10, &&(2) \\\\ a_6 + a_8 & = 14, &&(3) \\\\ a_8 + a_{10} & = 18, &&(4) \\\\ a_{10} + a_2 & = 2. &&(5) \\end{align*} Summing these five equations, we get $2(a_2 + a_4 + a_6 + a_8 + a_{10}) = 50,$ from which \\[a_2 + a_4 + a_6 + a_8 + a_{10} = 25. \\hspace{15mm} (\\bigstar)\\] Subtracting $(1)+(4)$ from $(\\bigstar),$ we obtain $a_6=\\boxed{1}.$", "For $i\\in\\{1,2,3,\\ldots,10\\},$ suppose Person $i$ announces the number $i.$\nLet $x$ be the number picked by Person $6.$ We construct the following table: \\[\\begin{array}{c|c|c||l} & & & \\\\ [-2.5ex] \\textbf{People} & \\textbf{Average of \\#s Picked} & \\textbf{Sum of \\#s Picked} & \\multicolumn{1}{c}{\\textbf{Conclusion}} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 6\\text{ and }8 & 7 & 14 & \\text{Person 8 picks } 14-x. \\\\ 8\\text{ and }10 & 9 & 18 & \\text{Person 10 picks } 4+x. \\\\ 10\\text{ and }2 & 1 & 2 & \\text{Person 2 picks } -2-x \\\\ 2\\text{ and }4 & 3 & 6 & \\text{Person 4 picks } 8+x \\\\ 4\\text{ and }6 & 5 & 10 & \\text{Person 6 picks } 2-x \\\\ \\end{array}\\] We have $x=2-x,$ from which $x=\\boxed{1}.$" ]

[ "Any subset of the ten points with three or more members can be made into exactly one such polygon. Thus, we need to count the number of such subsets. There are $2^{10} = 1024$ total subsets of a ten-member set , but of these ${10 \\choose 0} = 1$ have 0 members, ${10 \\choose 1} = 10$ have 1 member and ${10 \\choose 2} = 45$ have 2 members. Thus the answer is $1024 - 1 - 10 - 45 = \\boxed{968}$" ]

[ "Note that 4 points can NEVER form 2 triangles. Therefore, we just need to multiply the probability that the first three segments picked form a triangle by 4. We can pick any segment for the first choice, then only segments that share an endpoint with the first one, then the one segment that completes the triangle. Note that the fourth segment doesn't matter in this case.\nNote that there are $(9 - 1) \\times 2 = 16$ segments that share an endpoint with the first segment. The answer is then $4 \\times \\frac{16}{44} \\times \\frac{1}{43} = \\frac{16}{11} \\times \\frac{1}{43} = \\frac{16}{473} \\implies m + n = \\boxed{489}$ -whatRthose", "Instead of working with the four segments, let's focus on their endpoints. When we select these segments, we are working with $4, 5, 6, 7,$ or $8$ endpoints in total.\nIf we have $6, 7,$ or $8$ endpoints, it is easy to see that we cannot form a triangle by drawing four segments between them, because at least one point will be \"left out\".\nHowever, if we have $5$ endpoints, we can use three segments form a triangle with three of the points and connect the remaining two points with the last segment. There are ${10\\choose5}$ ways to select these $5$ points from the original $10,$ and ${5\\choose3}$ ways to decide which three points are in the triangle.\nFinally, if we have $4$ endpoints, we can also form a triangle with three of the points, then use the remaining segment to connect the last point to either of the previous three. We have ${10\\choose4}$ ways to select the $4$ points and ${4\\choose3}$ ways to choose three points for the triangle. Finally, we must connect the last point to one vertex of the triangle; we can do this in $3$ ways.\nAs in Solution 1, there are ${45\\choose4}$ total ways to select four segments. So, our desired probability is\n\\[\\dfrac{{10\\choose5}{5\\choose3}+{10\\choose4}{4\\choose3}\\cdot 3}{{45\\choose4}}=\\dfrac{16}{473} \\implies m + n = \\boxed{489}.\\]" ]

[ "Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is \\[\\binom{10}{0} + \\binom{9}{1} + \\binom{8}{2} + \\binom{7}{3} + \\binom{6}{4} + \\binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \\boxed{89}.\\]", "Recall that the number of ways to arrange 1x1 and 2x1 blocks to form a 1xn rectangle results in Fibonacci numbers . Clearly, $\\boxed{89}$ is the only fibonacci number, so no calculation is needed for this problem.", "Let $S_n$ be the number of possible seating arrangements with $n$ women. Consider $n \\ge 3,$ and focus on the rightmost woman. If she returns back to her seat, then there are $S_{n-1}$ ways to seat the remaining $n-1$ women. If she sits in the second to last seat, then the woman who previously sat there must now sit at the rightmost seat. This gives us $S_{n-2}$ ways to seat the other $n-2$ women, so we obtain the recursion \\[S_n = S_{n-1}+S_{n-2}.\\]\nStarting with $S_1=1$ and $S_2=2,$ we can calculate $S_{10}=\\boxed{89}.$" ]

[ "We could start by looking for a pattern.\n$98, 49, 44, 22, 11, 6, 54, 27, 22, 11, 6, \\ldots .$\nFrom here, we see that we have a pattern of $22, 11, 6, 54, 27, \\ldots .$ after $98, 49, 44$\nOur problem is now really\nFind the $95^\\text{th}$ term of the sequence that goes $22, 11, 6, 54, 27, 22, 11, 6, 54, 27, 22, \\ldots .$\nThere are 5 terms in each repetition of the pattern, and $95\\equiv0\\pmod{5}$ , so the answer is $\\boxed{27}$" ]

[ "Let $M$ and $N$ be midpoints of $\\overline{AB}$ and $\\overline{CD}$ . The given conditions imply that $\\triangle ABD\\cong\\triangle BAC$ and $\\triangle CDA\\cong\\triangle DCB$ , and therefore $MC=MD$ and $NA=NB$ . It follows that $M$ and $N$ both lie on the common perpendicular bisector of $\\overline{AB}$ and $\\overline{CD}$ , and thus line $MN$ is that common perpendicular bisector. Points $B$ and $C$ are symmetric to $A$ and $D$ with respect to line $MN$ . If $X$ is a point in space and $X'$ is the point symmetric to $X$ with respect to line $MN$ , then $BX=AX'$ and $CX=DX'$ , so $f(X) = AX+AX'+DX+DX'$\nLet $Q$ be the intersection of $\\overline{XX'}$ and $\\overline{MN}$ . Then $AX+AX'\\geq 2AQ$ , from which it follows that $f(X) \\geq 2(AQ+DQ) = f(Q)$ . It remains to minimize $f(Q)$ as $Q$ moves along $\\overline{MN}$\nAllow $D$ to rotate about $\\overline{MN}$ to point $D'$ in the plane $AMN$ on the side of $\\overline{MN}$ opposite $A$ . Because $\\angle DNM$ is a right angle, $D'N=DN$ . It then follows that $f(Q) = 2(AQ+D'Q)\\geq 2AD'$ , and equality occurs when $Q$ is the intersection of $\\overline{AD'}$ and $\\overline{MN}$ . Thus $\\min f(Q) = 2AD'$ . Because $\\overline{MD}$ is the median of $\\triangle ADB$ , the Length of Median Formula shows that $4MD^2 = 2AD^2 + 2BD^2 - AB^2 = 2\\cdot 28^2 + 2 \\cdot 44^2 - 52^2$ and $MD^2 = 684$ . By the Pythagorean Theorem $MN^2 = MD^2 - ND^2 = 8$\nBecause $\\angle AMN$ and $\\angle D'NM$ are right angles, \\[(AD')^2 = (AM+D'N)^2 + MN^2 = (2AM)^2 + MN^2 = 52^2 + 8 = 4\\cdot 678.\\] It follows that $\\min f(Q) = 2AD' = 4\\sqrt{678}$ . The requested sum is $4+678=\\boxed{682}$" ]

[ "Within the conference, there are 8 teams, so there are $\\dbinom{8}{2}=28$ pairings of teams, and each pair must play two games, for a total of $28\\cdot 2=56$ games within the conference.\nEach team also plays 4 games outside the conference, and there are 8 teams, so there are a total of $4\\cdot 8 =32$ games outside the conference.\nTherefore, the total number of games is $56+32 = \\boxed{88}$ is our answer." ]

[ "Suppose we undo each of the four folds, considering just the top square until we completely unfold the paper. $x$ will be marked in the square if the face that shows after all the folds is face up, $y$ if that face is facing down.\nStep 0: Step 1: Step 2: Step 3: Step 4: The marked square is in the same spot as the number $9\\rightarrow \\boxed{9}$" ]

[ "Sarah is leaving $3$ questions unanswered, guaranteeing her $3 \\times 1.5 = 4.5$ points. She will either get $6$ points or $0$ points for the rest of the questions. Let $x$ be the number of questions Sarah answers correctly. \\begin{align*} 6x+4.5 &\\ge 100\\\\ 6x &\\ge 95.5\\\\ x &\\ge 15.92 \\end{align*} The number of questions she answers correctly has to be a whole number, so round up to get $\\boxed{16}$" ]

[ "Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by $5$ is the total value per row. The sum of the $25$ integers is $-10+-9+...+14=11+12+13+14=50$ , and the common sum is $\\frac{50}{5}=\\boxed{10}$", "Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get $0 + 1 + 2 + 3 + 4 = \\boxed{10}$ as our answer.\n~Baolan", "Taking the average of the first and last terms, $-10$ and $14$ , we have that the mean of the set is $2$ . There are 5 values in each row, column or diagonal, so the value of the common sum is $5\\cdot2$ , or $\\boxed{10}$ . \n~Arctic_Bunny, edited by KINGLOGIC", "Let us consider the horizontal rows. Since there are five of them, each with constant sum $x$ , we can add up the 25 numbers in 5 rows for a sum of $5x$ . Since the sum of the 25 numbers used is $-10-9-8-\\cdots{}+12+13+14+15=11+12+13+14+15=50$ $5x=50$ and $x=\\boxed{10}$ . \n~cw357", "The mean of the set of numbers is $(14-10) \\div 2 = 2$ . The numbers around it must be equal (i.e. if the mean of $1$ $2$ $3$ $4$ , and $5$ is $3$ , then $2+4=1+5$ .)\nOne row of the square would be \\[\\square \\square 2 \\square \\square\\]\nAdding the numbers would be\n\\[0, 1, 2, 3, 4\\]\nwith a sum of $\\boxed{10}$" ]

[ "If the sum of each row, column, and diagonal is x, then we have a total of 12x for the sum. The sum of the rows and columns is the sum of all the numbers doubled, which is $50\\cdot2=100$ . Therefore $100+2x=12x$ $100=10x$ , and $x=\\boxed{10}$ .\n~MC413551" ]

[ "Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by $5$ is the total value per row. The sum of the $25$ integers is $-10+-9+...+14=11+12+13+14=50$ , and the common sum is $\\frac{50}{5}=\\boxed{10}$", "Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get $0 + 1 + 2 + 3 + 4 = \\boxed{10}$ as our answer.\n~Baolan", "Taking the average of the first and last terms, $-10$ and $14$ , we have that the mean of the set is $2$ . There are 5 values in each row, column or diagonal, so the value of the common sum is $5\\cdot2$ , or $\\boxed{10}$ . \n~Arctic_Bunny, edited by KINGLOGIC", "Let us consider the horizontal rows. Since there are five of them, each with constant sum $x$ , we can add up the 25 numbers in 5 rows for a sum of $5x$ . Since the sum of the 25 numbers used is $-10-9-8-\\cdots{}+12+13+14+15=11+12+13+14+15=50$ $5x=50$ and $x=\\boxed{10}$ . \n~cw357", "The mean of the set of numbers is $(14-10) \\div 2 = 2$ . The numbers around it must be equal (i.e. if the mean of $1$ $2$ $3$ $4$ , and $5$ is $3$ , then $2+4=1+5$ .)\nOne row of the square would be \\[\\square \\square 2 \\square \\square\\]\nAdding the numbers would be\n\\[0, 1, 2, 3, 4\\]\nwith a sum of $\\boxed{10}$" ]

[ "If the sum of each row, column, and diagonal is x, then we have a total of 12x for the sum. The sum of the rows and columns is the sum of all the numbers doubled, which is $50\\cdot2=100$ . Therefore $100+2x=12x$ $100=10x$ , and $x=\\boxed{10}$ .\n~MC413551" ]

[ "We use the property that the digits of a number must sum to a multiple of $9$ if it are divisible by $9$ . This means $2+0+1+8+U$ must be divisible by $9$ . The only possible value for $U$ then must be $7$ . Since we are looking for the remainder when divided by $8$ , we can ignore the thousands. The remainder when $187$ is divided by $8$ is $\\boxed{3}$" ]

[ "Once the two cards are drawn, there are $\\dbinom{50}{2} = 1225$ ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below $a$ , which occurs in $\\dbinom{a-1}{2}$ ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above $a+9$ , which occurs in $\\dbinom{43-a}{2}$ ways. Thus, \\[p(a)=\\frac{\\dbinom{43-a}{2}+\\dbinom{a-1}{2}}{1225}.\\] Simplifying, we get $p(a)=\\frac{(43-a)(42-a)+(a-1)(a-2)}{2\\cdot1225}$ , so we need $(43-a)(42-a)+(a-1)(a-2)\\ge (1225)$ . If $a=22+b$ , then \\begin{align*}(43-a)(42-a)+(a-1)(a-2)&=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\\ge (1225) \\\\ b^2\\ge \\frac{385}{2} &= 192.5 >13^2 \\end{align*} So $b> 13$ or $b< -13$ , and $a=22+b<9$ or $a>35$ , so $a=8$ or $a=36$ . Thus, $p(8) = \\frac{616}{1225} = \\frac{88}{175}$ , and the answer is $88+175 = \\boxed{263}$" ]

[ "Since there are five types of squares: $1 \\times 1, 2 \\times 2, 3 \\times 3, 4 \\times 4,$ and $5 \\times 5.$ We must find how many of each square contain the black shaded square in the center.\nIf we list them, we get that\nThus, the answer is $1+4+9+4+1=19\\Rightarrow\\boxed{19}$", "We use complementary counting. There are only $2\\times2$ and $1\\times1$ squares that do not contain the black square. Counting, there are $12$ $2\\times2$ squares, and $25-1 = 24$ $1\\times1$ squares that do not contain the black square. That gives $12+24=36$ squares that don't contain it. There are a total of $25+16+9+4+1 = 55$ squares possible $(25$ $1\\times1$ squares $16$ $2\\times2$ squares $9$ $3\\times3$ squares $4$ $4\\times4$ squares and $1$ $5\\times5$ square), therefore there are $55-36 = 19$ squares that contain the black square, which is $\\boxed{19}$" ]

[ "The number $\\mod{3}$ is congruent to sum of a number's digits $\\mod{3}$ is congruent to the number $\\mod{3}$ $74A52B1 \\pmod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 \\pmod{3}$ is also congruent to 0. $7+4+5+2+1 \\equiv 1 \\pmod{3}$ , so $A+B\\equiv 2 \\pmod{3}$ . As we know, $326AB4C\\equiv 0 \\pmod{3}$ , so $3+2+6+A+B+4+C =15+A+B+C\\equiv 0 \\pmod{3}$ , and therefore $A+B+C\\equiv 0 \\pmod{3}$ . We can substitute 2 for $A+B$ , so $2+C\\equiv 0 \\pmod{3}$ , and therefore $C\\equiv 1\\pmod{3}$ . This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is $\\boxed{1}$", "Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. $7 + 4 + 5 + 2 + 1 = 19$ . To be a multiple of $3$ $A + B$ has to be either $2$ or $5$ or $8$ ... and so on. We add up the numerical digits in the second number; $3 + 2 + 6 + 4 = 15$ . We then add two of the selected values, $5$ to $15$ , to get $20$ . We then see that C = $1, 4$ or $7, 10$ ... and so on, otherwise the number will not be divisible by three. We then add $8$ to $15$ , to get $23$ , which shows us that C = $1$ or $4$ or $7$ ... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1, 4,$ and $7$ . However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$ , but there is a $1$ , so $\\boxed{1}$ is our answer." ]

[ "Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\\cdot8=144$ . This means the square will have four sides of length 12. The only way to do this is shown below.\n\nAs you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \\frac{12}{2} = \\boxed{6}$", "As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.\n\nAs you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$\n\nFrom there we can say $3y = 18$ so $y = \\frac{18}{3} = \\boxed{6}$", "Because the two hexagons are congruent, we know that the perpendicular line to $A$ is half of $BC$ , or $4$ . Next, we plug the answer choices in to see which one works. Trying $A$ , we get the area of one hexagon is $72$ , as desired, so the answer is $\\boxed{6}$" ]

[ "Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\\cdot8=144$ . This means the square will have four sides of length 12. The only way to do this is shown below.\n\nAs you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \\frac{12}{2} = \\boxed{6}$", "As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.\n\nAs you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$\n\nFrom there we can say $3y = 18$ so $y = \\frac{18}{3} = \\boxed{6}$", "Because the two hexagons are congruent, we know that the perpendicular line to $A$ is half of $BC$ , or $4$ . Next, we plug the answer choices in to see which one works. Trying $A$ , we get the area of one hexagon is $72$ , as desired, so the answer is $\\boxed{6}$" ]

[ "Let us consider the number of points for a certain $x$ -coordinate. For any $x$ , the viable points are in the range $[-20 + |x|, 20 - |x|]$ . This means that our total sum is equal to \\begin{align*} 1 + 3 + 5 + \\cdots + 41 + 39 + 37 + \\cdots + 1 &= (1 + 3 + 5 + \\cdots + 39) + (1 + 3 + 5 + \\cdots + 41) \\\\ & = 20^2 + 21^2 \\\\ & = 29^2 \\\\ &= \\boxed{841} ~mathboy100", "Since the second point is the origin, this is equivalent to finding all points $(x, y)$ such that $|x| + |y| \\leq 20$ . Due to the absolute values, the set of all such points will be symmetric about the origin meaning we can focus on the first quadrant and multiply by $4$\nTo avoid overcounts, ignore points on the axes. This means $x, y > 0$ . If $x = 1$ , there are $19$ solutions for $y$ $y = 1, 2, 3, \\ldots, 19$ ). If $x = 2$ , there are $18$ solutions. This pattern repeats until $x = 19$ , at which point there is $1$ solution for $y$\nSo we get $19 + 18 + 17 + \\cdots + 1 = \\frac{19(20)}{2} = 190$ points in the first quadrant. Multiplying by $4$ gives $760$ . Now, the $x$ axis has $y = 0$ which gives $|x| \\leq 20$ , meaning there are $41$ solutions. This is the same with the $y$ axis, but we overcounted the origin by $1$\nOur final answer is $760 + 41 + 41 - 1 = \\boxed{841}$", "This solution refers to the Diagram section. The problem can be visualized as depicted on the right split equally into four \"triangular\" parts excluding the origin. The \"triangular\" parts are identical the ones that would be used in a visual proof of the formula for triangular numbers. Becuase of this the number of points in each part is equal to $\\frac{n(n+1)}{2}$ where $n$ is the length of a \"leg\" of the \"triangle\" which is $20$ for this problem. Substituting and computing, we get $210.$ Multiplying by $4$ and adding $1$ to account for all parts and the origin, we get $210\\cdot4 + 1 = \\boxed{841}.$", "This solution refers to the Diagram section.\nAs shown below, the taxicab distance between each red point and the origin is even, and the taxicab distance between each blue point and the origin is odd. Note that the red array consists of $21^2=441$ points, and the blue array consists of $20^2=400$ points.\nTogether, the answer is $441+400=\\boxed{841}.$", "Let $P = (x, y)$ . Since the problem asks for taxicab distances from the origin, we want $|x| + |y| \\le 20$ . The graph of all solutions to this equation on the $xy$ -plane is a square with vertices at $(0, \\pm 20)$ and $(\\pm 20, 0)$ (In order to prove this, one can divide the sections of this graph into casework on the four quadrants, and tie together the resulting branches.) We want the number of lattice points on the border of the square and inside the square. \nEach side of the square goes through an equal number of lattice points, so if we focus on one side going from $(0,20)$ to $(20, 0)$ , we can see that it goes through $21$ points in total. In addition, each of the vertices gets counted twice, so the total number of border points is $21\\cdot4 - 4 = 80$ . Also, the area of the square is $800$ , so when we plug this information inside Pick's theorem, we get $800 = i + \\frac{80}{2} - 1 \\implies i = 761$ . Then our answer is $761+80 = \\boxed{841}.$", "Instead of considering all points with integer coordinates, first consider points with nonnegative coordinates only. Then, we want $x + y \\le 20$ where $x$ and $y$ are nonnegative integers. We can introduce a third variable, say $z$ , such that $z = 20 - (x + y)$ . Note that counting the number of ways to have $x + y + z = 20$ is the same as counting the number of ways to have $x + y \\le 20$ . Therefore, by stars and bars, there are $\\dbinom{20 + 3 - 1}{3 - 1} = 231$ solutions with nonnegative integer coordinates.\nThen, we can copy our solutions over to the other four quadrants. First, so as not to overcount, we remove all points on the axes. There are $20 + 20 + 1 = 41$ such points with nonnegative integer coordinates. We multiply the $190$ remaining points by $4$ to get $760$ points that are not on the axes. Then, we can add back the $41$ nonnegative points on the axes, as well as the $40$ other points on the negative axes to get $760 + 41 + 40 = \\boxed{841}.$" ]

[ " Note that if the $y$ -intercepts have a sum of $0$ , the distance from the origin to each of the intercepts must be the same. Call this distance $a$ . Since the $\\angle PAQ = 90^\\circ$ , the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is $\\sqrt{6^2+8^2} = 10$ , this means $a=10$ , and the length of the hypotenuse is $2a = 20$ . Since the $x$ -coordinate of $A$ is the same as the altitude to the hypotenuse, $[APQ] = \\dfrac{20 \\cdot 6}{2} = \\boxed{60}$", "We can let the two lines be \\[y=mx+b\\] \\[y=-\\frac{1}{m}x-b\\] This is because the lines are perpendicular, hence the $m$ and $-\\frac{1}{m}$ , and the sum of the y-intercepts is equal to 0, hence the $b, -b$\nSince both lines contain the point $(6,8)$ , we can plug this into the two equations to obtain \\[8=6m+b\\] and \\[8=-6\\frac{1}{m}-b\\]\nAdding the two equations gives \\[16=6m+\\frac{-6}{m}\\] Multiplying by $m$ gives \\[16m=6m^2-6\\] \\[\\implies 6m^2-16m-6=0\\] \\[\\implies 3m^2-8m-3=0\\] Factoring gives \\[(3m+1)(m-3)=0\\]\nPlugging $m=3$ into one of our original equations, we obtain \\[8=6(3)+b\\] \\[\\implies b=8-6(3)=-10\\]\nSince $\\bigtriangleup APQ$ has hypotenuse $2|b|=20$ and the altitude to the hypotenuse is equal to the the x-coordinate of point $A$ , or 6, the area of $\\bigtriangleup APQ$ is equal to \\[\\frac{20\\cdot6}{2}=\\boxed{60}\\]", "Since their sum is $0$ , let the y intercepts be P $(0,a)$ and Q $(0,-a)$ . The slope of $AP$ is $\\frac{8-a}{6}$ . The slope of AQ is $\\frac{8+a}{6}$ . Since multiplying the slopes of perpendicular lines yields a product of $-1$ , we have $\\frac{64-a^2}{36}=-1$ , which results in $a^2=100$ . We can use either the positive or negative solution because if we choose $10$ , then the other y-intercept is $-10$ ; but if we choose $-10$ , then the other y-intercept is $10$ . For simplicity, we choose that $a=10$ in this solution.\nNow we have a triangle APQ with points A $(6,8)$ , P $(0,10)$ , and Q $(0,-10)$ . By the Pythagorean theorem, we have that $AP=\\sqrt{6^2+2^2}=2\\sqrt{10}$ , and that $AQ=\\sqrt{6^2+18^2}=6\\sqrt{10}$ $PQ$ is obviously $10-(-10)=20$ since they have the same $x$ coordinate. Now using Heron's formula, we have $\\sqrt{s(s-a)(s-b)(s-c)}=\\sqrt{(4\\sqrt{10}+10)(4\\sqrt{10}-10)(10+2\\sqrt{10})(10-2\\sqrt{10})}=\\sqrt{60^2}=60 \\implies \\boxed{60}$", "Using point-slope form, the first line has the equation \\[y-8=m\\left(x-6\\right) \\longrightarrow y=mx-6m+8\\] The second line has the equation \\[y-8=-\\frac{1}{m}\\left(x-6\\right) \\longrightarrow y=-\\frac{x}{m}+\\frac{6}{m}+8\\] At the y-intercept, the value of the x-coordinate is $0$ , hence: the first equation is $y=-6m+8$ and the second is $y=\\frac{6}{m}+8$ . Since the y-intercepts sum to $0$ , they are opposites, so: \\[-6m+8=-\\left(\\frac{6}{m}+8\\right)=-\\frac{6}{m}-8\\] \\[6m-\\frac{6}{m}=16\\] Multiply both sides by m: $6m^{2}-6=16m \\longrightarrow 3m^{2}-8m-3=0$ . The solution to this quadratic, using the quadratic formula, is: $\\frac{8\\pm\\sqrt{64-4\\left(3\\right)\\left(-3\\right)}}{6}=\\frac{8\\pm\\sqrt{100}}{6}=\\frac{8\\pm10}{6}$ This yields $m=-\\frac{1}{3}$ and $m=3$ . Plugging $m=3$ into the second equation, we get $y=\\frac{6}{3}+8=10$ . Plugging $m=-\\frac{1}{3}$ into the first equation, we get $y=-10$ So the base is $20$ and the height is $6$ , the area is $60 \\Longrightarrow \\boxed{60}$" ]

[ "Obviously $1$ is in the top left corner, $8$ is in the top right corner, and $64$ is in the bottom right corner. To find the bottom left corner, subtract $7$ from $64$ which is $57$ . Adding the results gives $1+8+57+64=130$ which is answer $\\boxed{130}$" ]

[ "The rectangle is divided into three smaller rectangles with a width of 50 mm and a length of $\\dfrac{80}{3}$ mm. According to the Pythagorean Theorem (or by noticing the 8-15-17 Pythagorean triple), the diagonal of the rectangle is $\\sqrt{50^2+\\left(\\frac{80}{3}\\right)^2}=\\frac{170}{3}$ mm. Since that on the lace, there are 6 of these diagonals, a width, and an extension of at least 200 mm on each side. Therefore, the minimum of the lace in millimeters is \\[6\\times \\dfrac{170}{3}+50+200\\times 2=\\boxed{790}.\\]" ]

[ "Let $r$ represent the rate Tom swims in miles per minute. Then we have\n$\\frac{1/2}{r} + \\frac{8}{5r} + \\frac{30}{10r} = 255$\nSolving for $r$ , we find $r = 1/50$ , so the time Tom spends biking is $\\frac{30}{(10)(1/50)} = \\boxed{150}$ minutes." ]

[ "Because the pencil costs a whole number of cents, the cost must be a factor of both $143$ and $195$ . They can be factored into $11\\cdot13$ and $3\\cdot5\\cdot13$ . The common factor cannot be $1$ or there would have to be more than $30$ sixth graders, so the pencil costs $13$ cents. The difference in costs that the sixth and seventh graders paid is $195-143=52$ cents, which is equal to $52/13 = \\boxed{4}$ sixth graders." ]

[ "Since the 5 members of each planet committee are distinct we get that the number of arrangement of sittings is in the form $N*(5!)^3$ because for each $M, V, E$ sequence we have $5!$ arrangements within the Ms, Vs, and Es.\nPretend the table only seats $3$ \"people\", with $1$ \"person\" from each planet. Counting clockwise, only the arrangement M, V, E satisfies the given constraints. Therefore, in the actual problem, the members must sit in cycles of M, V, E, but not necessarily with one M, one V, and one E in each cycle(for example, MMVVVE, MVVVEEE, MMMVVVEE all count as cycles). These cycles of MVE must start at seat $1$ , since an M is at seat $1$ . We simply count the number of arrangements through casework.\n1. The entire arrangement is one cycle- There is only one way to arrange this, MMMMMVVVVVEEEEE\n2. Two cycles - There are 3 Ms, Vs and Es left to distribute among the existing MVEMVE. Using stars and bars, we get $\\binom{4}{1}=4$ ways for the members of each planet. Therefore, there are $4^3=64$ ways in total.\n3. Three cycles - 2 Ms, Vs, Es left, so $\\binom{4}{2}=6$ , making there $6^3=216$ ways total.\n4. Four cycles - 1 M, V, E left, each M can go to any of the four MVE cycles and likewise for V and E, $4^3=64$ ways total\n5. Five cycles - MVEMVEMVEMVEMVE is the only possibility, so there is just $1$ way.\nCombining all these cases, we get $1+1+64+64+216= \\boxed{346}$", "The arrangements must follow the pattern MVEMVE....... where each MVE consists of some Martians followed by some Venusians followed by some Earthlings, for $1, 2, 3, 4,$ or $5$ MVE's. If there are $k$ MVE's, then by stars and bars, there are ${4 \\choose k-1}$ choices for the Martians in each block, and the same goes for the Venusians and the Earthlings. Thus, we have $N = 1^3+4^3+6^3+4^3+1^3 = \\boxed{346}$ - aops5234" ]

[ "$1^n$ will always be 1, so we can ignore those terms, and using the definition ( $2002 / 7 = 286$ ):\n\\[(10^{2002} + 1)^{\\frac {10}7} = 10^{2860}+\\dfrac{10}{7}10^{858}+\\dfrac{15}{49}10^{-1144}+\\cdots\\]\nSince the exponent of the $10$ goes down extremely fast, it suffices to consider the first few terms. Also, the $10^{2860}$ term will not affect the digits after the decimal, so we need to find the first three digits after the decimal in\n\\[\\dfrac{10}{7}10^{858}\\]\n(The remainder after this term is positive by the Remainder Estimation Theorem ). Since the repeating decimal of $\\dfrac{10}{7}$ repeats every 6 digits, we can cut out a lot of 6's from $858$ to reduce the problem to finding the first three digits after the decimal of\n$\\dfrac{10}{7}$\nThat is the same as $1+\\dfrac{3}{7}$ , and the first three digits after $\\dfrac{3}{7}$ are $\\boxed{428}$" ]

[ "There are $2! = 2$ ways to order the boys on the ends, and there are $3!=6$ ways to order the girls in the middle. We get the answer to be $2 \\cdot 6 = \\boxed{12}$" ]

[ "Let $x$ be the number of the children and the mom. The father, who is $48$ , plus the sum of the ages of the kids and mom divided by the number of kids and mom plus $1$ (for the dad) = $20$ . This is because the average age of the entire family is $20.$ This statement, written as an equation, is: \\[\\frac{48+16x}{x+1}=20\\] \\[48+16x=20x+20\\] \\[4x=28\\] \\[x=7\\]\n$7$ people - $1$ mom = $6$ children.\nTherefore, the answer is $\\boxed{6}$", "Let $m$ be the Mom's age.\nLet the number of children be $x$ and their average age be $y$ . Their age totaled up is simply $xy$\nWe have the following two equations:\n$\\frac{m+48+xy}{2+x}=20$ , where $m+48+xy$ is the family's total age and $2+x$ is the total number of people in the family.\n$\\frac{m+48+xy}{2+x}=20$\n$m+48+xy=40+20x$\nThe next equation is $\\frac{m+xy}{1+x}=16$ , where $m+xy$ is the total age of the Mom and the children, and $1+x$ is the number of children along with the Mom.\n$\\frac{m+xy}{1+x}=16$\n$m+xy=16+16x$\nWe know the value for $m+xy$ , so we substitute the value back in the first equation.\n$m+48+xy=40+20x$\n$(16+16x)+48=40+20x$\n$x=6$\nEarlier, we set $x$ to be the number of children. Therefore, there are $\\boxed{6}$ children." ]

[ "Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $\\bmod{10}$\n$1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$\nThe last digit to appear in the units position of a number in the Fibonacci sequence is $6 \\Longrightarrow \\boxed{6}$" ]

[ "Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $\\bmod{10}$\n$1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$\nThe last digit to appear in the units position of a number in the Fibonacci sequence is $6 \\Longrightarrow \\boxed{6}$" ]

[ "Let the number of two-legged birds be $x$ and the number of four-legged mammals be $y$ . We can now use systems of equations to solve this problem.\nWrite two equations:\n$2x + 4y = 522$\n$x + y = 200$\nNow multiply the latter equation by $2$\n$2x + 4y = 522$\n$2x + 2y = 400$\nBy subtracting the second equation from the first equation, we find that $2y = 122 \\implies y = 61$ . Since there were $200$ heads, meaning that there were $200$ animals, there were $200 - 61 = \\boxed{139}$ two-legged birds.", "Let's add 2 legs for each bird and assume each bird has four-legged as each mammal. So the total legs of these birds and mammals would be $4*200=800$ . Actually, there were only $522$ legs. The difference between these two numbers exactly gives us the number of all the legs we added for all birds: $800 - 522 = 278$ . Because each bird was added by 2 legs, so the total number of birds would be $278/2 = \\boxed{139}$ two-legged birds. ---LarryFlora" ]

[ "This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that $2^{10}=1024$\nHowever, because the first term is $2^0=1$ and not $2^1=2$ , the solution to the problem is $10-0+1=\\boxed{11}$" ]

[ "Within each division, there are $\\binom {6}{2} = 15$ pairings, and each of these games happens twice. The same goes for the other division so that there are $4(15)=60$ games within their own divisions. The number of games between the two divisions is $(6)(6)=36$ . Together there are $60+36=\\boxed{96}$ conference games." ]

[ "Each team plays 10 games in its own division and 6 games against teams in the other division. So each of the 12 teams plays 16 conference games. Because each game involves two teams, there are $\\frac{12\\times 16}{2}=\\boxed{96}$ games scheduled. ~aopsav (Credit to AoPS Alcumus)" ]

[ "Draw the following four lines as shown: \nWe see these lines split the figure into five squares with side length $\\sqrt2$ . Thus, the area is $5\\cdot\\left(\\sqrt2\\right)^2=5\\cdot 2 = \\boxed{10}$", "There are $5$ lattice points in the interior of the logo and $12$ lattice points on the boundary of the logo. Because of Pick's Theorem, the area of the logo is $5+\\frac{12}{2}-1=\\boxed{10}$", "Notice that the area of the figure is equal to the area of the $4 \\times 4$ square subtracted by the $12$ triangles that are half the area of each square, which is $1$ . The total area of the triangles not in the figure is $12 \\cdot \\frac{1}{2} = 6$ , so the answer is $16-6 = \\boxed{10}$", "Draw the following four lines as shown:\n\nThe area of the big square is $4$ , and the area of each triangle is $0.5$ . There are $12$ of these triangles, so the total area of all the triangles is $0.5\\cdot12=6$ . Therefore, the area of the entire figure is $4+6=\\boxed{10}$", "The coordinates are $(1,2), (2,1), (3,2), (4,1), (5,2), (4,3), (5,4), (4,5), (3,4), (2,5), (1,4), (2,3)$ Use the Shoelace Theorem to get $\\boxed{10}$", "If the triangles are rearranged such that the gaps are filled, there would be a $4$ by $2$ rectangle, and two $1$ by $1$ squares are present. Thus, the answer is $\\boxed{10}$" ]

[ "There are $100+80-60=120$ females in either band or orchestra, so there are $230-120=110$ males in either band or orchestra. Suppose $x$ males are in both band and orchestra. \\[80+100-x=110\\Rightarrow x=70.\\] Thus, the number of males in band but not orchestra is $80-70=10\\rightarrow \\boxed{10}$" ]

[ "Since the three angles of a triangle add up to $180^{\\circ}$ and one of the angles is $90^{\\circ}$ because it's a right triangle, $a^{\\circ} + b^{\\circ} = 90^{\\circ}$\nThe greatest prime number less than $90$ is $89$ . If $a=89^{\\circ}$ , then $b=90^{\\circ}-89^{\\circ}=1^{\\circ}$ , which is not prime.\nThe next greatest prime number less than $90$ is $83$ . If $a=83^{\\circ}$ , then $b=7^{\\circ}$ , which IS prime, so we have our answer $\\boxed{7}$ ~quacker88", "Looking at the answer choices, only $7$ and $11$ are coprime to $90$ . Testing $7$ , the smaller angle, makes the other angle $83$ which is prime, therefore our answer is $\\boxed{7}$", "It is clear that $\\gcd(a,b)=1.$ By the Euclidean Algorithm, we have \\[\\gcd(a,b)=\\gcd(a+b,b)=\\gcd(90,b)=1,\\] so $90$ and $b$ are relatively prime.\nThe least such prime number $b$ is $7,$ from which $a=90-b=83$ is also a prime number. Therefore, the answer is $\\boxed{7}.$" ]

[ "Since the three angles of a triangle add up to $180^{\\circ}$ and one of the angles is $90^{\\circ}$ because it's a right triangle, $a^{\\circ} + b^{\\circ} = 90^{\\circ}$\nThe greatest prime number less than $90$ is $89$ . If $a=89^{\\circ}$ , then $b=90^{\\circ}-89^{\\circ}=1^{\\circ}$ , which is not prime.\nThe next greatest prime number less than $90$ is $83$ . If $a=83^{\\circ}$ , then $b=7^{\\circ}$ , which IS prime, so we have our answer $\\boxed{7}$ ~quacker88", "Looking at the answer choices, only $7$ and $11$ are coprime to $90$ . Testing $7$ , the smaller angle, makes the other angle $83$ which is prime, therefore our answer is $\\boxed{7}$", "It is clear that $\\gcd(a,b)=1.$ By the Euclidean Algorithm, we have \\[\\gcd(a,b)=\\gcd(a+b,b)=\\gcd(90,b)=1,\\] so $90$ and $b$ are relatively prime.\nThe least such prime number $b$ is $7,$ from which $a=90-b=83$ is also a prime number. Therefore, the answer is $\\boxed{7}.$" ]

[ "If we change $0$ , the units column would be incorrect.\nIf we change $1$ , then the leading $1$ in the sum would be incorrect.\nHowever, looking at the $2$ in the hundred-thousands column, it would be possible to change the $2$ to either a $5$ (no carry) or a $6$ (carry) to create a correct statement.\nChanging the $2$ to a $5$ would give $745586 + 859430$ on top, which equals $1605016$ . This does not match up to the bottom.\nChanging the $2$ to a $6$ gives $746586 + 869430$ on top, which has a sum of $1616016$ . This is the number on the bottom if the $2$ s were changed to $6$ s.\nThus $d=2$ and $e=6$ . so $d+e= 8 \\boxed{8}$" ]

[ "Doing the addition as is, we get $641 + 852 + 973 = 2466$ . This number is $10$ larger than the desired sum of $2456$ . Therefore, we must make one of the three numbers $10$ smaller.\nWe may either change $641 \\rightarrow 631$ $852 \\rightarrow 842$ , or $973 \\rightarrow 963$ . Either change results in a valid sum. The largest digit that could be changed is thus the $7$ in the number $973$ , and the answer is $\\boxed{7}$" ]

[ "The three unlabelled vertical sides have the same sum as the two labelled vertical sides, which is $10$\nThe four unlabelled horizontal sides have the same sum as the one large horizontal side, which is $12$\nThus, the perimeter is $2(12+10) = 44$ , which is option $\\boxed{44}$" ]

[ "Let $M$ be the midpoint of the chord $BC$ . From right triangle $OMB$ , we have $OM = \\sqrt{OB^2 - BM^2} =4$ . This gives $\\tan \\angle BOM = \\frac{BM}{OM} = \\frac 3 4$\nNotice that the distance $OM$ equals $PN + PO \\cos \\angle AOM = r(1 + \\cos \\angle AOM)$ , where $r$ is the radius of circle $P$\nHence \\[\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5\\] (where $R$ represents the radius, $5$ , of the large circle given in the question). Therefore, since $\\angle AOM$ is clearly acute, we see that \\[\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3\\]\nNext, notice that $\\angle AOB = \\angle AOM - \\angle BOM$ . We can therefore apply the subtraction formula for $\\tan$ to obtain \\[\\tan \\angle AOB =\\frac{\\tan \\angle AOM - \\tan \\angle BOM}{1 + \\tan \\angle AOM \\cdot \\tan \\angle BOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}\\] It follows that $\\sin \\angle AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$ , such that the answer is $7 \\cdot 25=\\boxed{175}$", "This solution, while similar to Solution 2, is arguably more motivated and less contrived.\nFirstly, we note the statement in the problem that \" $AD$ is the only chord starting at $A$ and bisected by $BC$ \" – what is its significance? What is the criterion for this statement to be true?\nWe consider the locus of midpoints of the chords from $A$ . It is well-known that this is the circle with diameter $AO$ , where $O$ is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor $\\frac{1}{2}$ and center $A$ . Thus, the locus is the result of the dilation with scale factor $\\frac{1}{2}$ and centre $A$ of circle $O$ . Let the center of this circle be $P$\nNow, $AD$ is bisected by $BC$ if they cross at some point $N$ on the circle. Moreover, since $AD$ is the only chord, $BC$ must be tangent to the circle $P$\nThe rest of this problem is straightforward.\nOur goal is to find $\\sin \\angle AOB = \\sin{\\left(\\angle AOM - \\angle BOM\\right)}$ , where $M$ is the midpoint of $BC$ . We have $BM=3$ and $OM=4$ .\nLet $R$ be the projection of $A$ onto $OM$ , and similarly let $Q$ be the projection of $P$ onto $OM$ . Then it remains to find $AR$ so that we can use the addition formula for $\\sin$\nAs $PN$ is a radius of circle $P$ $PN=2.5$ , and similarly, $PO=2.5$ . Since $OM=4$ , we have $OQ=OM-QM=OM-PN=4-2.5=1.5$ . Thus $PQ=\\sqrt{2.5^2-1.5^2}=2$\nFurther, we see that $\\triangle OAR$ is a dilation of $\\triangle OPQ$ about center $O$ with scale factor $2$ , so $AR=2PQ=4$\nLastly, we apply the formula: \\[\\sin{\\left(\\angle AOM - \\angle BOM\\right)} = \\sin \\angle AOM \\cos \\angle BOM - \\sin \\angle BOM \\cos \\angle AOM = \\left(\\frac{4}{5}\\right)\\left(\\frac{4}{5}\\right)-\\left(\\frac{3}{5}\\right)\\left(\\frac{3}{5}\\right)=\\frac{7}{25}\\] Thus the answer is $7\\cdot25=\\boxed{175}$", "Let the center of the circle be $O$ . Fix $B,C,$ and $A$ . Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$ , it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$\nNow, let the intersection of $BC$ and $AD$ be $E$ and let the midpoint of $AO$ (the center of the circle tangent to $BC$ that we described beforehand) be $F$ . Drop the altitude from $O$ to $BC$ and call its intersection with $BC$ $K$ . Drop the perpendicular from $F$ to $KO$ and call its intersection with $KO$ $L$ . Clearly, $KO = \\sqrt{OC^2-KC^2} = \\sqrt{5^2-3^2} = 4$ and since $EF$ is radius, it equals $\\frac{5}{2}$ . The same applies for $FO$ , which also equals $\\frac{5}{2}$ . By the Pythagorean theorem, we deduce that $FL = 2$ , so $EK = 2$ . This is very important information! Now we know that $BE = 1$ , so by Power of a Point, $AE = ED = \\sqrt{5}$\nWe’re almost there! Since by the Pythagorean theorem, $ED^2 + EO^2 = 25$ , we deduce that $EO = 2\\sqrt{5}$ $EC=OC=5$ , so $\\sin (CEO) = \\frac{2\\sqrt{5}}{5}$ . Furthermore, since $\\sin (CEO) = \\cos(DEC)$ , we know that $\\cos (DEC) = \\frac{2\\sqrt{5}}{5}$ . By the law of cosines, \\[DC^2 = (\\sqrt{5})^2 + 5^2 -10\\sqrt{5} \\cdot \\frac{2\\sqrt{5}}{5} = 10\\] Therefore, $DC = \\sqrt{10} \\Longleftrightarrow BA = \\sqrt{2}$ . Now, drop the altitude from $O$ to $BA$ and call its intersection with $BA$ $Z$ . Then, by the Pythagorean theorem, $OZ = \\frac{7\\sqrt{2}}{2}$ . Thus, $\\sin (BOZ) = \\frac{\\sqrt{2}}{10}$ and $\\cos (BOZ) = \\frac{7\\sqrt{2}}{10}$ . As a result, $\\sin (BOA) = \\sin (2 BOZ) = 2\\sin(BOZ)\\cos(BOZ) = \\frac{7}{25}$ $7 \\cdot 25 = \\boxed{175}$", "Let I be the intersection of AD and BC.\nLemma: $AI = ID$ if and only if $\\angle AIO = 90$\nProof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If $\\angle AIO = 90$ , We can get $\\triangle AIO \\cong \\triangle OID$\nLet be this the circle with diameter AO.\nThus, we get $\\angle AIO = 90$ , implying I must lie on $\\omega$ . I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC.\nNow, create a coordinate system such that the origin is O and the y-axis is perpendicular to BC. The positive x direction is from B to C.\nLet Z be (0,5).\nLet Y be (-5,0).\nLet X be the center of $\\omega$ . Since $\\omega$ 's radius is $\\frac{5}{2}$ , the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so $sin(XOY) = sin(AOY) = \\frac{3}{5}$ $sin(BOZ) = \\frac{3}{5}$ . If we let $sin(\\theta) = \\frac{3}{5}$ , we can find that what we are looking for is $sin(90 - 2\\theta)$ , which we can evaluate and get $\\frac{7}{25} \\implies \\boxed{175}$", "Let $O$ be the center of the circle. The locus of midpoints of chords with $A$ as a endpoint is a circle with diameter $\\overline{AO}$ . Additionally, this circle must be tangent to $\\overline{BC}$ . Let the center of this circle be $P$ . Let $M$ be the midpoint of $BC$ $N$ be the foot of the perpendicular from $P$ to $\\overline{BM}$ , and $K$ be the foot of the perpendicular from $B$ to $\\overline{AP}$ . Let $x=BK$\nFrom right triangle $BKO$ , we get $KO = \\sqrt{25-x^2}$ . Thus, $KP = \\sqrt{25-x^2}-\\frac52$\nSince $BO = 5$ $BM = 3$ , and $\\angle BMO$ is right, $MO=4$ . From quadrilateral $MNPO$ , we get $MN = \\sqrt{PO^2 - (MO - NP)^2} = \\sqrt{(5/2)^2 - (4 - 5/2)^2} = \\sqrt{(5/2)^2 - (3/2)^2} = 2$ . Thus, $BN = 1$\nSince angles $BNP$ and $BKP$ are right, we get \\[BK^2+KP^2 = BN^2 + NP^2 \\implies x^2 + \\left(\\sqrt{25-x^2}-\\frac52\\right)^2 = \\left(\\frac52\\right)^2 + 1\\] \\[25 - 5\\sqrt{25-x^2} = 1\\] \\[5\\sqrt{25-x^2} = 24\\] \\[25(25-x^2) = 24^2\\] \\[25x^2 = 25^2 - 24^2 = 49\\] \\[x = \\frac75\\] Thus, $\\sin \\angle AOB = \\frac{x}{5} = \\frac{7}{25}\\implies \\boxed{175}$" ]

[ "First look at the two cousins' ages that multiply to $24$ . Since the ages must be single-digit, the ages must either be $3 \\text{ and } 8$ or $4 \\text{ and } 6.$\nNext, look at the two cousins' ages that multiply to $30$ . Since the ages must be single-digit, the only ages that work are $5 \\text{ and } 6.$ Remembering that all the ages must all be distinct, the only solution that works is when the ages are $3, 8$ and $5, 6$\nWe are required to find the sum of the ages, which is \\[3 + 8 + 5 + 6 = \\boxed{22}.\\]" ]

[ "Let $a, b,$ and $c$ denote the bases of altitudes $12, 15,$ and $20,$ respectively. Since they are all altitudes and bases of the same triangle, they have the same area, so $\\frac{12a}{2}=\\frac{15b}{2}=\\frac{20c}{2}.$ Multiplying by $2$ , we get $12a=15b=20c.$ Notice that a simple solution to the equation is if all of them equal $12 \\cdot 15 \\cdot 20.$ That means $a=15 \\cdot 20, b=12 \\cdot 20,$ and $c=12 \\cdot 15.$ Simplifying our solution to check for Pythagorean triples we see that this is just a Pythagorean triple, namely a $3-4-5$ triangle. Since the other two angles of a right triangle must be acute, the right angle must be the greatest angle. Therefore, our answer is $\\boxed{90}.$" ]

[ "If the angles are in an arithmetic progression, they can be expressed as $a$ $a+n$ $a+2n$ $a+3n$ , and $a+4n$ for some real numbers $a$ and $n$ .\nNow we know that the sum of the degree measures of the angles of a pentagon is $180(5-2)=540$ .\nAdding our expressions for the five angles together, we get $5a+10n=540$ .\nWe now divide by 5 to get $a+2n=108$ . It so happens that $a+2n$ is one of the angles we defined earlier, so that angle must have a measure of $\\boxed{108}$ .\n(In fact, for any arithmetic progression with an odd number of terms,\nthe middle term is equal to the average of all the terms.)" ]

[ "The sum of the interior angles of any quadrilateral is $360^\\circ.$ \\begin{align*} 360 &= \\angle A + \\angle B + \\angle C + \\angle D\\\\ &= \\angle A + \\frac{1}{2}A + \\frac{1}{3}A + \\frac{1}{4}A\\\\ &= \\frac{12}{12}A + \\frac{6}{12}A + \\frac{4}{12}A + \\frac{3}{12}A\\\\ &= \\frac{25}{12}A \\end{align*} \\[\\angle A = 360 \\cdot \\frac{12}{25} = 172.8 \\approx \\boxed{173}\\]" ]

[ "The sum of the interior angles of any quadrilateral is $360^\\circ.$ \\begin{align*} 360 &= \\angle A + \\angle B + \\angle C + \\angle D\\\\ &= \\angle A + \\frac{1}{2}A + \\frac{1}{3}A + \\frac{1}{4}A\\\\ &= \\frac{12}{12}A + \\frac{6}{12}A + \\frac{4}{12}A + \\frac{3}{12}A\\\\ &= \\frac{25}{12}A \\end{align*} \\[\\angle A = 360 \\cdot \\frac{12}{25} = 172.8 \\approx \\boxed{173}\\]" ]

[ "Let $S$ be the sum of all the incomes but the largest one. For the actual data, the mean is $\\frac{S+98000}{1000}$ , and for the incorrect data the mean is $\\frac{S+980000}{1000}$ . The difference is $882, or \\rightarrow \\boxed{882}$" ]

[ "The base is $\\overline{BC}=4$ . The height has a length of the difference of the y-coordinates of A and B, which is 2. Therefore the area is $4\\cdot 2=8\\Rightarrow \\boxed{8}$" ]

[ " We can see that the radius of the circle is $4\\sqrt{3}$ . We know that the radius is $\\frac{1}{3}$ of each median line of the triangle; each median line is therefore $12\\sqrt{3}$ . Since the median line completes a $30$ $60$ $90$ triangle, we can conclude that one of the sides of the triangle is $24$ . Triple the side length and we get our answer, $\\boxed{72}$" ]

[ "If the circumference of a circle is $24\\pi$ , the radius would be $12$ . Since the area of a circle is $\\pi r^2$ , the area is $144\\pi$ . The answer is $\\boxed{144}$", "By definition, $\\pi$ is the ratio of the circumference to the diameter. Since the circumference is $24\\pi$ , the diameter must be $24$ and the radius is $12$ . Therefore, by the area of circle formula $A=\\pi r^{2}$ the area is $12^{2}\\pi=144\\pi$ and $k=144 \\Longrightarrow \\boxed{144}$" ]

[ "The area of the larger pizza is $16\\pi$ , while the area of the smaller pizza is $9\\pi$ . Therefore, the larger pizza is $\\frac{7\\pi}{9\\pi} \\cdot 100\\%$ bigger than the smaller pizza. $\\frac{7\\pi}{9\\pi} \\cdot 100\\% = 77.777....$ , which is closest to $\\boxed{78}$" ]

[ "\nObviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we do know how to find the area of.\nIf we continue segment $\\overline{FE}$ until it reaches the right side at $G$ , we create two rectangles - one on the top and one on the bottom.\nWe know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of $ABGF$ is $6\\times5 = 30$ . For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other.\nNote that $GC+GB=9$ , and $GB=AF=5$ , so we must have \\[GC+5=9\\Rightarrow GC=4\\]\nThe area of the bottom rectangle is then \\[(DC)(GC)=4\\times 4=16\\]\nFinally, we just add the areas of the rectangles together to get $16 + 30 = 46$\n$\\boxed{46}$", "\nLet $\\langle ABCDEF \\rangle$ be the area of polygon $ABCDEF$ . Also, let $G$ be the intersection of $DC$ and $AF$ when both are extended.\nClearly, \\[\\langle ABCDEF \\rangle = \\langle ABCG \\rangle - \\langle GFED \\rangle\\]\nSince $AB=6$ and $BC=9$ $\\langle ABCG \\rangle =6\\times 9=54$\nTo compute the area of $GFED$ , note that \\[AB=GD+DC\\] \\[BC=GF+FA\\]\nWe know that $AB=6$ $DC=4$ $BC=9$ , and $FA=5$ , so \\[6=GD+4\\Rightarrow GD=2\\] \\[9=GF+5\\Rightarrow GF=4\\]\nThus $\\langle GFED \\rangle = 4\\times 2=8$\nFinally, we have \\begin{align*} \\langle ABCDEF \\rangle &= \\langle ABCG \\rangle - \\langle GFED \\rangle \\\\ &= 54-8 \\\\ &= 46 \\\\ \\end{align*}\nThis is answer choice $\\boxed{46}$" ]

[ "Notice that $AF + DE = BC$ , so $DE=4$ . Let $O$ be the intersection of the extensions of $AF$ and $DC$ , which makes rectangle $ABCO$ . The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$\n\\[\\text{Area} = 52 = 8 \\cdot 9- EF \\cdot 4\\]\nSolving for the unknown, $EF=5$ , therefore $DE+EF=4+5=\\boxed{9}$" ]