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[ "To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\\choose 2} = 36$ . Similarly, there are ${9\\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles.\nFor $s$ , there are $8^2$ unit squares $7^2$ of the $2\\times2$ squares, and so on until $1^2$ of the $8\\times 8$ squares. Using the sum of squares formula, that gives us $s=1^2+2^2+\\cdots+8^2=\\dfrac{(8)(8+1)(2\\cdot8+1)}{6}=12*17=204$\nThus $\\frac sr = \\dfrac{204}{1296}=\\dfrac{17}{108}$ , and $m+n=\\boxed{125}$", "First, to find the number of squares, we can look case by case by the side length of the possible squares on the checkerboard. We see that there are $8^2$ ways to place a $1$ $1$ square and $7^2$ for a $2$ $2$ square. This pattern can be easily generalized and we see that the number of squares is just $\\sum^8_{i=1}{i^2}$ . This can be simplified by using the well-known formula for the sum of consecutive squares $\\frac{n(n+1)(2n+1)}{6}$ to get $204$\nThen, to find the number of rectangles, first note that a square falls under the definition of a rectangle. We can break up the rectangles into cases for the length x width. As we note down the cases for $1$ $1, 1$ $2 , 2$ $1, 2$ $2,...,$ we see they are respectively $8$ $8, 8$ $7, 7$ $8, 7$ $7, ...$ . We can quickly generalize this pattern to basically just ${\\sum^8_{i=1}{i}}\\cdot{\\sum^8_{i=1}{i}}$ . This gets us ${(\\frac{9\\cdot8}{2})}^2,$ which is just $1296.$\nNow, to calculate the ratio of $s/r,$ we divide $204$ by $1296$ to get a simplified fraction of $\\frac{17}{108}.$\nThus, our answer is just $s + r = 17+108 = \\boxed{125}$ ~MathWhiz35" ]

[ "The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.\n\\[\\begin{tabular}{|c|c|c|}\\hline 1 & 3 &\\\\ \\hline & 2 & A\\\\ \\hline & & B\\\\ \\hline\\end{tabular}\\]\nThe last square in the top row cannot be either 1 or 3, so it must be 2.\n\\[\\begin{tabular}{|c|c|c|}\\hline 1 & 3 & 2\\\\ \\hline & 2 & A\\\\ \\hline & & B\\\\ \\hline\\end{tabular}\\]\nThe other two squares in the rightmost column with A and B cannot be two, so they must be 1 and 3 and therefore have a sum of $1+3=\\boxed{4}$" ]

[ "$\\frac 2{\\log_4{2000^6}} + \\frac 3{\\log_5{2000^6}}$\n$=\\frac{\\log_4{16}}{\\log_4{2000^6}}+\\frac{\\log_5{125}}{\\log_5{2000^6}}$\n$=\\frac{\\log{16}}{\\log{2000^6}}+\\frac{\\log{125}}{\\log{2000^6}}$\n$=\\frac{\\log{2000}}{\\log{2000^6}}$\n$=\\frac{\\log{2000}}{6\\log{2000}}$\n$=\\frac{1}{6}$\nTherefore, $m+n=1+6=\\boxed{007}$", "Alternatively, we could've noted that, because $\\frac 1{\\log_a{b}} = \\log_b{a}$\n\\begin{align*} \\frac 2{\\log_4{2000^6}} + \\frac 3{\\log_5{2000^6}} &= 2 \\cdot \\frac{1}{\\log_4{2000^6}} + 3\\cdot \\frac {1}{\\log_5{2000^6} }\\\\ &=2{\\log_{2000^6}{4}} + 3{\\log_{2000^6}{5}} \\\\ &={\\log_{2000^6}{4^2}} + {\\log_{2000^6}{5^3}}\\\\ &={\\log_{2000^6}{4^2 \\cdot 5^3}}\\\\ &={\\log_{2000^6}{2000}}\\\\ &= {\\frac{1}{6}}.\\end{align*}\nTherefore our answer is $1 + 6 = \\boxed{007}$", "We know that $2 = \\log_4{16}$ and $3 = \\log_5{125}$ , and by base of change formula, $\\log_a{b} = \\frac{\\log_c{b}}{\\log_c{a}}$ . Lastly, notice $\\log a + \\log b = \\log ab$ for all bases. \\begin{align*} \\frac 2{\\log_4{2000^6}} + \\frac 3{\\log_5{2000^6}} = \\log_{2000^6}{16} + \\log_{2000^6}{125} = \\log_{2000^6}{2000} = \\frac16 \\implies \\boxed{007}", "\\[\\frac{2}{\\log_4 2000^6} + \\frac{3}{\\log_5 2000^6}\\] \\[= \\frac{1}{3\\log_4 2000} + \\frac{1}{2\\log_5 2000}\\] \\[= \\frac{1}{3} \\log_{2000} 4 + \\frac{1}{2} \\log_{2000} 5\\] \\[= \\log_{2000} (\\sqrt[3]{4} \\cdot \\sqrt{5}) = x\\] \\[\\implies 2^{4x} \\cdot 5^{3x} = 2^{\\frac{2}{3}} \\cdot 5^{\\frac{1}{2}}\\] \\[\\implies 4x + (3\\log_2 5)x = \\frac{2}{3}+\\frac{1}{2} \\log_2 5\\] \\[\\implies x = \\frac{\\frac{2}{3} + \\frac{1}{2} \\log_2 5}{4 + 3\\log_2 5}\\] \\[\\implies 6x = \\frac{4 + 3\\log_2 5}{4 + 3\\log_2 5}\\] \\[\\implies x = \\frac{1}{6}\\] \\[\\implies m + n = \\boxed{007}\\]" ]

[ "We take cases on the thousands digit, which must be either $1$ or $2$ :\nIf the number is of the form $\\overline{1bcd},$ where $b, c, d$ are digits, then we must have $d = 1 + b + c.$ Since $d \\le 9,$ we must have $b + c \\le 9 - 1 = 8.$ By casework on the value of $b$ , we find that there are $1 + 2 + \\dots + 9 = 45$ possible pairs $(b, c)$ , and each pair uniquely determines the value of $d$ , so we get $45$ numbers with the given property.\nIf the number is of the form $\\overline{2bcd},$ then it must be one of the numbers $2000, 2001, \\dots, 2012.$ Checking all these numbers, we find that only $2002$ has the given property.\nTherefore, the number of numbers with the property is $45 + 1 = \\boxed{46}$", "Let's start with the case that starts with $200$ . We have only one number, which is $2002$ . If we look at the $1900s$ , we have no solutions because $1+9 = 10$ , and because we can only use digits from $1$ through $9$ , it is impossible. If we looks at the $1800s$ , we do have one solution, which is $1809$ . If we look a the $1700s$ , we have $2$ solutions, namely, $1708$ and $1719$\nWe can see a pattern here. The pattern is every hundred you go down, you have $1$ more solution. Therefore, we have $1+0+1+2+3+4+5+6+7+8+9$ which is = $\\boxed{46}$" ]

[ "The prime factorization of $2013$ is $61\\cdot11\\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and will lead to an extra factor in the numerator. The highest prime less than $61$ is $59$ , so there must be a factor of $59$ in the denominator. It follows that $b_1 = 59$ (to minimize $b_1$ as well), so the answer is $|61-59| = \\boxed{2}.\\]" ]

[ "The prime factorization of $2013$ is $61\\cdot11\\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and will lead to an extra factor in the numerator. The highest prime less than $61$ is $59$ , so there must be a factor of $59$ in the denominator. It follows that $b_1 = 59$ (to minimize $b_1$ as well), so the answer is $|61-59| = \\boxed{2}.\\]" ]

[ "Note that $2014\\equiv -3 \\mod2017$ . We have for $k\\ge1$ \\[\\dbinom{2014}{k}\\equiv \\frac{(-3)(-4)(-5)....(-2-k)}{k!}\\mod 2017\\] \\[\\equiv (-1)^k\\dbinom{k+2}{k} \\mod 2017\\] Therefore \\[\\sum \\limits_{k=0}^{62} \\dbinom{2014}{k}\\equiv \\sum \\limits_{k=0}^{62}(-1)^k\\dbinom{k+2}{2} \\mod 2017\\] This is simply an alternating series of triangular numbers that goes like this: $1-3+6-10+15-21....$ After finding the first few sums of the series, it becomes apparent that \\[\\sum \\limits_{k=1}^{n}(-1)^k\\dbinom{k+2}{2}\\equiv -\\left(\\frac{n+1}{2} \\right) \\left(\\frac{n+1}{2}+1 \\right) \\mod 2017 \\textnormal{ if n is odd}\\] and \\[\\sum \\limits_{k=1}^{n}(-1)^k\\dbinom{k+2}{2}\\equiv \\left(\\frac{n}{2}+1 \\right)^2 \\mod 2017 \\textnormal{ if n is even}\\] Obviously, $62$ falls in the second category, so our desired value is \\[\\left(\\frac{62}{2}+1 \\right)^2 = 32^2 = \\boxed{1024}\\]" ]

[ "Taking the root, we get $N=\\sqrt{25^{64}\\cdot 64^{25}}=5^{64}\\cdot 8^{25}$\nNow, we have $N=5^{64}\\cdot 8^{25}=5^{64}\\cdot (2^{3})^{25}=5^{64}\\cdot 2^{75}$\nCombining the $2$ 's and $5$ 's gives us $(2\\cdot 5)^{64}\\cdot 2^{(75-64)}=(2\\cdot 5)^{64}\\cdot 2^{11}=10^{64}\\cdot 2^{11}$\nThis is the number $2048$ with a string of sixty-four $0$ 's at the end. Thus, the sum of the digits of $N$ is $2+4+8=14\\Longrightarrow\\boxed{14}$" ]

[ "Casework by the units digit $u$ will help organize the answer.\n$u=0$ gives no solutions, since no real numbers are divisible by $0$\n$u=1$ has $4$ solutions, since all numbers are divisible by $1$\n$u=2$ has $4$ solutions, since every number ending in $2$ is even (ie divisible by $2$ ).\n$u=3$ has $1$ solution: $33$ $\\pm 10$ or $\\pm 20$ will retain the units digit, but will stop the number from being divisible by $3$ $\\pm 30$ is the smallest multiple of $10$ that will keep the number divisible by $3$ , but those numbers are $3$ and $63$ , which are out of the range of the problem.\n$u=4$ has $2$ solutions: $24$ and $44$ . Adding or subtracting $10$ will kill divisibility by $4$ , since $10$ is not divisible by $4$\n$u=5$ has $4$ solutions: every number ending in $5$ is divisible by $5$\n$u=6$ has $1$ solution: $36$ $\\pm 10$ or $\\pm 20$ will kill divisibility by $3$ , and thus kill divisibility by $6$\n$u=7$ has no solutions. The first multiples of $7$ that end in $7$ are $7$ and $77$ , but both are outside of the range of this problem.\n$u=8$ has $1$ solution: $48$ $\\pm 10, \\pm 20, \\pm 30$ will all kill divisibility by $8$ since $10, 20,$ and $30$ are not divisible by $8$\n$u=9$ has no solutions. $9$ and $99$ are the smallest multiples of $9$ that end in $9$\nTotalling the solutions, we have $0 + 4 + 4 + 1 + 2 + 4 + 1 + 0 + 1 + 0 = 17$ solutions, giving the answer $\\boxed{17}$ , which is 17." ]

[ "The prime factorization of $6545$ is $5\\cdot7\\cdot11\\cdot17 =385\\cdot17$ , which contains a three digit number, but we want 6545 to be expressed as ab x cd. Now we do trial and error: \\[5\\cdot7=35 \\text{, } 11\\cdot17=187 \\text{ X}\\] \\[5\\cdot11=55 \\text{, } 7\\cdot17=119 \\text{ X}\\] \\[5\\cdot17=85 \\text{, } 7\\cdot11=77 \\text{ }\\surd\\] \\[85+77= \\boxed{162}\\]" ]

[ "This problem can be approached similarly to other base number problems.\nSince $120 < 695 < 720$ , divide $695$ by $120$ . The quotient is $5$ and the remainder is $95$ , so rewrite the number as \\[695 = 5 \\cdot 120 + 95\\] Similarly, dividing $95$ by $24$ results in a quotient of $3$ and a remainder of $23$ , so the number can be rewritten as \\[695 = 5 \\cdot 120 + 3 \\cdot 24 + 23\\] Repeat the steps to get \\[695 = 5 \\cdot 120 + 3 \\cdot 24 + 3 \\cdot 6 + 2 \\cdot 2 + 1\\] The answer is $\\boxed{3}$ . One can also stop at the second step by noting $23 < 24$" ]

[ "From the second bullet point, we know that the second digit must be $3$ , for a number divisible by $10$ ends in zero. Since there is a remainder of $1$ when $N$ is divided by $9$ , the multiple of $9$ must end in a $2$ for it to have the desired remainder $\\pmod {10}.$ We now look for this one:\n$9(1)=9\\\\ 9(2)=18\\\\ 9(3)=27\\\\ 9(4)=36\\\\ 9(5)=45\\\\ 9(6)=54\\\\ 9(7)=63\\\\ 9(8)=72$\nThe number $72+1=73$ satisfies both conditions. We subtract the biggest multiple of $11$ less than $73$ to get the remainder. Thus, $73-11(6)=73-66=\\boxed{7}$", "This two digit number must take the form of $10x+y,$ where $x$ and $y$ are integers $0$ to $9.$ However, if x is an integer, we must have $y=3.$ So, the number's new form is $10x+3.$ This needs to have a remainder of $1$ when divided by $9.$ Because of the $9$ divisibility rule, we have \\[10x+3 \\equiv 1 \\pmod 9.\\] We subtract the three, getting \\[10x \\equiv -2 \\pmod 9.\\] which simplifies to \\[10x \\equiv 7 \\pmod 9.\\] However, $9x \\equiv 0 \\pmod 9,$ so \\[10x - 9x \\equiv 7 - 0 \\pmod 9\\] and \\[x \\equiv 7 \\pmod 9.\\]\nLet the quotient of $9$ in our modular equation be $c,$ and let our desired number be $z,$ so $x=9c+7$ and $z = 10x+3.$ We substitute these values into $z = 10x+3,$ and get \\[z = 10(9c+7) + 3\\] so \\[z = 90c+73.\\] As a result, $z \\equiv 73 \\pmod {90}.$\nTo prove generalization vigorously, we can let $a$ be the remainder when $z$ is divided by $11.$ Setting up a modular equation, we have \\[90c + 73 \\equiv a \\pmod {11}.\\] Simplifying, \\[90c+7 \\equiv a \\pmod {11}\\] If $c = 1,$ then we don't have a 2 digit number! Thus, $c=0$ and $a=\\boxed{7}$", "We know that the number has to be one more than a multiple of $9$ , because of the remainder of one, and the number has to be $3$ more than a multiple of $10$ , which means that it has to end in a $3$ . Now, if we just list the first few multiples of $9$ adding one to the number we get: $10, 19, 28, 37, 46, 55, 64, 73, 82, 91$ . As we can see from these numbers, the only one that has a three in the denominator is $73$ , thus we divide $73$ by $11$ , getting $6$ $R7$ , hence, $\\boxed{7}$ .\n-fn106068" ]

[ "We begin by equating the two expressions:\n\\[a\\sqrt{2}+b\\sqrt{3}+c\\sqrt{5} = \\sqrt{104\\sqrt{6}+468\\sqrt{10}+144\\sqrt{15}+2006}\\]\nSquaring both sides yields:\n\\[2ab\\sqrt{6} + 2ac\\sqrt{10} + 2bc\\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\\sqrt{6}+468\\sqrt{10}+144\\sqrt{15}+2006\\]\nSince $a$ $b$ , and $c$ are integers, we can match coefficients:\n\\begin{align*} 2ab\\sqrt{6} &= 104\\sqrt{6} \\\\ 2ac\\sqrt{10} &=468\\sqrt{10} \\\\ 2bc\\sqrt{15} &=144\\sqrt{15}\\\\ 2a^2 + 3b^2 + 5c^2 &=2006 \\end{align*}\nSolving the first three equations gives: \\begin{eqnarray*}ab &=& 52\\\\ ac &=& 234\\\\ bc &=& 72 \\end{eqnarray*}\nMultiplying these equations gives $(abc)^2 = 52 \\cdot 234 \\cdot 72 = 2^63^413^2 \\Longrightarrow abc = \\boxed{936}$", "We realize that the quantity under the largest radical is a perfect square and attempt to rewrite the radicand as a square. Start by setting $x=\\sqrt{2}$ $y=\\sqrt{3}$ , and $z=\\sqrt{5}$ . Since\n\\[(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)\\]\nwe attempt to rewrite the radicand in this form:\n\\[2006+2(52xy+234xz+72yz)\\]\nFactoring, we see that $52=13\\cdot4$ $234=13\\cdot18$ , and $72=4\\cdot18$ . Setting $p=13$ $q=4$ , and $r=18$ , we see that\n\\[2006=13^2x^2+4^2y^2+18^2z^2=169\\cdot2+16\\cdot3+324\\cdot5\\]\nso our numbers check. Thus $104\\sqrt{2}+468\\sqrt{3}+144\\sqrt{5}+2006=(13\\sqrt{2}+4\\sqrt{3}+18\\sqrt{5})^2$ . Square rooting gives us $13\\sqrt{2}+4\\sqrt{3}+18\\sqrt{5}$ and our answer is $13\\cdot4\\cdot18=\\boxed{936}$" ]

[ "We know that $9<N<17$ and we wish to bound $\\frac{6+10+N}{3}=\\frac{16+N}{3}$\nFrom what we know, we can deduce that $25<N+16<33$ , and thus \\[8.\\overline{3}<\\frac{N+16}{3}<11\\]\nThe only answer choice that falls in this range is choice $\\boxed{10}$" ]

[ "Let $w=\\lfloor x \\rfloor$ and $f=\\{x\\}$ denote the whole part and the fractional part of $x,$ respectively, for which $0\\leq f<1$ and $x=w+f.$\nWe rewrite the given equation as \\[w\\cdot f=a\\cdot(w+f)^2. \\hspace{38.75mm}(1)\\] Since $a\\cdot(w+f)^2\\geq0,$ it follows that $w\\cdot f\\geq0,$ from which $w\\geq0.$\nWe expand and rearrange $(1)$ as \\[af^2+(2a-1)wf+aw^2=0, \\hspace{23mm}(2)\\] which is a quadratic with either $f$ or $w.$\nFor simplicity purposes, we will treat $w$ as some fixed nonnegative integer so that $(2)$ is a quadratic with $f.$ By the Quadratic Formula, we have \\[f=w\\Biggl(\\frac{1-2a\\pm\\sqrt{1-4a}}{2a}\\Biggr). \\hspace{25mm}(3)\\] If $w=0,$ then $f=0.$ We get $x=w+f=0,$ which does not affect the sum of the solutions. Therefore, we consider the case for $w\\geq1:$\nRecall that $0\\leq f<1,$ so $\\frac{1-2a\\pm\\sqrt{1-4a}}{2a}\\geq0.$ From the discriminant, we require that $0\\leq1-4a<1,$ or \\[0<a\\leq\\frac14. \\hspace{54mm}(4)\\]\nWe consider each part of $0\\leq f<1$ separately:\nNow, we express $x$ in terms of $w$ and $k:$ \\[x=w+f=w+wk=w(1+k).\\] The sum of all solutions to the original equation is \\[\\sum_{w=1}^{W}w(1+k)=(1+k)\\cdot\\sum_{w=1}^{W}w=(1+k)\\cdot\\frac{W(W+1)}{2}=420. \\hspace{10mm}(\\bigstar)\\] As $1+k<1+\\frac1W,$ we conclude that $1+k$ is slightly above $1$ so that $\\frac{W(W+1)}{2}$ is slightly below $420,$ or $W(W+1)$ is slightly below $840.$ By observations, we get $W=28.$ Substituting this into $(\\bigstar)$ produces $k=\\frac{1}{29},$ which satisfies $\\frac{1}{W+1}\\leq k<\\frac1W,$ as required.\nFinally, we solve for $a$ in $k=\\frac{1}{2a}-1-\\frac{\\sqrt{1-4a}}{2a}:$ \\begin{align*} \\frac{1}{29}&=\\frac{1}{2a}-1-\\frac{\\sqrt{1-4a}}{2a} \\\\ \\frac{2}{29}a&=1-2a-\\sqrt{1-4a} \\\\ \\frac{60}{29}a-1&=-\\sqrt{1-4a} \\\\ \\frac{60^2}{29^2}a^2-\\frac{120}{29}a+1&=1-4a \\\\ \\frac{60^2}{29^2}a^2-\\frac{4}{29}a&=0 \\\\ a\\left(\\frac{60^2}{29^2}a-\\frac{4}{29}\\right)&=0. \\end{align*} Since $a>0,$ we obtain $\\frac{60^2}{29^2}a-\\frac{4}{29}=0,$ from which \\[a=\\frac{4}{29}\\cdot\\frac{29^2}{60^2}=\\frac{29}{900}.\\] The answer is $29+900=\\boxed{929}.$", "Let $x_n$ be a root in the interval $(n,n+1)$ . In this interval, $\\lfloor x_n \\rfloor = n$ and $\\{x_n\\}=x_n-n$ , so we must have $ax_n^2 = nx_n-n^2$ , i.e., $ax_n^2-nx_n+n^2=0$ . We can homogenize this equation by setting $x_n=n\\zeta$ ; then $x_1=\\zeta$ , and $\\zeta$ is a root of $a\\zeta^2-\\zeta+1=0$\nSuppose $N$ is the largest integer for which there is such a root; we have, for $n=1,2,\\ldots , N$ \\[n < x_n = n\\zeta < n+1\\] Summing over $n\\in \\{1,2,\\ldots , N\\}$ we get \\[\\tfrac 12 N(N+1) < 420 = \\tfrac 12 N(N+1)\\zeta < \\tfrac 12 N(N+3)\\] From the right inequality we get $27< N$ and from the left one we get $N<29$ . Thus $N=28$ . Using this in the middle equality we get $\\zeta = \\tfrac{30}{29}$ . Since $\\zeta$ satisfies $a\\zeta^2-\\zeta+1=0$ , we get \\[a = \\zeta^{-2}(\\zeta-1)= \\tfrac{29^2}{30^2}\\cdot \\tfrac 1{29}= \\tfrac{29}{900}.\\] The answer is $29+900=\\boxed{929}.$", "First note that $\\lfloor x\\rfloor \\cdot \\{x\\}<0$ when $x<0$ while $ax^2\\ge 0\\forall x\\in \\mathbb{R}$ . Thus we only need to look at positive solutions ( $x=0$ doesn't affect the sum of the solutions). \nNext, we break $\\lfloor x\\rfloor\\cdot \\{x\\}$ down for each interval $[n,n+1)$ , where $n$ is a positive integer. Assume $\\lfloor x\\rfloor=n$ , then $\\{x\\}=x-n$ . This means that when $x\\in [n,n+1)$ $\\lfloor x\\rfloor \\cdot \\{x\\}=n(x-n)=nx-n^2$ . Setting this equal to $ax^2$ gives \\[nx-n^2=ax^2\\implies ax^2-nx+n^2=0 \\implies x=\\frac{n\\pm \\sqrt{n^2-4an^2}}{2a}\\] We're looking at the solution with the positive $x$ , which is $x=\\frac{n-n\\sqrt{1-4a}}{2a}=\\frac{n}{2a}\\left(1-\\sqrt{1-4a}\\right)$ . Note that if $\\lfloor x\\rfloor=n$ is the greatest $n$ such that $\\lfloor x\\rfloor \\cdot \\{x\\}=ax^2$ has a solution, the sum of all these solutions is slightly over $\\sum_{k=1}^{n}k=\\frac{n(n+1)}{2}$ , which is $406$ when $n=28$ , just under $420$ . Checking this gives \\begin{align*} \\sum_{k=1}^{28}\\frac{k}{2a}\\left(1-\\sqrt{1-4a}\\right)&=\\frac{1-\\sqrt{1-4a}}{2a}\\cdot 406=420 \\\\ \\frac{1-\\sqrt{1-4a}}{2a}&=\\frac{420}{406}=\\frac{30}{29} \\\\ 29-29\\sqrt{1-4a}&=60a \\\\ 29\\sqrt{1-4a}&=29-60a \\\\ 29^2-4\\cdot 29^2a&=29^2+3600a^2-120\\cdot 29a \\\\ 3600a^2&=116a \\\\ a&=\\frac{116}{3600}=\\frac{29}{900} \\implies \\boxed{929} ~ktong" ]

[ "We have these equations: $196a+14b+c=225a+15c+b=222a+37c$ .\nTaking the last two we get $3a+b=22c$ . Because $c \\neq 0$ otherwise $a \\ngtr 0$ , and $a \\leq 5$ $c=1$\nThen we know $3a+b=22$ .\nTaking the first two equations we see that $29a+14c=13b$ . Combining the two gives $a=4, b=10, c=1$ . Then we see that $222 \\times 4+37 \\times1=\\boxed{925}$", "We know that $196a+14b+c=225a+15c+b=222a+37c$ . Combining the first and third equations give that $196a+14b+c=222a+37c$ , or \\[7b=13a+18c\\] The second and third gives $222a+37c=225a+15c+b$ , or \\[22c-3a=b\\] \\[154c-21a=7b=13a+18c\\] \\[4c=a\\] We can have $a=4,8,12$ , but only $a=4$ falls within the possible digits of base $6$ . Thus $a=4$ $c=1$ , and thus you can find $b$ which equals $10$ . Thus, our answer is $4\\cdot225+1\\cdot15+10=\\boxed{925}$", "We're given that $196a+14b+c=225a+15c+b=222a+37c.$ By taking the difference of the first $2$ equalities, we receive $29a+14c=13b.$ Taking $\\pmod{13}$ , we receive $3a+c \\equiv 0 \\pmod{13}.$ We receive the following cases: $(a,c)=(4,1)$ or $(3,4).$ (Note that $(2,7)$ doesn't work since $a,c<6$ by third condition). We can just check these two, and find that $(a,b,c)=(4,10,1),$ and just plugging in $(a,c)$ into the third expression we receive $888+37=\\boxed{925}.$" ]

[ "The nearest fractions to $\\frac 27$ with numerator $1$ are $\\frac 13, \\frac 14$ ; and with numerator $2$ are $\\frac 26, \\frac 28 = \\frac 13, \\frac 14$ anyway. For $\\frac 27$ to be the best approximation for $r$ , the decimal must be closer to $\\frac 27 \\approx .28571$ than to $\\frac 13 \\approx .33333$ or $\\frac 14 \\approx .25$\nThus $r$ can range between $\\frac{\\frac 14 + \\frac{2}{7}}{2} \\approx .267857$ and $\\frac{\\frac 13 + \\frac{2}{7}}{2} \\approx .309523$ . At $r = .2678, .3096$ , it becomes closer to the other fractions, so $.2679 \\le r \\le .3095$ and the number of values of $r$ is $3095 - 2679 + 1 = \\boxed{417}$" ]

[ "Let the value in the empty box in the middle row be $x$ , and the value in the empty box in the top row be $y$ $y$ is the answer we're looking for.\n\nFrom the diagram, $600 = 30x$ , making $x = 20$\n\nIt follows that $20 = 5y$ , so $y = \\boxed{4}$", "Another way to do this problem is to realize what makes up the bottommost number. This method doesn't work quite as well for this problem, but in a larger tree, it might be faster. Again, let the value in the empty box in the middle row be $x$ , and the value in the empty box in the top row be $y$ $y$ is the answer we're looking for.\n\nWe can write some equations:\n$600 = 30x\\\\ 30 = 6\\cdot 5\\\\ x = 5y$\nNow we can substitute into the first equation using the two others:\n$600 = (6\\cdot5)(5y)\\\\ 600= 6\\cdot5\\cdot5\\cdot y\\\\ 600=6\\cdot25\\cdot y\\\\ 600 = 150y\\\\ \\boxed{4} = y$" ]

[ "The function $f(x) = \\sin x$ has roots in the form of $\\pi n$ for all integers $n$ . Therefore, we want $\\frac{1}{x} = \\pi n$ on $\\frac{1}{10000} \\le x \\le \\frac{1}{1000}$ , so $1000 \\le \\frac 1x = \\pi n \\le 10000$ . There are $\\frac{10000-1000}{\\pi} \\approx \\boxed{2900}$ solutions for $n$ on this interval.", "We know that $x$ belongs to the interval $(0.0001,0.001)$ for $\\sin(1/x)$ . We see that when we plug in $x$ into $\\sin(1/x)$ , the argument $(1/x)$ is always from the range $(1000, 10000)$ . Therefore, the problem simply asks for all the zeros of $\\sin(x)$ with $x$ values between $(1000, 10000)$ . We know that the $x$ values of any sine graph is $\\pi(n-1)$ so, we see that values of $n$ are any integer value from $320$ to $3184$ and therefore gives us an answer of approximately $2865$ which is answer $\\boxed{2900}$" ]

[ "It must be assumed that the pipes have an equal height.\nWe can represent the amount of water carried per unit time by cross sectional area.\nCross sectional of Pipe with diameter $6 in$ \\[\\pi r^2 = \\pi \\cdot 3^2 = 9\\pi\\]\nCross sectional area of pipe with diameter $1 in$\n\\[\\pi r^2 = \\pi \\cdot 0.5 \\cdot 0.5 = \\frac{\\pi}{4}\\]\nSo number of 1 in pipes required is the number obtained by dividing their cross sectional areas\n\\[\\frac{9\\pi}{\\frac{\\pi}{4}} = 36\\]\nSo the answer is $\\boxed{36}$" ]

[ "Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be $\\binom{100}{2}=4950$ . However this also counts the 100 sides of the polygon, so the actual answer is $4950-100=\\boxed{4850}$", "The formula for the number of diagonals of a polygon with $n$ sides is $n(n-3)/2$ . Taking $n=100$ , we see that the number of diagonals that may be drawn in this polygon is $100(97)/2$ or $\\boxed{4850}$" ]

[ "We can rewrite this as $2^{32}5^{25}$ . We can also combine some of the factors to introduce factors of $10$ , whose digit count is simple to evaluate because it simply adds $0$ s. Thus, we have $2^{32}5^{25}=2^72^{25}5^{25}=2^710^{25}$ . We can see that this final number is $2^7$ with $25$ $0$ s annexed onto it. $2^7=128$ , which has $3$ digits, so the entire number has $25+3=28$ digits, $\\boxed{28}$" ]

[ "\nAs shown in the diagram above, all nine altitudes, medians, and interior angle bisectors are distinct, except for the three coinciding lines from the vertex opposite to the base. Thusly, there are $7$ distinct lines, so our answer is $\\boxed{7}$ , and we are done." ]

[ "If $x=x^2+y^2$ and $y=2xy$ , then we can break this into two cases.\nCase 1: $y = 0$\nIf $y = 0$ , then $x = x^2$ and $0 = 0$\nTherefore, $x = 0$ or $x = 1$\nThis yields 2 solutions\nCase 2: $x = \\frac{1}{2}$\nIf $x = \\frac{1}{2}$ , this means that $y = y$ , and $\\frac{1}{2} = \\frac{1}{4} + y^2$\nBecause y can be negative or positive, this yields $y = \\frac{1}{2}$ or $y = -\\frac{1}{2}$\nThis yields another 2 solutions.\n$2+2 = \\boxed{4}$" ]

[ "We can simplify $\\sqrt{1984}$ to $8\\sqrt{31}$ . Therefore, the only solutions are $a\\sqrt{31}+b\\sqrt{31}$ such that $a+b=8$ and $0<a<b$ . The only solutions to this are $a=1, b=7; a=2, b=6; a=3, b=5$ . Each of these gives distinct pairs of $(x, y)$ , so there are $3$ pairs, $\\boxed{3}$" ]

[ "Let $x$ be the population in $1996$ , and let $k$ be the constant of proportionality.\nIf $n=1994$ , then the difference in population between $1996$ and $1994$ is directly proportional to the population in $1995$\nTranslating this sentence, $(x - 39) = k(60)$\nSimilarly, letting $n=1995$ gives the sentence $(123 - 60) = kx$\nSince $kx = 63$ , we have $k = \\frac{63}{x}$\nPlugging this into the first equation, we have:\n$(x - 39) = \\frac{60\\cdot 63}{x}$\n$x - 39 = \\frac{3780}{x}$\n$x^2 - 39x - 3780 = 0$\n$(x - 84)(x + 45) = 0$\nSince $x>0$ , we must have $x=84$ , and the answer is $\\boxed{84}$" ]

[ "Say we add $N$ ounces of water to the shaving lotion. Since half of a $9$ ounce bottle of shaving lotion is alcohol, we know that we have $\\frac{9}{2}$ ounces of alcohol. We want $\\frac{9}{2}=0.3(9+N)$ (because we want the amount of alcohol, $\\frac{9}{2}$ , to be $30\\%$ , or $0.3$ , of the total amount of shaving lotion, $9+N$ ). Solving, we find that \\[9=0.6(9+N)\\implies9=5.4+0.6N\\implies3.6=0.6N\\implies6=N.\\] So, the total amount of water we need to add is $\\boxed{6}$", "The concentration of alcohol after adding $n$ ounces of water is $\\frac{4.5}{9+n}$\nTo get a solution of 30% alcohol, we solve $\\frac{4.5}{9+n}=\\frac{3}{10}$\n$45=27+3n$\n$18=3n$\n$6=n \\implies \\boxed{6}$" ]

[ "By the Zero Product Property $x-y+2=0$ or $3x+y-4=0$ in the first equation and $x+y-2=0$ or $2x-5y+7=0$ in the second equation. Thus, from the first equation, $y = x+2$ or $y =-3x+4$ , and from the second equation, $y=-x+2$ or $y = \\frac{2}{5}x + \\frac{7}{5}$\nIf a point is common to the two graphs, then the point must be in one of the lines in the first equation as well as one of the lines in the second equation. Since the slopes of the lines are different, none of the lines are parallel. Thus, there are $2 \\cdot 2 = \\boxed{4}$ points of intersection in the two graphs." ]

[ " The distance between the two parallel tangents is the length of the circle's diameter, so the distance from a point that satisfies the conditions and the two tangents is the length of the circle's radius. From the diagram, there are $\\boxed{3}$ points that satisfies the conditions." ]

[ "Consider the cases $x>0$ and $x<0$ , and also note that by AM-GM, for any positive number $a$ , we have $a+\\frac{17}{a} \\geq 2\\sqrt{17}$ , with equality only if $a = \\sqrt{17}$ . Thus, if $x>0$ , considering each equation in turn, we get that $y \\geq \\sqrt{17}, z \\geq \\sqrt{17}, w \\geq \\sqrt{17}$ , and finally $x \\geq \\sqrt{17}$\nNow suppose $x > \\sqrt{17}$ . Then $y - \\sqrt{17} = \\frac{x^{2}+17}{2x} - \\sqrt{17} = (\\frac{x-\\sqrt{17}}{2x})(x-\\sqrt{17}) < \\frac{1}{2}(x-\\sqrt{17})$ , so that $x > y$ . Similarly, we can get $y > z$ $z > w$ , and $w > x$ , and combining these gives $x > x$ , an obvious contradiction.\nThus we must have $x \\geq \\sqrt{17}$ , but $x \\ngtr \\sqrt{17}$ , so if $x > 0$ , the only possibility is $x = \\sqrt{17}$ , and analogously from the other equations we get $x = y = z = w = \\sqrt{17}$ ; indeed, by substituting, we verify that this works.\nAs for the other case, $x < 0$ , notice that $(x,y,z,w)$ is a solution if and only if $(-x,-y,-z,-w)$ is a solution, since this just negates both sides of each equation and so they are equivalent. Thus the only other solution is $x = y = z = w = -\\sqrt{17}$ , so that we have $2$ solutions in total, and therefore the answer is $\\boxed{2}$" ]

[ "Notice that $a^0=1, a>0$ . So $2^0=1$ . So $2x^2-7x+5=0$ . Evaluating the discriminant, we see that it is equal to $7^2-4*2*5=9$ . So this means that the equation has two real solutions. Therefore, select $\\boxed{1}$" ]

[ "Solution by e_power_pi_times_i\nTake the logarithm with a base of $2$ to both sides, resulting in the equation $2x^2-7x+5 = 0$ . Factoring results in $(2x-5)(x-1) = 0$ , so there are $\\boxed{2}$ real solutions." ]

[ "We know that $2^{6x+3}\\cdot4^{3x+6}=2^{6x+3}\\cdot(2^2)^{3x+6}=2^{6x+3}\\cdot2^{6x+12}=2^{12x+15}$ .\nWe also know that $8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}$ .\nThere are infinite solutions to the equation $2^{12x+15}=2^{12x+15}$ , so the answer is $\\boxed{3}$" ]

[ "We know that the radius of the wheel is $3$ feet, so the total circumference of the wheel is $6\\pi$ feet. We also know that one mile is equivalent to $5280$ feet. It takes $\\frac{5280}{6\\pi}$ revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is $\\boxed{880}$" ]

[ "We can write all possible triangles adding up to 12 or less \\[(2, 4, 5)=11\\] \\[(3, 4, 5)=12\\] \\[(2, 3, 4)=9\\]\nThis leaves $\\boxed{3}$ scalene triangles." ]

[ "If the first number of a group of $n$ consecutive numbers is $a$ , the $n^\\text{th}$ number is $a+n-1$ . We know that the sum of the group of numbers is $100$ , so \\[\\frac{n(2a+n-1)}{2} = 100\\] \\[2a+n-1=\\frac{200}{n}\\] \\[2a = 1-n + \\frac{200}{n}\\] We know that $n$ and $a$ are positive integers, so we check values of $n$ that are a factor of $200$ . Of these values, the only ones that result in a positive integer $a$ is when $n = 5$ or when $n = 8$ , so there are $\\boxed{2}$ sets of two or more consecutive positive integers whose sum is $100$" ]

[ "There are 5 significant digits, $1$ $1$ $0$ $2$ , and $5$ . The answer is $\\boxed{5}$" ]

[ "Solving for $x$ in the equation yields $x =rfthe meaning of theta 0$ . Solving the inequality results in $y \\le 254 \\frac{1}{3}$ . From the two conditions, $y$ can be an odd number from $1$ to $253$ , so there are $127$ solutions where $x$ and $y$ are integers. The answer is $\\boxed{127}$", "We can solve for first solution by applying extended euclids division algorithm or we can apply hit and trial for the first solution to get $x_0$ $380$ and $y_0$ $1$ . then the general solution of the given diophanitine equation will be $x$ $x_0$ $3t$ and $y$ $y_0$ $2t$ . Since we need only positive integer solutions So we solve $380$ $3t$ $>$ $0$ and $1$ $2t$ $>$ $0$ to get $t$ $>$ $0$ (applying Greatest integer function) also we can clearly see that $t_{(min)}$ $=$ $0$ so,t $<$ $GIF$ $383$ $3$ ). That implies $t$ ranges from $0$ to $127$ . Hence,the correct answer is $127$ $\\boxed{127}$" ]

[ "$X$ has to contain $\\{1,~2\\}$ , so only $\\{3,~4,~5\\}$ matters. There are two choices for the elements; the element is either in $X$ or outside of $X$ . With this combinatorics in mind, the answer is simply $2^3=\\boxed{8}.$ ~lopkiloinm" ]

[ "Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference. That means the last term is $a+r(n-1)$\nWe can write an equation on the difference between the last and first term based on the conditions. \\[a+r(n-1)-a =10.5\\] \\[rn-r=10.5\\] Also, half of the terms add up to $24$ while the other half of the terms add up to $30$ , so \\[24 + r\\frac{n}{2} = 30\\] \\[nr = 12\\] Substituting the value back to a previous equation, \\[12-r=10.5\\] \\[r=1.5\\] Substituting to a previous equation again, \\[1.5n-1.5=10.5\\] \\[n=8\\] Thus, there are $\\boxed{8}$ terms in the arithmetic sequence." ]

[ "Use properties of exponents to move the squares outside the brackets use difference of squares.\n\\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\\]\nUsing the binomial theorem, we can see that the number of terms is $\\boxed{5}$" ]

[ "We can create a tree of possibilities, progressively eliminating the absolute value signs by creating different cases.\n\nSo we have $4$ possible solutions: $-2, \\frac{2}{3}, 2,$ and $-\\frac{4}{3}$ . Checking for extraneous solutions, we find that the only ones that work are $2$ and $-\\frac{4}{3}$ , so there are $2$ solutions, $\\boxed{2}$" ]

[ "We can factor the second equation to get $c(a+b)=23$ , so we see that $c$ must be a factor of $23$ , and since this is prime $c=1$ or $c=23$ . However, if $c=23$ , then $a+b=1$ , which is impossible for the field of positive integers. Therefore, $c=1$ for all possible solutions. Substituting this into the original equations gives\n$ab+b=44$\nand\n$a+b=23$\nFrom the second equation, $a=23-b$ , and substituting this into the first equation yields $b(23-b)+b=44$ , or $b^2-24b+44=0$ . Factoring this gives $(b-2)(b-22)=0$ , so $b=2$ or $b=22$ . Both of these yield integer solutions for $a$ , giving $a=21$ or $a=1$ , respectively.\nTherefore, the only solutions are $(21, 2, 1)$ and $(1, 22, 1)$ , yielding $2$ solutions, $\\boxed{2}$" ]

[ "From rule 1, the largest number, $12$ , can be second or third. From rule 2, because there are five places, the smallest number $-2$ can either be third or fourth. The median, $6$ can be second, third, or fourth. Because we know the middle three numbers, the first and last numbers are $4$ and $9$ , disregarding their order. Their average is $(4+9)/2 = \\boxed{6.5}$" ]

[ "It is helpful to consider the cube $ABCDEFGH$ , where the vertices of the cube represent the faces of the octahedron, and the edges of the cube represent adjacent octahedral faces. Each assignment of the numbers $1,2,3,4,5,6,7$ , and $8$ to the faces of the octahedron corresponds to a permutation of $ABCDEFGH$ , and thus to an octagonal circuit of these vertices. The cube has 16 diagonal segments that join nonadjacent vertices. In effect, the problem asks one to count octagonal circuits that can be formed by eight of these diagonals. Six of the diagonals are edges of tetrahedron $ACFH$ , six are edges of tetrahedron $DBEG$ , and four are $\\textit{long}$ , joining a vertex of one tetrahedron to the diagonally opposite point from the other. Notice that each vertex belongs to exactly one long diagonal.\nIt follows that an octagon cannot have two successive long diagonals. Also notice that an octagonal path can jump from one tetrahedron to the other only along one of the long diagonals. It follows that an octagon must contain either 2 long diagonals separated by 3 tetrahedron edges or 4 long diagonals alternating with tetrahedron edges.\nTo form an octagon that contains four long diagonals, choose two opposite edges from tetrahedron $ACFH$ and two opposite edges from tetrahedron $DBEG$ . For each of the three ways to choose a pair of opposite edges from tetrahedron $ACFH$ , there are two possible ways to choose a pair of opposite edges from tetrahedron $DBEG$ . There are 6 distinct octagons of this type and $8\\cdot 2$ ways to describe each of them, making 96 permutations.\nTo form an octagon that contains exactly two of the long diagonals, choose a three-edge path along tetrahedron $ACFH$ , which can be done in $4! = 24$ ways. Then choose a three-edge path along tetrahedron $DBEG$ which, because it must start and finish at specified vertices, can be done in only 2 ways. Since this counting method treats each path as different from its reverse, there are $8\\cdot 24\\cdot 2=384$ permutations of this type.\nIn all, there are $96 +384 = 480$ permutations that correspond to octagonal circuits formed exclusively from cube diagonals. The probability of randomly choosing such a permutation is $\\tfrac{480}{8!}=\\tfrac{1}{84}$ , and $m+n=\\boxed{085}$", "Choose one face of the octahedron randomly and label it with $1$ . There are three faces adjacent to this one, which we will call A-faces. There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face.\nClearly, the labels for the A-faces must come from the set $\\{3,4,5,6,7\\}$ , since these faces are all adjacent to $1$ . There are thus $5 \\cdot 4 \\cdot 3 = 60$ ways to assign the labels for the A-faces.\nThe labels for the B-faces and C-face are the two remaining numbers from the above set, plus $2$ and $8$ . The number on the C-face must not be consecutive to any of the numbers on the B-faces.\nFrom here it is easiest to brute force the $10$ possibilities for the $4$ numbers on the B and C faces:\nThere is a total of $10$ possibilities. There are $3!=6$ permutations (more like \"rotations\") of each, so $60$ acceptable ways to fill in the rest of the octahedron given the $1$ . There are $7!=5040$ ways to randomly fill in the rest of the octahedron. So the probability is $\\frac {60}{5040} = \\frac {1}{84}$ . The answer is $\\boxed{085}$", "Consider the cube formed from the face centers of the regular octahedron. Color the vertices in a checker board fashion. We seek the number of circuits traversing the cube entirely composed of diagonals. Notice for any vertex, it can be linked to at most one different-colored vertex, i.e. its opposite vertex. Thus, there are only two possible types of circuits ( $B$ for black and $W$ for white).\nType I: $BB-WWWW-BB$ . There are $4!$ ways to arrange the black vertices and consequently two of the white vertices and $2!$ ways to arrange the other two white vertices. Since the template has a period of $8$ , there are $4!\\cdot 2!\\cdot 8 = 384$ circuits of type I.\nType II: $B-WW-BB-WW-B$ . There are $4!$ ways to arrange the black vertices and consequently the white vertices. Since the template has a period of $4$ , there are $4! \\cdot 4 = 96$ circuits of type II.\nThus, there are $384+96=480$ circuits satisfying the given condition, out of the $8!$ possible circuits. Therefore, the desired probability is $\\frac{480}{8!} = \\frac{1}{84}$ . The answer is $\\boxed{085}$", "As in the previous solution, consider the cube formed by taking each face of the octahedron as a vertex. Let one fixed vertex be A. Then each configuration (letting each vertex have a number value from 1-8) of A and the three vertices adjacent to A uniquely determine a configuration that satisfies the conditions, i.e. no two vertices have consecutive numbers. Thus, the number of desired configurations is equivalent to the number of ways of choosing the values of A and its three adjacent vertices.\nThe value of A can be chosen in 8 ways, and the 3 vertices adjacent to A can be chosen in $5\\cdot4\\cdot3=60$ ways, since they aren't adjacent to each other, but they can't, after all, be consecutive values to A. For example, if A=1, then the next vertex can't be 1,2 or 8, so there are 5 choices. However, the next vertex also adjacent to A can be chosen in 4 ways; it can't be equal to 1,2,8, or the value of the previously chosen vertex. With the same reasoning, the last such vertex has 3 possible choices.\nThe total number of ways to choose the values of the vertices of the cube independently is 8!, so our probability is thus $\\frac{8\\cdot60}{8!}=\\frac{8\\cdot5\\cdot4\\cdot3}{8!}=\\frac{1}{84}$ , from which the answer is $\\boxed{085}$", "\nThe probability is equivalent to counting the number of Hamiltonian cycles in this 3D graph over $7!.$ This is because each Hamiltonian cycle corresponds to eight unique ways to label the faces. Label the vertices $AR,BR,CR,DR,AX,BX,CX,DX$ where vertices $ab$ and $cd$ are connected if $a=c$ or $b=d.$\nCase 1: Four of the vertical edges are used. $6\\cdot 2=12.$\nCase 2: Two of the vertical edges are used. $4\\cdot 3 \\cdot 2\\cdot 2=48.$\nSo, the probability is $\\frac{60}{5040}=\\frac{1}{84}.$ Therefore, our answer is $\\boxed{085}$", "As with some of the previous solutions, consider the cube formed by connecting the centroids of the faces on the octahedron. We choose a random vertex(hence fixing the diagram), giving us $7!$ ways as our denominator. WLOG, we color this start vertex red, and we color all $3$ vertices adjacent to it blue. We repeat this for the other vertices.\nNote that there is a 1-1 correspondence between the number of valid face numberings with rotational symmetry and the number of ways to move a particle to every vertex of the cube and returning to the start vertex with only diagonal moves. We can move along either short diagonals and long diagonals.\nNext, note that we can only move from the red vertices to the blue vertices and back with long diagonals(since short diagonals keep us on the color we are currently on). Thus, it is easy to check that the only possible sequences of long and short diagonals are cyclic permutations of $SSSLSSSL$ and $SLSLSLSL$\nCase 1: $SSSLSSSL$ .\nThere are $4$ cyclic permutations, and we can traverse the entirety of the red tetrahedron in $4$ steps in $3!$ ways. Then, after the move to the blue tetrahedron from one of the non-starting red vertices, we realize that our first step cannot be to the blue vertex opposite the starting red vertex. Hence, there are $2$ possibilities for our first step. However, once we make our first move, the path is fixed. This leaves us with $4! \\cdot 2$ ways.\nCase 2: $SLSLSLSL$ There are $2$ cyclic permutations in this case. After we choose one of the $3$ red vertices to go to from the starting vertex and move to the blue tetrahedron, we are once again left with $2$ choices to move to by similar logic to the first case. However, after we move back to the red tetrahedron, our choices are fixed. This leaves us with $2 \\cdot 3 \\cdot 2$ ways.\nFinally, summing and dividing by $7!$ gives us $\\frac{60}{7!} = \\frac{1}{84} \\rightarrow \\boxed{085}$ , as desired. - Spacesam" ]

[ "Suppose that the two identical digits are both $1$ . Since the thousands digit must be $1$ , only one of the other three digits can be $1$ . This means the possible forms for the number are\nBecause the number must have exactly two identical digits, $x\\neq y$ $x\\neq1$ , and $y\\neq1$ . Hence, there are $3\\cdot9\\cdot8=216$ numbers of this form.\nNow suppose that the two identical digits are not $1$ . Reasoning similarly to before, we have the following possibilities:\nAgain, $x\\neq y$ $x\\neq 1$ , and $y\\neq 1$ . There are $3\\cdot9\\cdot8=216$ numbers of this form.\nThus the answer is $216+216=\\boxed{432}$", "Consider a sequence of $4$ digits instead of a $4$ -digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is $\\frac{1}{10}$ . This means we can find all possible sequences with one digit repeated twice, and then divide by $10$\nIf we let the three distinct digits of the sequence be $a, b,$ and $c$ , with $a$ repeated twice, we can make a table with all possible sequences:\n\\[\\begin{tabular}{ccc} aabc & abac & abca \\\\ baac & baca & \\\\ bcaa && \\\\ \\end{tabular}\\]\nThere are $6$ possible sequences.\nNext, we can see how many ways we can pick $a$ $b$ , and $c$ . This is $10(9)(8) = 720$ , because there are $10$ digits, from which we need to choose $3$ with regard to order. This means there are $720(6) = 4320$ sequences of length $4$ with one digit repeated. We divide by 10 to get $\\boxed{432}$ as our answer.", "We'll use complementary counting. We will split up into $3$ cases: (1) no number is repeated, (2) $2$ numbers are repeated, and $2$ other numbers are repeated, (3) $3$ numbers are repeated, or (4) $4$ numbers are repeated.\nCase 1:\nThere are $9$ choices for the hundreds digit (it cannot be $1$ ), $8$ choices for the tens digit (it cannot be $1$ or what was chosen for the hundreds digit), and $7$ for the units digit. This is a total of $9\\cdot8\\cdot7=504$ numbers.\nCase 2:\nOne of the three numbers must be $1$ , and the other two numbers must be the same number, but cannot be $1$ (That will be dealt with in case 4). There are $3$ choices to put the $1$ , and there are $9$ choices (any number except for $1$ ) to pick the other number that is repeated, so a total of $3\\cdot9=27$ numbers.\nCase 3: \nWe will split it into $2$ subcases: one where $1$ is repeated $3$ times, and one where another number is repeated $3$ times.\nWhen $1$ is repeated $3$ times, then one of the digits is not $1$ . There are $9$ choices for that number, and $3$ choices for its location,so a total of $9\\cdot3=27$ numbers.\nWhen a number other than $1$ is repeated $3$ times, then there are $9$ choices for the number, and you don't have any choices on where to put that number. \nSo in Case 3 there are $27+9=36$ numbers\nCase 4:\nThere is only $1$ number: $1111$\nThere are a total of $1000$ $4$ -digit numbers that begin with $1$ (from $1000$ to $1999$ ), so by complementary counting you get $1000-(504+27+36+1)=\\boxed{432}$ numbers.", "Let us proceed by casework.\nCase 1:\nWe will count the amount of numbers that have two identical digits that are not one. The thousands digit is fixed, and we are choosing two spots to hold two identical digits that are chosen from $0, 2-9$ , which is $9$ options. For the last digit, their are $8$ possibilities since it can be neither $1$ or the other number that is chosen. The final outcome is ${3}\\choose{2}$ $* 9 * 8 = 216$ possibilities for this case.\nCase 2:\nThe last case will be the amount of numbers that have two identical digits thare are $1$ . There are ${3}\\choose{1}$ places to pick the $1$ . For the other $2$ digits, there are $9$ and $8$ options respectively. Thus, we have $3 * 9 * 8 = 216$ .\nSumming the two cases, we get $216 + 216 = \\boxed{432}$" ]

[ "Let x be the square root of the smaller of the two perfect squares. Then, $(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \\le 2023$ . Thus, $x \\le 1011$ . So there are $\\boxed{1011}$ numbers that satisfy the equation.", "The smallest number that can be expressed as the difference of a pair of consecutive positive squares is $3$ , which is $2^2-1^2$ . The largest number that can be expressed as the difference of a pair of consecutive positive squares that is less than or equal to $2023$ is $2023$ , which is $1012^2-1011^2$ . These numbers are in the form $(x+1)^2-x^2$ , which is just $2x+1$ . These numbers are just the odd numbers from 3 to 2023, so there are $[(2023-3)/2]+1=1011$ such numbers. The answer is $\\boxed{1011}$" ]

[ "The arithmetic mean of the numbers $3, 5, 7, a,$ and $b$ is equal to $\\frac{3+5+7+a+b}{5}=\\frac{15+a+b}{5}=15$ . Solving for $a+b$ , we get $a+b=60$ . Dividing by $2$ to find the average of the two numbers $a$ and $b$ gives $\\frac{60}{2}=\\boxed{30}$" ]

[ "Let $A = \\log(a)$ and $B = \\log(b)$\nThe first three terms of the arithmetic sequence are $3A + 7B$ $5A + 12B$ , and $8A + 15B$ , and the $12^\\text{th}$ term is $nB$\nThus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \\Rightarrow A = 2B$\nSince the first three terms in the sequence are $13B$ $22B$ , and $31B$ , the $k$ th term is $(9k + 4)B$\nThus the $12^\\text{th}$ term is $(9\\cdot12 + 4)B = 112B = nB \\Rightarrow n = 112\\Rightarrow \\boxed{112}$", "If $\\log(a^3b^7)$ $\\log(a^5b^{12})$ , and $\\log(a^8b^{15})$ are in arithmetic progression , then $a^3b^7$ $a^5b^{12}$ , and $a^8b^{15}$ are in geometric progression . Therefore,\n\\[a^2b^5=a^3b^3 \\Rightarrow a=b^2\\]\nTherefore, $a^3b^7=b^{13}$ $a^5b^{12}=b^{22}$ , therefore the 12th term in the sequence is $b^{13+9*11}=b^{112} \\Rightarrow \\boxed{112}$", "Given the first three terms form an arithmetic progression, we have: \\[a = \\log(a^3b^7)\\] \\[a+d = \\log(a^5b^{12})\\] \\[a+2d = \\log(a^8b^{15}).\\] Subtracting the first equation from the second and the third from the second, respectively, gives us these two expressions for $d$ \\[d = \\log \\left( \\frac{a^5b^{12}}{a^3b^7} \\right) = \\log a^2b^5\\] \\[d = \\log \\left( \\frac{a^8b^{15}}{a^5b^{12}} \\right) = \\log a^3b^3.\\] The desired $12$ th term in the sequence is $a+11d$ , so we can substitute our values for $a$ and $d$ (using either one of our two expressions for $d$ ): \\[a+11d = \\log a^3b^7 + 11\\log(a^2b^5)\\] \\[= \\log a^3b^7 + \\log(a^{22}b^{55})\\] \\[= \\log a^{25}b^{62}.\\] The answer must be expressed as $\\log(b^n)$ , however. We're in luck: the two different yet equal expressions for $d$ allow us to express $a$ and $b$ in terms of each other: \\[\\log a^2b^5 = \\log a^3b^3\\] \\[a^2b^5 = a^3b^3\\] \\[a=b^2.\\] Plugging in $a=b^2$ , we have: \\[a+11d = \\log b^{50}b^{62}\\] \\[= \\log b^{112} \\Rightarrow \\boxed{112}.\\] ~ Jingwei325 $\\smiley$" ]

[ "In order to solve the problem, we first need to find each of the three variables. We can use the proportions the problem gives us to find the value of one part, and, by extension, the values of the variables (as $x$ would have $2$ parts, $y$ would have $3$ , and $z$ would have $5$ ). One part, after some algebra, equals $10$ , so $x$ $y$ , and $z$ are $20$ $30$ , and $50$ , respectively.\nWe can plug $x$ and $y$ into the equation given to us: $30 = 20a-10$ , and then solve to get $a = \\boxed{2}$" ]

[ "We fill out the grid, as shown below: From the four numbers that appear in the shaded squares, $\\boxed{3}$ of them are prime: $19,23,$ and $47.$", "Note that given time constraint, it's better to only count from perfect squares (in pink), as shown below: From the four numbers that appear in the shaded squares, $\\boxed{3}$ of them are prime: $19,23,$ and $47.$" ]

[ "If $(x,y)$ denotes the greatest common divisor of $x$ and $y$ , then we have $d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)$ . Now assuming that $d_n$ divides $100+n^2$ , it must divide $2n+1$ if it is going to divide the entire expression $100+n^2+2n+1$\nThus the equation turns into $d_n=(100+n^2,2n+1)$ . Now note that since $2n+1$ is odd for integral $n$ , we can multiply the left integer, $100+n^2$ , by a power of two without affecting the greatest common divisor. Since the $n^2$ term is quite restrictive, let's multiply by $4$ so that we can get a $(2n+1)^2$ in there.\nSo $d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)$ . It simplified the way we wanted it to!\nNow using similar techniques we can write $d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)$ . Thus $d_n$ must divide $\\boxed{401}$ for every single $n$ . This means the largest possible value for $d_n$ is $401$ , and we see that it can be achieved when $n = 200$", "We know that $a_n = 100+n^2$ and $a_{n+1} = 100+(n+1)^2 = 100+ n^2+2n+1$ . Since we want to find the GCD of $a_n$ and $a_{n+1}$ , we can use the Euclidean algorithm\n$a_{n+1}-a_n = 2n+1$\nNow, the question is to find the GCD of $2n+1$ and $100+n^2$ .\nWe subtract $2n+1$ $100$ times from $100+n^2$ \\[(100+n^2)-100(2n+1)\\] \\[=n^2+100-200n-100\\] This leaves us with \\[n^2-200n.\\] Factoring, we get \\[n(n-200)\\] Because $n$ and $2n+1$ will be coprime, the only thing stopping the GCD from being $1$ is $n-200.$ We want this to equal 0, because that will maximize our GCD. \nSolving for $n$ gives us $n=200$ . The last remainder is 0, thus $200*2+1 = \\boxed{401}$ is our GCD." ]

[ "In the arithmetic sequence, let $a$ be the first term and $d$ be the common difference, where $d>0.$ The sum of the first six terms is \\[a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.\\] We are given that \\begin{align*} 6a+15d &= 990, \\\\ 2a &= a+5d. \\end{align*} The second equation implies that $a=5d.$ Substituting this into the first equation, we get \\begin{align*} 6(5d)+15d &=990, \\\\ 45d &= 990 \\\\ d &= 22. \\end{align*} It follows that $a=110.$ Therefore, the greatest number of apples growing on any of the six trees is $a+5d=\\boxed{220}.$", "Let the terms in the sequence be defined as \\[a_1, a_2, ..., a_6.\\]\nSince this is an arithmetic sequence, we have $a_1+a_6=a_2+a_5=a_3+a_4.$ So, \\[\\sum_{i=1}^6 a_i=3(a_1+a_6)=990.\\] Hence, $(a_1+a_6)=330.$ And, since we are given that $a_6=2a_1,$ we get $3a_1=330\\implies a_1=110$ and $a_6=\\boxed{220}.$" ]

[ "The only possibilities for the numbers are $11,12,13,14,15,16$ and $10,11,12,13,14,15$\nIn the second case, the common sum would be $(10+11+12+13+14+15)/3=25$ , so $11$ must be paired with $14$ , which\nit isn't.\nThus, the only possibility is the first case and the sum of the six numbers is $81\\rightarrow \\boxed{81}$" ]

[ "We can execute Dijkstra's algorithm by hand to find the shortest path from $A$ to every other town, including $Z$ . This effectively proves that, assuming we execute the algorithm correctly, that we will have found the shortest distance. The distance estimates for each step of the algorithm (from $A$ to each node) are shown below:\n\\[\\begin{array}{|c|c|c|c|c|c|c|} \\hline \\text{Current node} & A & M & C & X & Y & Z \\\\ \\hline A & 0 & 8 & \\infty & 5 & \\infty & \\infty \\\\ X & 0 & 7 & \\infty & 5 & 15 & \\infty \\\\ M & 0 & 7 & 21 & 5 & 13 & 32 \\\\ Y & 0 & 7 & 18 & 5 & 13 & 30 \\\\ C & 0 & 7 & 18 & 5 & 13 & 28 \\\\ Z & 0 & 7 & 18 & 5 & 13 & \\textbf{28} \\\\ \\hline \\end{array}\\] The steps are as follows: starting with the initial node $A$ , set $d(A)=0$ and $d(v)=\\infty$ for all $v \\in \\{M,C,X,Y,Z\\}$ where $d(v)$ indicates the distance from $A$ to $v$ . Consider the outgoing edges $(A,X)$ and $(A,M)$ and update the distance estimates $d(X)=5$ and $d(M)=8$ , completing the first row of the table.\nThe node $X$ is the unvisited node with the lowest distance estimate, so we will consider $X$ and its outgoing edges $(X,Y)$ and $(X,M)$ . The distance estimate $d(Y)$ equals $d(X)+10=15$ , and the distance estimate $d(M)$ updates to $d(X)+2=7$ , because $7 < 8$ . This completes the second row of the table. Repeating this process for each unvisited node (in order of its distance estimate) yields the correct distance $d(Z) = \\boxed{28}$ once the algorithm is complete.", "From $A$ , we want to find the shortest route to $Z$ , so we want to try to find the shortest path through each node (not necessarily all of them). We should follow the arrows, since all of them lead to $Z$ . From $A$ , there are $2$ paths we can take, to $M$ $(8)$ , or to $X$ $(5)$ . We travel to $X$ , since $5 < 8$ . From $X$ , we go to $M$ $(2 < 10)$ , we go to $Y$ $6 < 14 < 25$ , we go to $C$ $(5 < 17)$ and finally go to $Z$ . Adding up gives $5 + 2 + 6 + 5 + 10 = 28 = \\boxed{28}$", "We can cross off a few routes:\nFinally, we are left with a single path AXMYCZ from A to Z which adding it up gives $28 = \\boxed{28}.$" ]

[ "We can simply see that path $A \\rightarrow X \\rightarrow M \\rightarrow Y \\rightarrow C \\rightarrow Z$ will give us the smallest value. Adding, $5+2+6+5+10 = \\boxed{28}$ . This is nice as it’s also the smallest value, solidifying our answer." ]

[ "Denote the page number as $x$ , with $x < n$ . The sum formula shows that $\\frac{n(n + 1)}{2} + x = 1986$ . Since $x$ cannot be very large, disregard it for now and solve $\\frac{n(n+1)}{2} = 1986$ . The positive root for $n \\approx \\sqrt{3972} \\approx 63$ . Quickly testing, we find that $63$ is too large, but if we plug in $62$ we find that our answer is $\\frac{62(63)}{2} + x = 1986 \\Longrightarrow x = \\boxed{033}$", "Use the same method as above where you represent the sum of integers from $1$ to $n$ expressed as $\\frac{n(n + 1)}{2}$ , plus the additional page number $k$ . We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.)\n$\\frac{n(n + 1)}{2} = 1986$ is the quadratic you must solve to obtain the upper bound of $n$ and $\\frac{n(n + 1)}{2} + n = 1986$ is the quadratic you must solve to obtain the lower bound of $n$\nSolving the two equations gives values that are respectively around $62.5$ and $61.5$ with the quadratic formula, and the only integer between the two is $62$\nThis implies that we can plug in $62$ and come to the same conclusion as the above solution where $x = \\boxed{033}$" ]

[ "The axis of $P$ is inclined at an angle $\\theta$ relative to the coordinate axis, where $\\tan\\theta = \\tfrac 34$ . We rotate the coordinate axis by angle $\\theta$ anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let $(\\widetilde{x}, \\widetilde{y})$ be the coordinates in the rotated system. Then $(x,y)$ and $(\\widetilde{x}, \\widetilde{y})$ are related by \\begin{align} \\nonumber x = \\widetilde{x}\\cos\\theta -\\widetilde{y}\\sin\\theta &= \\tfrac 45 \\widetilde{x} - \\tfrac 35 \\widetilde{y}, \\\\ y = \\widetilde{x}\\sin\\theta +\\widetilde{y}\\cos\\theta &= \\tfrac 35 \\widetilde{x} + \\tfrac 45 \\widetilde{y} \\end{align} In the rotated coordinate system, the parabola has focus at $(0,0)$ and the two points on it are at $(5,0)$ and $(-5,0)$ . Therefore, the directrix is $\\widetilde{y}=\\pm 5$ ; we can, WLOG, choose $\\widetilde{y}=-5$ . For a point on the parabola, it is equidistant from the focus and directrix, so the equation of the parabola is \\begin{align}\\tag{2} \\widetilde{x}^2+\\widetilde{y}^2 = (\\widetilde{y}+5)^2 \\qquad &\\Longrightarrow\\qquad \\widetilde{y} = \\tfrac{1}{10}(\\widetilde{x}^2-25) \\end{align} From $(1)$ we have $|4x+3y|=5\\widetilde{x}$ , so we need $|\\widetilde{x}|<200$ . Substituting $(2)$ in $(1)$ , we get \\begin{align*} 50x &= 40 \\widetilde{x} - 3 \\widetilde{x}^2 + 75, \\\\ 50y &= 30 \\widetilde{x} + 4 \\widetilde{x}^2 - 100 \\end{align*} For $x$ to be an integer $\\widetilde{x}$ must be a multiple of 5; setting $\\widetilde{x}=5a$ we get \\[2x = 8a - 3 a^2 + 3\\] Now we need $a$ to be odd, i.e. $\\widetilde{x}=5a$ is an odd multiple of $5$ , in which case we get $y = 3 a + 2 a^2 - 2$ , which is also an integer. The values that satisfy the given conditions correspond to $\\widetilde{x}= \\{\\pm 5\\cdot (2k-1)\\mid k = 0, 1, \\ldots , 19 \\}={-195, -185, -175, ..., 195}$ , and there are $\\boxed{40}$ such numbers." ]

[ "A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p)^2+p$ . Because the $y$ -intercept is $(0,-p)$ and $p\\ne 0$ , it follows that $a=-2/p$ . Thus \\[y=-\\frac{2}{p}(x^2-2px+p^2)+p=-\\frac{2}{p}x^2+4x-p,\\] so $\\boxed{4}$" ]

[ "Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by $4 - (-2)$ (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are $(2, 0), (-2, 0).$ Then $0 = 4a - 2 \\rightarrow a = 0.5$ , and $0 = 4 - 4b \\rightarrow b = 1.$ Then $a + b = \\boxed{1.5}$", "The parabolas must intersect the x-axis at the same two points for the kite to form. We find the x values at which they intersect by equating them and solving for x as shown below. $y = ax^2-2$ and $y = 4 - bx^2\\rightarrow ax^2-2 = 4-bx^2\\rightarrow (a+b)x^2 = 6 \\rightarrow x = +\\sqrt{\\dfrac{6}{a+b}}$ or $-\\sqrt{\\dfrac{6}{a+b}}$ . The x-values of the y-intercepts is 0, so we plug in zero in each of them and get $-2$ and $4$ . The area of a kite is $\\dfrac{d_1*d_2}{2}$ . The $d_1$ is $2+4 = 6$ . The $d_2$ is $2\\sqrt{\\dfrac{6}{a+b}}$ . Solving for the area $\\rightarrow \\dfrac{1}{2}*(6)*(2*\\sqrt{\\dfrac{6}{a+b}}) = 12 \\rightarrow (2*\\sqrt{\\dfrac{6}{a+b}}) = 4 \\rightarrow (\\sqrt{\\dfrac{6}{a+b}}) = 2 \\rightarrow \\dfrac{6}{a+b} = 4 \\rightarrow \\dfrac{6}{4} = (a+b)$ , therefore $a + b = \\boxed{1.5}$" ]

[ "Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\\left(\\frac{d-c}{a-b},\\frac{ad-bc}{a-b}\\right),\\left(\\frac{c-d}{a-b},\\frac{bc-ad}{a-b}\\right)$ . Because $72= 4\\cdot 18$ , we have that $4(c-d)\\left(\\frac{c-d}{a-b}\\right) = (c+d)\\left(\\frac{c+d}{a-b}\\right)$ or that $2(c-d)=c+d$ , which gives $c=3d$ (consider a homothety , or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by $4\\times$ , it follows that the stretch along the diagonal, or the ratio of side lengths, is $2\\times$ ). The area of the triangular half of the parallelogram on the right side of the y-axis is given by $9 = \\frac{1}{2} (c-d)\\left(\\frac{d-c}{a-b}\\right)$ , so substituting $c = 3d$\nThus $3|d$ , and we verify that $d = 3$ $a-b = 2 \\Longrightarrow a = 3, b = 1$ will give us a minimum value for $a+b+c+d$ . Then $a+b+c+d = 3 + 1 + 9 + 3 = \\boxed{16}$", "The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines $c,d,(b-a)x+c,(b-a)x+d$ and $c,-d,(b-a)x+c,(b-a)x-d$ . Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides $d-c$ and $\\frac{d-c}{b-a}$ $\\frac{(d-c)^2}{b-a}=18$ , and the area contained by the latter is $\\frac{(c+d)^2}{b-a}=72$ . Thus, $d=3c$ and $b-a$ must be even if the former quantity is to equal $18$ $c^2=18(b-a)$ so $c$ is a multiple of $3$ . Putting this all together, the minimal solution for $(a,b,c,d)=(3,1,3,9)$ , so the sum is $\\boxed{16}$", "Let $a$ and $b$ be the slopes of the lines such that $b > a$ (i.e. the line $bx+c$ is steeper than $ax+c$ ) and $c > d$ (i.e. the point $(0, c)$ is higher than the point $(0, d)$ . Upon drawing a diagram, we see that both the smaller and the larger parallelogram can be split along the x-axis, such that both of their areas are the combinations of two corresponding triangles. The area of a triangle is $\\frac{bh}{2}$ , but since a parallelogram is two such triangles, the area becomes $bh$\nLet $b_1$ and $h_1$ denote the base length and height, respectively, of the triangles pertaining to the smaller parallelogram and $b_2$ and $h_2$ denote those of the larger parallelogram. Notice that $b_1$ is simple the distance from $(0, d)$ to $(0, c)$ , or $(c-d)$ . Also notice that $h_1$ is the distance from the $x$ -axis to the intersection of lines $ax+c$ and $bx+d$ . This is equivalent to the value of the $x$ -coordinate of intersection, so we solve for $x$\n$ax+c=bx+d$\n$\\Rightarrow bx-ax = c-d$\n$\\Rightarrow (b-a)x = c-d$\n$\\Rightarrow x = \\frac{c-d}{b-a}$\nThe area of the smaller parallelogram is $b_1*h_1$ , or\n$(c-d) * \\frac{c-d}{b-a}$\n$\\Rightarrow \\frac{(c-d)^2}{b-a}$\n$b_2$ is the distance from $(0, -d)$ to $(0, c)$ , or $(c+d)$ $h_2$ is the $x$ -coordinate of the intersection of the lines $ax+c$ and $bx-d$ . Again, we solve for x:\n$ax+c=bx-d$\n$\\Rightarrow bx-ax = c+d$\n$\\Rightarrow (b-a)x = c+d$\n$\\Rightarrow x = \\frac{c+d}{b-a}$\nThe area of the larger parallelogram is $b-1*h_1$ , or\n$(c+d) * \\frac{c+d}{b-a}$\n$\\Rightarrow \\frac{(c+d)^2}{b-a}$\nThe areas of the parallelograms are given to us: $18$ and $72$ . Therefore we can set up a ratio:\n$\\frac{18}{72} = \\frac{\\frac{(c-d)^2}{b-a}}{\\frac{(c+d)^2}{b-a}}$\n$\\Rightarrow 18(c+d)^2 = 72(c-d)^2$\n$\\Rightarrow (c+d)^2 = 4(c-d)^2$\n$\\Rightarrow c^2 + 2cd + d^2 = 4c^2 - 8cd + 4d^2$\n$\\Rightarrow 3c^2 - 10cd +3d^2 = 0$\n$\\Rightarrow (3c-d)(c-3d)=0$\n$\\Rightarrow c=3d, c=\\frac{d}{3}$\nWe established earlier that $c>d$ , so $c=3d$ . Plugging this into the intial equations yields\n$\\frac{16d^2}{b-a} = 72$\nand\n$\\frac{4d^2}{b-a} = 18$\nSolving for $d$ , we get\n$d = 3\\sqrt{\\frac{b-a}{2}}$\nWe want the sum of $a$ $b$ $c$ , and $d$ . We can now rewrite this\n$a + b + 12\\sqrt{\\frac{b-a}{2}}$\nWe are told that $a$ $b$ $c$ , and $d$ are all positive integers. Therefore the value under the radical must be a perfect square greater than 0. We can rewrite this\n$\\frac{b-a}{2} = k^2$\nwhere $k$ is some positive integer.\nRearranging, we get\n$b = a + 2k^2$\nNow we can rewrite the sum as\n$a + a + 2k^2 +12k$\nSince both $a$ and $k$ must be at least $1$ , the minimum value is\n$1 + 1 + 2(1)^2 + 12(1) = 16 \\Rightarrow \\boxed{16}$" ]

[ "$M-N$ is the amount by which $M$ is greater than $N$ . We divide this by $N$ to get the percent by which $N$ is increased in the form of a decimal, and then multiply by $100$ to make it a percentage. Therefore, the answer is $\\boxed{100}$" ]

[ "Let $a$ be the sidelength of one square, and $b$ be the sidelength of the other, where $a>b$ . If the perimeter of one is $3$ times the other's, then $a=3b$ . The area of the larger square over the area of the smaller square is\n\\[\\frac{a^2}{b^2} = \\frac{(3b)^2}{b^2} = \\frac{9b^2}{b^2} = \\boxed{9}\\]" ]

[ "You might have not seen this coming but there is a very simple way to do this. If we try to make a rectangle out of this, we have to take out both of the lines that are taking out part of the rectangle we want to make. but now we see that to finish the rectangle, we have to use those same irregular lines!\n\nSo all we have to do is find the perimeter of the shape as if it would be a rectangle. After that, we get $\\boxed{28}$" ]

[ "Let the circle intersect $\\overline{PM}$ at $B$ . Then note $\\triangle OPB$ and $\\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point . Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have \\[\\frac{19}{AM} = \\frac{152-2AM-19+19}{152} = \\frac{152-2AM}{152}\\] Solving, $AM = 38$ . So the ratio of the side lengths of the triangles is 2. Therefore, \\[\\frac{PB+38}{OP}= 2 \\text{ and } \\frac{OP+19}{PB} = 2\\] so $2OP = PB+38$ and $2PB = OP+19.$ Substituting for $PB$ , we see that $4OP-76 = OP+19$ , so $OP = \\frac{95}3$ and the answer is $\\boxed{098}$", "Reflect triangle $PAM$ across line $AP$ , creating an isoceles triangle. Let $x$ be the distance from the top of the circle to point $P$ , with $x + 38$ as $AP$ . Given the perimeter is 152, subtracting the altitude yields the semiperimeter $s$ of the isoceles triangle, as $114 - x$ . The area of the isoceles triangle is:\n$[PAM] = r \\cdot s$\n$[PAM] = 19 \\cdot (114 - x)$\nNow use similarity, draw perpendicular from $O$ to $PM$ , name the new point $D$ . Triangle $PDO$ is similar to triangle $PAM$ , by AA Similarity. Equating the legs, we get:\n$\\frac{\\sqrt{x}}{19} = \\frac{\\sqrt{x + 38}}{AM}$\nSolving for $AM$ , it yields $19 \\cdot \\sqrt{\\frac{x + 38}{x}}$\n$19 \\cdot (114 - x) = AM \\cdot AP = 19 \\cdot (x + 38) \\cdot \\sqrt{\\frac{x + 38}{x}}$\nThe $x^3$ cancels, yielding a quadratic. Solving yields $x = \\frac{38}{3}$ .\nAdd $19$ to find $OP$ , yielding $\\frac{95}{3}$ or $\\boxed{098}$", "Let the foot of the perpendicular from $O$ to $PM$ be $D;$ now $OD=19.$ Also let $AM=x$ and $PM=y.$ This means that $OP=\\frac{y}{x}\\cdot 19$ , since $O$ is on the angle bisector of $\\angle M.$\nWe have that $\\tan(\\angle AMO)=\\frac{19}{x},$ so \\[\\tan(\\angle M)=\\tan (2\\cdot \\angle AMO)=\\frac{38x}{x^{2}-361}.\\]\nHowever $\\tan(\\angle M)=\\frac{PA}{AM}=\\frac{PO+OA}{AM}=\\frac{\\frac{y}{x}\\cdot 19 + 19}{x}$ , so \\[\\frac{38x}{x^{2}-361}=19\\cdot \\frac{\\frac{y}{x}+1}{x}\\] \\[\\frac{2x^{2}}{x^{2}-361}=\\frac{y}{x}+1\\] \\[\\frac{x^{2}+361}{x^{2}-361}=\\frac{y}{x}.\\] \\[x\\cdot \\frac{x^{2}+361}{x^{2}-361}=y\\]\nWe now use the fact that the perimeter of $\\triangle PAM$ is $152$ \\[PO+OA+AM+MP=152\\] \\[\\frac{y}{x}\\cdot 19+19+x+y=152\\] \\[19\\left(\\frac{x^{2}+361}{x^{2}-361}\\right)+x\\cdot \\left(\\frac{x^{2}+361}{x^{2}-361}\\right)+x+19=152\\] \\[(x+19)\\left(\\frac{x^{2}+361}{x^{2}-361}+\\frac{x^{2}-361}{x^{2}-361}\\right)=152\\] \\[\\frac{2x^{2}}{x-19}=152\\] \\[x^{2}-76x+19\\cdot 76=0.\\] This quadratic factors as $(x-38)^{2}=0,$ so $x=38$ , and \\[\\frac{y}{x}=\\frac{38^{2}+361}{38^{2}-361}=\\frac{5}{3}\\] \\[OP=\\frac{y}{x}\\cdot 19=\\frac{95}{3}\\to \\boxed{98.}\\]" ]

[ "In the first school, $2000 \\cdot 22\\% = 2000 \\cdot 0.22 = 440$ students prefer tennis.\nIn the second school, $2500 \\cdot 40\\% = 2500 \\cdot 0.40 = 1000$ students prefer tennis.\nIn total, $440 + 1000 = 1440$ students prefer tennis out of a total of $2000 + 2500 = 4500$ students\nThis means $\\frac{1440}{4500}\\cdot 100\\% = \\frac{32}{100} \\cdot 100\\% = 32\\%$ of the students in both schools prefer tennis, giving answer $\\boxed{32}$" ]

[ "For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a $3:2$ ratio. Therefore, assume they made $3x$ and $2x$ two- and three- point shots, respectively, and thus $3x+1$ free throws. The total number of points is \\[2 \\times (3x) + 3 \\times (2x) + 1 \\times (3x+1) = 15x+1\\]\nSet that equal to $61$ , we get $x = 4$ , and therefore the number of free throws they made $3 \\times 4 + 1 = 13 \\Rightarrow \\boxed{13}$", "Let $x$ be the number of free throws. Then the number of points scored by two-pointers is $2(x-1)$ and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is $x+4(x-1) = 61 \\Rightarrow x=13$ , giving us $\\boxed{13}$ for an answer.", "We let $a$ be the number of $2$ -point shots, $b$ be the number of $3$ -point shots, and $x$ be the number of free throws. We are looking for $x.$ We know that $2a=3b$ , and that $x=a+1$ . Also, $2a+3b+1x=61$ . We can see\n\\begin{align*} a&=x-1 \\\\ 2a &= 2x-2 \\\\ 3a &= 2x-2. \\\\ \\end{align*}\nPlugging this into $2a+3b+1x=61$ , we see\n\\begin{align*} 2x-2+2x-2+x &= 61 \\\\ 5x-4 &= 61 \\\\ 5x &= 65 \\\\ x &= \\boxed{13}" ]

[ "For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a $3:2$ ratio. Therefore, assume they made $3x$ and $2x$ two- and three- point shots, respectively, and thus $3x+1$ free throws. The total number of points is \\[2 \\times (3x) + 3 \\times (2x) + 1 \\times (3x+1) = 15x+1\\]\nSet that equal to $61$ , we get $x = 4$ , and therefore the number of free throws they made $3 \\times 4 + 1 = 13 \\Rightarrow \\boxed{13}$", "Let $x$ be the number of free throws. Then the number of points scored by two-pointers is $2(x-1)$ and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is $x+4(x-1) = 61 \\Rightarrow x=13$ , giving us $\\boxed{13}$ for an answer.", "We let $a$ be the number of $2$ -point shots, $b$ be the number of $3$ -point shots, and $x$ be the number of free throws. We are looking for $x.$ We know that $2a=3b$ , and that $x=a+1$ . Also, $2a+3b+1x=61$ . We can see\n\\begin{align*} a&=x-1 \\\\ 2a &= 2x-2 \\\\ 3a &= 2x-2. \\\\ \\end{align*}\nPlugging this into $2a+3b+1x=61$ , we see\n\\begin{align*} 2x-2+2x-2+x &= 61 \\\\ 5x-4 &= 61 \\\\ 5x &= 65 \\\\ x &= \\boxed{13}" ]

[ "$(-1,-2)$ is $4$ units west and $3$ units south of $(3,1)$ . Performing a counterclockwise rotation of $270^{\\circ}$ , which is equivalent to a clockwise rotation of $90^{\\circ}$ , the answer is $3$ units west and $4$ units north of $(3,1)$ , or $\\boxed{0,5}$", "We write $(-1, -2)$ as $-1-2i.$ We'd like to rotate about $(3, 1),$ which is $3+i$ in the complex plane, by an angle of $270^{\\circ} = \\frac{3\\pi}{2} \\text{ rad}$ counterclockwise.\nThe formula for rotating the complex number $z$ about the complex number $w$ by an angle of $\\theta$ counterclockwise is given as $(z-w)e^{\\theta i} + w.$ Plugging in our values $z = -1-2i, w = 3+i, \\theta = \\frac{3\\pi}{2}$ , we evaluate the expression as $((-1-2i) - (3+i))e^{\\frac{3\\pi i}{2}} + (3+i) = (-4-3i)(-i) + (3+i) = 4i-3i^2+3+i = 4i-3+3+i = 5i,$ which corresponds to $\\boxed{0,5}$ on the Cartesian plane." ]

[ "The entire situation is in the picture below. The correct answer is $\\boxed{3,2}$" ]

[ "Because all the central angles of a circle add up to $360^\\circ,$\n\\begin{align*} \\angle BOC + \\angle AOB + \\angle AOC &= 360\\\\ 120 + 140 + \\angle AOC &= 360\\\\ \\angle AOC &= 100. \\end{align*}\nTherefore, the measure of $\\text{arc}AC$ is also $100^\\circ.$ Since the measure of an inscribed angle is equal to half the measure of the arc it intercepts, $\\angle ABC = \\boxed{50}$" ]

[ "The final image of $P$ is $(-6,3)$ . We know the reflection rule for reflecting over $y=-x$ is $(x,y) \\rightarrow (-y, -x)$ . So before the reflection and after rotation the point is $(-3,6)$\nBy definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$ . The first slope is $\\frac{5-6}{1-(-3)} = \\frac{-1}{4}$ . This means the slope of $P$ and $(1,5)$ is $4$\nRotations also preserve distance to the center of rotation, and since we only \"travelled\" up and down by the slope once to get from $(3,-6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$\nTherefore point $P$ is located at $(1+1, 5+4) = (2,9)$ . The answer is $9-2 = 7 = \\boxed{7}$", "Let us reconstruct that coordinate plane as the complex plane. Then, the point $P(a, b)$ becomes $a+b\\cdot{i}$ . \nA $90^\\circ$ rotation around the point $(1, 5)$ can be done by translating the point $(1, 5)$ to the origin, rotating around the origin by $90^\\circ$ , and then translating the origin back to the point $(1, 5)$ \\[a+b\\cdot{i} \\implies (a-1)+(b-5)\\cdot{i} \\implies ((a-1)+(b-5)\\cdot{i})\\cdot{i} = 5-b+(a-1)i \\implies 5+1-b+(a-1+5)i = 6-b+(a+4)i.\\] By basis reflection rules, the reflection of $(-6, 3)$ about the line $y = -x$ is $(-3, 6)$ . \nHence, we have \\[6-b+(a+4)i = -3+6i \\implies b=9, a=2,\\] from which $b-a = 9-2 = \\boxed{7}$", "Using the same method as in Solution 1, we can obtain that the point before the reflection is $(-3,6)$ . \nIf we let the original point be $(x, y)$ , then we can use that the starting point is $(1,5)$ to obtain two vectors $\\langle -4,1 \\rangle$ and $\\langle x-1, y-5 \\rangle$ .\nWe know that two vectors are perpendicular if their dot product is equal to $0$ , and that both points are the same distance ( $\\sqrt {17}$ ) from $(1,5)$\nTherefore, we can write two equations using these vectors: $(x-1)^2 + (y-5)^2 = 17$ (from distance and pythagorean theorem) and $-4x+y-1 = 0$ (from dot product)\nSolving, we simplify the second equation to $y=4x+1$ , and plug it into the first equation.\nWe obtain $(x-1)^2 + (4x-4)^2 = 17$ . We can simplify this to the quadratic $17x^2-34x=0$ . \nWhen we factor out $17x$ , we find that $x = 2$ or $x = 0$ . However, $x$ cannot equal $0$ .\nTherefore, $x = 2$ , and plugging this into the second equation gives us that $y = 9$ . \nSince the point is $(9, 2)$ , we compute $9-2 = \\boxed{7}$", "Using the same method as in Solution 1 reflecting $(-6,3)$ about the line $y = -x$ gives us $(-3,6).$\nLet the original point be $\\langle x,y \\rangle.$ From point $(1,5),$ we form the vectors $\\langle -4,1 \\rangle$ and $\\langle x-1, y-5 \\rangle$ that extend out from the initial point. If they are perpendicular, we know that their dot product has to equal zero. Therefore, \\[\\langle -4,1 \\rangle \\cdot \\langle x-1, y-5 \\rangle = 0 \\implies -4x+y-1= 0.\\] Now, we have to do some guess and check from the multiple choices. Let $y - x = A$ where $A$ is one of the answer choices. Then, $A -3x = 1.$ By intuition and logical reasoning we deduce that $A$ must be $1 \\pmod 3$ so that brings our potential answers down to $\\text{\\textbf{(A)}}$ and $\\text{\\textbf{(C)}}.$ If $A = 1$ from $\\text{\\textbf{(A)}},$ then $x = 0,$ which we can quickly rule out since we know thar $P$ rotated counterclockwise not clockwise. Hence, $\\boxed{7}$ is the answer." ]

[ "The final image of $P$ is $(-6,3)$ . We know the reflection rule for reflecting over $y=-x$ is $(x,y) \\rightarrow (-y, -x)$ . So before the reflection and after rotation the point is $(-3,6)$\nBy definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$ . The first slope is $\\frac{5-6}{1-(-3)} = \\frac{-1}{4}$ . This means the slope of $P$ and $(1,5)$ is $4$\nRotations also preserve distance to the center of rotation, and since we only \"travelled\" up and down by the slope once to get from $(3,-6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$\nTherefore point $P$ is located at $(1+1, 5+4) = (2,9)$ . The answer is $9-2 = 7 = \\boxed{7}$", "Let us reconstruct that coordinate plane as the complex plane. Then, the point $P(a, b)$ becomes $a+b\\cdot{i}$ . \nA $90^\\circ$ rotation around the point $(1, 5)$ can be done by translating the point $(1, 5)$ to the origin, rotating around the origin by $90^\\circ$ , and then translating the origin back to the point $(1, 5)$ \\[a+b\\cdot{i} \\implies (a-1)+(b-5)\\cdot{i} \\implies ((a-1)+(b-5)\\cdot{i})\\cdot{i} = 5-b+(a-1)i \\implies 5+1-b+(a-1+5)i = 6-b+(a+4)i.\\] By basis reflection rules, the reflection of $(-6, 3)$ about the line $y = -x$ is $(-3, 6)$ . \nHence, we have \\[6-b+(a+4)i = -3+6i \\implies b=9, a=2,\\] from which $b-a = 9-2 = \\boxed{7}$", "Using the same method as in Solution 1, we can obtain that the point before the reflection is $(-3,6)$ . \nIf we let the original point be $(x, y)$ , then we can use that the starting point is $(1,5)$ to obtain two vectors $\\langle -4,1 \\rangle$ and $\\langle x-1, y-5 \\rangle$ .\nWe know that two vectors are perpendicular if their dot product is equal to $0$ , and that both points are the same distance ( $\\sqrt {17}$ ) from $(1,5)$\nTherefore, we can write two equations using these vectors: $(x-1)^2 + (y-5)^2 = 17$ (from distance and pythagorean theorem) and $-4x+y-1 = 0$ (from dot product)\nSolving, we simplify the second equation to $y=4x+1$ , and plug it into the first equation.\nWe obtain $(x-1)^2 + (4x-4)^2 = 17$ . We can simplify this to the quadratic $17x^2-34x=0$ . \nWhen we factor out $17x$ , we find that $x = 2$ or $x = 0$ . However, $x$ cannot equal $0$ .\nTherefore, $x = 2$ , and plugging this into the second equation gives us that $y = 9$ . \nSince the point is $(9, 2)$ , we compute $9-2 = \\boxed{7}$", "Using the same method as in Solution 1 reflecting $(-6,3)$ about the line $y = -x$ gives us $(-3,6).$\nLet the original point be $\\langle x,y \\rangle.$ From point $(1,5),$ we form the vectors $\\langle -4,1 \\rangle$ and $\\langle x-1, y-5 \\rangle$ that extend out from the initial point. If they are perpendicular, we know that their dot product has to equal zero. Therefore, \\[\\langle -4,1 \\rangle \\cdot \\langle x-1, y-5 \\rangle = 0 \\implies -4x+y-1= 0.\\] Now, we have to do some guess and check from the multiple choices. Let $y - x = A$ where $A$ is one of the answer choices. Then, $A -3x = 1.$ By intuition and logical reasoning we deduce that $A$ must be $1 \\pmod 3$ so that brings our potential answers down to $\\text{\\textbf{(A)}}$ and $\\text{\\textbf{(C)}}.$ If $A = 1$ from $\\text{\\textbf{(A)}},$ then $x = 0,$ which we can quickly rule out since we know thar $P$ rotated counterclockwise not clockwise. Hence, $\\boxed{7}$ is the answer." ]

[ "The line $y = 2000$ is a horizontal line located $12$ units beneath the point $(1000, 2012)$ . When a point is reflected about a horizontal line, only the $y$ - coordinate will change. The $x$ - coordinate remains the same. Since the $y$ -coordinate of the point is $12$ units above the line of reflection, the new $y$ - coordinate will be $2000 - 12 = 1988$ . Thus, the coordinates of the reflected point are $(1000, 1988)$ $\\boxed{1000,1988}$" ]

[ "Consider the points on the complex plane . The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:\n\\[(a+11i)\\left(\\mathrm{cis}\\,60^{\\circ}\\right) = (a+11i)\\left(\\frac 12+\\frac{\\sqrt{3}i}2\\right)=b+37i.\\]\nEquating the real and imaginary parts, we have:\n\\begin{align*}b&=\\frac{a}{2}-\\frac{11\\sqrt{3}}{2}\\\\37&=\\frac{11}{2}+\\frac{a\\sqrt{3}}{2} \\end{align*}\nSolving this system, we find that $a=21\\sqrt{3}, b=5\\sqrt{3}$ . Thus, the answer is $\\boxed{315}$", "Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? $\\sqrt{3}$ and perpendiculars inspires this solution:\nFirst, drop a perpendicular from $O$ to $AB$ . Call this midpoint of $AB M$ . Thus, $M=\\left(\\frac{a+b}{2}, 24\\right)$ . The vector from $O$ to $M$ is $\\left[\\frac{a+b}{2}, 24\\right]$ . Meanwhile from point $M$ we can use a vector with $\\frac{\\sqrt{3}}{3}$ the distance; we have to switch the $x$ and $y$ and our displacement is $\\left[8\\sqrt{3}, \\frac{(a+b)\\sqrt{3}}{6}\\right]$ . (Do you see why we switched $x$ and $y$ due to the rotation of 90 degrees?)\nWe see this displacement from $M$ to $A$ is $\\left[\\frac{a-b}{2}, 13\\right]$ as well. Equating the two vectors, we get $a+b=26\\sqrt{3}$ and $a-b=16\\sqrt{3}$ . Therefore, $a=21\\sqrt{3}$ and $b=5\\sqrt{3}$ . And the answer is $\\boxed{315}$", "Plot this equilateral triangle on the complex plane.\nTranslate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives $\\left(\\frac{a+b}{3}, 16i\\right)$ . The new coordinates of the equilateral triangle are $\\left(-\\frac{a+b}{3}-16i\\right), \\left(a-\\frac{a+b}{3}-5i\\right), \\left(b-\\frac{a+b}{3}+21i\\right)$ . These three vertices are solutions of a cubic polynomial of form $x^3 + C$ . By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots $r_1, r_2,$ and $r_3,$ $\\, r_1r_2 + r_2r_3 + r_3r_1 = 0$ .) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation $5a = 21b.$ Now use the equation with only real parts. This should give you a quadratic $a^2 - ab + b^2 = 1083$ . Use your previously obtained equation to plug in for $a$ and solve for $b$ , which should yield $5\\sqrt{3}$ $a$ is then $\\frac{21}{5}\\sqrt{3}$ . Multiplying $a$ and $b$ yields $\\boxed{315}$", "Just using the Pythagorean Theorem, we get that $a^2 + 11^2 = (b-a)^2 + 26^2 = b^2 = 37^2$\n$a^2 + 121 = b^2 + 1369 ==> a^2 = b^2 + 1248$ . Expanding the second and subtracting the first equation from it we get $b^2 = 2ab - 555$ $b^2 = 2ab - 555 ==> a^2 = 2ab + 693$\nWe have $b^2 + 1248 = 2b\\sqrt{b^2+1248} + 693$\nMoving the square root to one side and non square roots to the other we eventually get $b^4 + 1110b^2 + 308025 = 4b^4 + 4992b^2$\n$3b^4 + 3882b^2 - 308025 = 0$\nThis factors to $(3b^2-225)(b^2+1369)$ , so $3b^2 = 225, b = 5\\sqrt 3$\nPlugging it back in, we find that a = $\\sqrt{1323}$ which is $21\\sqrt3$ , so the product $ab$ is $\\boxed{315}$" ]

[ "The slope of this line is $\\frac{y_2-y_1}{x_2-x_1}=\\frac{12+6}{6-0}=3$ . Hence, its equation is $y=3x-6$ . The only given point which satisfies these conditions is $\\boxed{3,3}$" ]

[ "Let $D$ be the foot of the perpendicular from $O$ to the plane of $ABC$ . By the Pythagorean Theorem on triangles $\\triangle OAD$ $\\triangle OBD$ and $\\triangle OCD$ we get:\n\\[DA^2=DB^2=DC^2=20^2-OD^2\\]\nIt follows that $DA=DB=DC$ , so $D$ is the circumcenter of $\\triangle ABC$\nBy Heron's Formula the area of $\\triangle ABC$ is (alternatively, a $13-14-15$ triangle may be split into $9-12-15$ and $5-12-13$ right triangles ):\n\\[K = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{21(21-15)(21-14)(21-13)} = 84\\]\nFrom $R = \\frac{abc}{4K}$ , we know that the circumradius of $\\triangle ABC$ is:\n\\[R = \\frac{abc}{4K} = \\frac{(13)(14)(15)}{4(84)} = \\frac{65}{8}\\]\nThus by the Pythagorean Theorem again,\n\\[OD = \\sqrt{20^2-R^2} = \\sqrt{20^2-\\frac{65^2}{8^2}} = \\frac{15\\sqrt{95}}{8}.\\]\nSo the final answer is $15+95+8=\\boxed{118}$", "We know the radii to $A$ $B$ , and $C$ form a triangular pyramid $OABC$ . We know the lengths of the edges $OA = OB = OC = 20$ . First we can break up $ABC$ into its two component right triangles $5-12-13$ and $9-12-15$ . Let the $y$ axis be perpendicular to the base and $x$ axis run along $BC$ , and $z$ occupy the other dimension, with the origin as $C$ . We look at vectors $OA$ and $OC$ . Since $OAC$ is isoceles we know the vertex is equidistant from $A$ and $C$ , hence it is $7$ units along the $x$ axis. Hence for vector $OC$ , in form $<x,y,z>$ it is $<7, h, l>$ where $h$ is the height (answer) and $l$ is the component of the vertex along the $z$ axis. Now on vector $OA$ , since $A$ is $9$ along $x$ , and it is $12$ along $z$ axis, it is $<-2, h, 12- l>$ . We know both vector magnitudes are $20$ . Solving for $h$ yields $\\frac{15\\sqrt{95} }{8}$ , so Answer = $\\boxed{118}$" ]

[ "2003amc10a10solution.gif\nLet the squares be labeled $A$ $B$ $C$ , and $D$\nWhen the polygon is folded, the \"right\" edge of square $A$ becomes adjacent to the \"bottom edge\" of square $C$ , and the \"bottom\" edge of square $A$ becomes adjacent to the \"bottom\" edge of square $D$\nSo, any \"new\" square that is attatched to those edges will prevent the polygon from becoming a cube with one face missing.\nTherefore, squares $1$ $2$ , and $3$ will prevent the polygon from becoming a cube with one face missing.\nSquares $4$ $5$ $6$ $7$ $8$ , and $9$ will allow the polygon to become a cube with one face missing when folded.\nThus the answer is $\\boxed{6}$", "Another way to think of it is that a cube missing one face has $5$ of its $6$ faces. Since the shape has $4$ faces already, we need another face. The only way to add another face is if the added square does not overlap any of the others. $1$ $2$ , and $3$ overlap, while squares $4$ to $9$ do not. The answer is $\\boxed{6}$", "If you're good at visualizing, you can imagine each box and fold up the shape into a 3D shape. This solution is only recommended if you are either in a hurry or extremely skilled at visualizing. We find out that $4,5,6,7,8$ and $9$ work. Therefore, the answer is $\\boxed{6}$ . ~Sophia866" ]

[ "2003amc10a10solution.gif\nLet the squares be labeled $A$ $B$ $C$ , and $D$\nWhen the polygon is folded, the \"right\" edge of square $A$ becomes adjacent to the \"bottom edge\" of square $C$ , and the \"bottom\" edge of square $A$ becomes adjacent to the \"bottom\" edge of square $D$\nSo, any \"new\" square that is attatched to those edges will prevent the polygon from becoming a cube with one face missing.\nTherefore, squares $1$ $2$ , and $3$ will prevent the polygon from becoming a cube with one face missing.\nSquares $4$ $5$ $6$ $7$ $8$ , and $9$ will allow the polygon to become a cube with one face missing when folded.\nThus the answer is $\\boxed{6}$", "Another way to think of it is that a cube missing one face has $5$ of its $6$ faces. Since the shape has $4$ faces already, we need another face. The only way to add another face is if the added square does not overlap any of the others. $1$ $2$ , and $3$ overlap, while squares $4$ to $9$ do not. The answer is $\\boxed{6}$", "If you're good at visualizing, you can imagine each box and fold up the shape into a 3D shape. This solution is only recommended if you are either in a hurry or extremely skilled at visualizing. We find out that $4,5,6,7,8$ and $9$ work. Therefore, the answer is $\\boxed{6}$ . ~Sophia866" ]

[ "Using the geometric series formula, $1 - x + x^2 + \\cdots - x^{17} = \\frac {1 - x^{18}}{1 + x} = \\frac {1-x^{18}}{y}$ . Since $x = y - 1$ , this becomes $\\frac {1-(y - 1)^{18}}{y}$ . We want $a_2$ , which is the coefficient of the $y^3$ term in $-(y - 1)^{18}$ (because the $y$ in the denominator reduces the degrees in the numerator by $1$ ). By the Binomial Theorem , this is $(-1) \\cdot (-1)^{15}{18 \\choose 3} = \\boxed{816}$", "Again, notice $x = y - 1$ . So\n\\begin{align*}1 - x + x^2 + \\cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \\cdots - (y - 1)^{17} \\\\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \\cdots + (1 - y)^{17}\\end{align*}.\nWe want the coefficient of the $y^2$ term of each power of each binomial, which by the binomial theorem is ${2\\choose 2} + {3\\choose 2} + \\cdots + {17\\choose 2}$ . The Hockey Stick Identity tells us that this quantity is equal to ${18\\choose 3} = \\boxed{816}$", "Again, notice $x=y-1$ . Substituting $y-1$ for $x$ in $f(x)$ gives: \\begin{align*}1 - x + x^2 + \\cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \\cdots - (y - 1)^{17} \\\\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \\cdots + (1 - y)^{17}\\end{align*}. From binomial theorem, the coefficient of the $y^2$ term is ${2\\choose 0} + {3\\choose 1} + \\cdots + {17\\choose 15}$ . This is actually the sum of the first 16 triangular numbers, which evaluates to $\\frac{(16)(17)(18)}{6} = \\boxed{816}$", "Let $f(x)=1-x+x^2-x^3+\\cdots+x^{16}-x^{17}$ and $g(y)=a_0+a_1y+a_2y^2+\\cdots +a_{16}y^{16}+a_{17}y^{17}$\nThen, since $f(x)=g(y)$ \\[\\frac{d^2f}{dx^2}=\\frac{d^2g}{dy^2}\\] $\\frac{d^2f}{dx^2} = 2\\cdot 1 - 3\\cdot 2x+\\cdots-17\\cdot 16x^{15}$ by the power rule.\nSimilarly, $\\frac{d^2g}{dy^2} = a_2(2\\cdot 1) + a_3(3\\cdot 2y)+\\cdots+a_{17}(17\\cdot 16y^{15})$\nNow, notice that if $x = -1$ , then $y = 0$ , so $f^{''}(-1) = g^{''}(0)$\n$g^{''}(0)= 2a_2$ , and $f^{''}(-1) = 2\\cdot 1 + 3\\cdot 2 +\\cdots + 16\\cdot 17$\nNow, we can use the hockey stick theorem to see that $2\\cdot 1 + 3\\cdot 2 +\\cdots + 16\\cdot 17 = 2\\binom{18}{3}$\nThus, $2a_2 = 2\\binom{18}{3}\\rightarrow a_2 = \\binom{18}{3}=\\boxed{816}$", "Let $V$ be the vector space of polynomials of degree $\\leq 17,$ and let $B = \\{1, x, x^2, ..., x^{17} \\}$ and $C = \\{1, (x+1), (x+1)^2, ..., (x+1)^{17} \\}$ be two bases for $V$ .\nLet $\\vec{v} \\in V$ be the polynomial given in the problem, and it is easy to see that $[ \\vec{v} ]_B = \\langle 1, -1, 1, -1, ... , 1, -1 \\rangle.$\nNote that the transformation matrix from $C$ to $B$ can be easily found to be $P_{C \\to B} = [ [\\vec{c_1}]_B [\\vec{c_2}]_B ... [\\vec{c_3}]_B ] = \\begin{bmatrix} \\tbinom{0}{0} & \\tbinom{1}{0} & \\tbinom{2}{0} & \\cdots & \\tbinom{17}{0} \\\\ 0 & \\tbinom{1}{1} & \\tbinom{2}{1} & \\cdots & \\tbinom{17}{1} \\\\ 0 & 0 & \\tbinom{2}{2} & \\cdots & \\tbinom{17}{2} \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & 0 & \\cdots & \\tbinom{17}{17} \\end{bmatrix} .$\nI claim that $P_{B \\to C} = \\begin{bmatrix} \\tbinom{0}{0} & -\\tbinom{1}{0} & \\tbinom{2}{0} & \\cdots & -\\tbinom{17}{0} \\\\ 0 & \\tbinom{1}{1} & -\\tbinom{2}{1} & \\cdots & \\tbinom{17}{1} \\\\ 0 & 0 & \\tbinom{2}{2} & \\cdots & -\\tbinom{17}{2} \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & 0 & \\cdots & \\tbinom{17}{17} \\end{bmatrix} ,$ where the term $\\dbinom{n}{k}$ is negated if $n+k$ is odd.\nOne can prove that the $i$ th row of $P_{C \\to B}$ dotted with the $j$ th column of $P_{B \\to C}$ is $\\delta_{i, j}$ by using combinatorial identities, which is left as an exercise for the reader. Thus, since the two matrices multiply to form $\\mathbb{I}_{18},$ we have proved that $P_{B \\to C} = \\begin{bmatrix} \\tbinom{0}{0} & -\\tbinom{1}{0} & \\tbinom{2}{0} & \\cdots & -\\tbinom{17}{0} \\\\ 0 & \\tbinom{1}{1} & -\\tbinom{2}{1} & \\cdots & \\tbinom{17}{1} \\\\ 0 & 0 & \\tbinom{2}{2} & \\cdots & -\\tbinom{17}{2} \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & 0 & \\cdots & \\tbinom{17}{17} \\end{bmatrix} .$\nTo find the coordinates of $\\vec{v}$ under basis $C,$ we compute the product $[ \\vec{v} ]_C = P_{B \\to C} [\\vec{v} ]_B = \\begin{bmatrix} \\tbinom{0}{0} & -\\tbinom{1}{0} & \\tbinom{2}{0} & \\cdots & -\\tbinom{17}{0} \\\\ 0 & \\tbinom{1}{1} & -\\tbinom{2}{1} & \\cdots & \\tbinom{17}{1} \\\\ 0 & 0 & \\tbinom{2}{2} & \\cdots & -\\tbinom{17}{2} \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & 0 & \\cdots & \\tbinom{17}{17} \\end{bmatrix} \\begin{bmatrix} 1 \\\\ -1 \\\\ 1 \\\\ \\vdots \\\\ -1 \\end{bmatrix} = \\begin{bmatrix} \\sum_{n=0}^{17} \\tbinom{n}{0} \\\\ -\\sum_{n=1}^{17} \\tbinom{n}{1} \\\\ \\sum_{n=2}^{17} \\tbinom{n}{2} \\\\ \\vdots \\\\ -\\sum_{n=17}^{17} \\tbinom{n}{17} \\end{bmatrix} = \\begin{bmatrix} \\tbinom{18}{1} \\\\ - \\tbinom{18}{2} \\\\ \\tbinom{18}{3} \\\\ \\vdots \\\\ -\\tbinom{18}{18} \\end{bmatrix},$ where the last equality was obtained via Hockey Stick Identity.\nThus, our answer is $a_2 = [ [ \\vec{v} ]_C ]_3 = \\dbinom{18}{3} = \\boxed{816}.$" ]

[ "We can see that $Q_1$ and $Q_2$ must have a root in common for them to both be factors of the same cubic.\nLet this root be $a$\nWe then know that $a$ is a root of $Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$ , so $x = \\frac{-k}{5}$\nWe then know that $\\frac{-k}{5}$ is a root of $Q_{1}$ so we get: $\\frac{k^{2}}{25}+(k-29)\\left(\\frac{-k}{5}\\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k$ or $k^{2}=30k$ , so $k=30$ is the highest.\nWe can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $\\boxed{030}$", "Again, let the common root be $a$ ; let the other two roots be $m$ and $n$ . We can write that $(x - a)(x - m) = x^2 + (k - 29)x - k$ and that $2(x - a)(x - n) = 2\\left(x^2 + \\left(k - \\frac{43}{2}\\right)x + \\frac{k}{2}\\right)$\nTherefore, we can write four equations (and we have four variables ), $a + m = 29 - k$ $a + n = \\frac{43}{2} - k$ $am = -k$ , and $an = \\frac{k}{2}$\nThe first two equations show that $m - n = 29 - \\frac{43}{2} = \\frac{15}{2}$ . The last two equations show that $\\frac{m}{n} = -2$ . Solving these show that $m = 5$ and that $n = -\\frac{5}{2}$ . Substituting back into the equations, we eventually find that $k = \\boxed{030}$", "Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$ , which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$ , where $R(x) = ax + b$ and $S(x) = cx + d$ . Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$ . Equating coefficients, we get the following system of equations:\n\\begin{align} a = 2c \\\\ b = -d \\\\ 2c(k - 29) - d = c(2k - 43) + 2d \\\\ -d(k - 29) - 2ck = d(2k - 43) + ck \\end{align}\nUsing equations $(1)$ and $(2)$ to make substitutions into equation $(3)$ , we see that the $k$ 's drop out and we're left with $d = -5c$ . Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $\\boxed{030}$", "Notice that if the roots of $Q_1(x)$ and $Q_2(x)$ are all distinct, then $P(x)$ would have four distinct roots, which is a contradiction since it's cubic. Thus, $Q_1(x)$ and $Q_2(x)$ must share a root. Let this common value be $r.$\nThus, we see that we have \\[r^2 + (k - 29)r - k = 0,\\] \\[2r^2 + (2k - 43)r + k = 0.\\] Adding the two equations gives us \\[3r^2 + (3k - 72)r = 0 \\implies r = 0, 24 - k.\\] Now, we have two cases to consider. If $r = 0,$ then we have that $Q_1(r) = 0 = r^2 + (k - 29)r - k \\implies k = 0.$ On the other hand, if $r = 24 - k,$ we see that \\[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \\implies k = \\boxed{030}.\\] This can easily be checked to see that it does indeed work, and we're done!" ]

[ "We see that the expression for the polynomial $P$ is very difficult to work with directly, but there is one obvious transformation to make: sum the geometric series\n\\begin{align*} P(x) &= \\left(\\frac{x^{18} - 1}{x - 1}\\right)^2 - x^{17} = \\frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17}\\\\ &= \\frac{x^{36} - x^{19} - x^{17} + 1}{(x - 1)^2} = \\frac{(x^{19} - 1)(x^{17} - 1)}{(x - 1)^2} \\end{align*}\nThis expression has roots at every $17$ th root and $19$ th roots of unity , other than $1$ . Since $17$ and $19$ are relatively prime , this means there are no duplicate roots. Thus, $a_1, a_2, a_3, a_4$ and $a_5$ are the five smallest fractions of the form $\\frac m{19}$ or $\\frac n {17}$ for $m, n > 0$\n$\\frac 3 {17}$ and $\\frac 4{19}$ can both be seen to be larger than any of $\\frac1{19}, \\frac2{19}, \\frac3{19}, \\frac 1{17}, \\frac2{17}$ , so these latter five are the numbers we want to add.\n$\\frac1{19}+ \\frac2{19}+ \\frac3{19}+ \\frac 1{17}+ \\frac2{17}= \\frac6{19} + \\frac 3{17} = \\frac{6\\cdot17 + 3\\cdot19}{17\\cdot19} = \\frac{159}{323}$ and so the answer is $159 + 323 = \\boxed{482}$" ]

[ "We have $\\frac{1+\\sqrt{3}i}{2} = \\omega$ where $\\omega = e^{\\frac{i\\pi}{3}}$ is a primitive 6th root of unity. Then we have\n\\begin{align*} f(\\omega) &= a\\omega^{2018} + b\\omega^{2017} + c\\omega^{2016}\\\\ &= a\\omega^2 + b\\omega + c \\end{align*}\nWe wish to find $f(1) = a+b+c$ . We first look at the real parts. As $\\text{Re}(\\omega^2) = -\\frac{1}{2}$ and $\\text{Re}(\\omega) = \\frac{1}{2}$ , we have $-\\frac{1}{2}a + \\frac{1}{2}b + c = 2015 \\implies -a+b + 2c = 4030$ . Looking at imaginary parts, we have $\\text{Im}(\\omega^2) = \\text{Im}(\\omega) = \\frac{\\sqrt{3}}{2}$ , so $\\frac{\\sqrt{3}}{2}(a+b) = 2019\\sqrt{3} \\implies a+b = 4038$ . As $a$ and $b$ do not exceed 2019, we must have $a = 2019$ and $b = 2019$ . Then $c = \\frac{4030}{2} = 2015$ , so $f(1) = 4038 + 2015 = 6053 \\implies f(1) \\pmod{1000} = \\boxed{053}$", "Denote $\\frac{1+\\sqrt{3}i}{2}$ with $\\omega$\nBy using the quadratic formula ( $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$ ) in reverse, we can find that $\\omega$ is the solution to a quadratic equation of the form $ax^2+bx+c=0$ such that $2a=2$ $-b=1$ , and $b^2-4ac=-3$ . This clearly solves to $a=1$ $b=-1$ , and $c=1$ , so $\\omega$ solves $x^2-x+1=0$\nMultiplying $x^2-x+1=0$ by $x+1$ on both sides yields $x^3+1=0$ . Muliplying this by $x^3-1$ on both sides yields $x^6-1=0$ , or $x^6=1$ . This means that $\\omega^6=1$\nWe can use this to simplify the equation $a\\omega^{2018}+b\\omega^{2017}+c\\omega^{2016}=f(\\omega)=2015+2019\\sqrt{3}i$ to $a\\omega^2+b\\omega+c=2015+2019\\sqrt{3}i.$\nAs in Solution 1, we use the values $\\omega=\\frac{1}{2}+\\frac{\\sqrt{3}}{2}i$ and $\\omega^2=-\\frac{1}{2}+\\frac{\\sqrt{3}}{2}i$ to find that $-\\frac{1}{2}a+\\frac{1}{2}b+c=2015$ and $\\frac{\\sqrt{3}}{2}a+\\frac{\\sqrt{3}}{2}b=2019\\sqrt{3} \\implies a+b=4038.$ Since neither $a$ nor $b$ can exceed $2019$ , they must both be equal to $2019$ . Since $a$ and $b$ are equal, they cancel out in the first equation, resulting in $c=2015$\nTherefore, $f(1)=a+b+c=2019+2019+2015=6053$ , and $6053\\bmod1000=\\boxed{053}$ . ~ emerald_block", "Calculate the first few powers of $\\frac{1+\\sqrt{3}i}{2}$\n$(\\frac{1+\\sqrt{3}i}{2})^1=\\frac{1+\\sqrt{3}i}{2}$\n$(\\frac{1+\\sqrt{3}i}{2})^2=\\frac{-1+\\sqrt{3}i}{2}$\n$(\\frac{1+\\sqrt{3}i}{2})^3=-1$\n$(\\frac{1+\\sqrt{3}i}{2})^4=\\frac{-1-\\sqrt{3}i}{2}$\n$(\\frac{1+\\sqrt{3}i}{2})^5=\\frac{1-\\sqrt{3}i}{2}$\n$(\\frac{1+\\sqrt{3}i}{2})^6=1$\nWe figure that the power of $\\frac{1+\\sqrt{3}i}{2}$ repeats in a cycle 6.\n$f(\\frac{1+\\sqrt{3}i}{2})=(a(\\frac{1+\\sqrt{3}i}{2})^2+b(\\frac{1+\\sqrt{3}i}{2})+c)(\\frac{1+\\sqrt{3}i}{2})^{2016}$\nSince 2016 is a multiple of 6, $(\\frac{1+\\sqrt{3}i}{2})^{2016}=1$\n$f(\\frac{1+\\sqrt{3}i}{2})=a(\\frac{-1+\\sqrt{3}i}{2})+b(\\frac{1+\\sqrt{3}i}{2})+c$\n$f(\\frac{1+\\sqrt{3}i}{2})=(-\\frac{1}{2}a+\\frac{1}{2}b+c)+(\\frac{\\sqrt{3}i}{2}a+\\frac{\\sqrt{3}i}{2}b)=2015+2019\\sqrt{3}i$\nTherefore, $-\\frac{1}{2}a+\\frac{1}{2}b+c=2015$ and $\\frac{\\sqrt{3}i}{2}a+\\frac{\\sqrt{3}i}{2}b=2019\\sqrt{3}i$\nUsing the first equation, we can get that $-a+b+2c=4030$ , and using the second equation, we can get that $a+b=4038$\nSince all coefficients are less than or equal to $2019$ $a=b=2019$\nTherefore, $2c=4030$ and $c=2015$\n$f(1)=a+b+c=2019+2019+2015=6053$ , and the remainder when it divides $1000$ is $\\boxed{053}.$" ]

[ "Let the roots be $r,s,r + s$ , and let $t = rs$ . Then\nand by matching coefficients, $2(r + s) = 2004 \\Longrightarrow r + s = 1002$ . Then our polynomial looks like \\[x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0\\] and we need the number of possible products $t = rs = r(1002 - r)$ . Because $m=t+1002^2$ is an integer, we also note that $t$ must be an integer.\nSince $r > 0$ and $t > 0$ , it follows that $0 < t = r(1002-r) < 501^2 = 251001$ , with the endpoints not achievable because the roots must be distinct and positive. Because neither $r$ nor $1002-r$ can be an integer, there are $251,000 - 500 = \\boxed{250,500}$ possible values of $n = -1002t$", "Letting the roots be $r$ $s$ , and $t$ , where $t = r+s$ , we see that by Vieta's Formula's, $2004 = r+s+t = t + t = 2t$ , and so $t = 1002$ . Therefore, $x-1002$ is a factor of $x^3 - 2004x^2 + mx + n$ . Letting $x = 0$ gives that $1002 \\mid n$ because $x - 1002 \\mid x^3 - 2004x^2 + mx + n$ . Letting $n = -1002a$ and noting that $x^3 - 2004x^2 + mx + n = (x-1002)(x^2 - bx + a)$ for some $b$ , we see that $b$ is the sum of the roots of $x^2 - bx + a$ $r$ and $s$ , and so $b = 1002$ . Now, we have that $x^2 - 1002x + a$ has roots $r$ and $s$ , and we wish to find the number of possible values of $a$ . By the quadratic formula, we see that \\[\\frac{1002 \\pm \\sqrt{1002^2 - 4a}}{2} = 501 \\pm \\sqrt{501^2 - a}\\] are the two values of noninteger positive real numbers $r$ and $s$ , neither of which is equal to $1002$ . This information gives us that $0 < 501^2 - a < 501^2$ , and so since $501^2 - a$ is evidently not a square, we have $501^2 - 1 - 500 = 251,001 - 500 - 1 = \\boxed{250,500}$ possible values of $n = 1002a$" ]

[ "By Vieta's Formulas , we know that $a$ is the sum of the three roots of the polynomial $x^3-ax^2+bx-2010$ . Again Vieta's Formulas tell us that $2010$ is the product of the three integer roots. Also, $2010$ factors into $2\\cdot3\\cdot5\\cdot67$ . But, since the polynomial has only three roots, two of the four prime factors must be multiplied so that we are left with three roots. To minimize $a$ $2$ and $3$ should be multiplied, which means $a$ will be $6+5+67=78$ and the answer is $\\boxed{78}$ .\n~JimPickens", "We can expand $(x+a)(x+b)(x+c)$ as $(x^2+ax+bx+ab)(x+c)$\n$(x^2+ax+bx+ab)(x+c)=x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc=x^3+x^2(a+b+c)+x(ab+ac+bc)+abc$\nWe do not care about $+bx$ in this case, because we are only looking for $a$ . We know that the constant term is $-2010=-(2\\cdot 3\\cdot 5\\cdot 67)$ We are trying to minimize a, such that we have $-ax^2$ Since we have three positive solutions, we have $(x-a)(x-b)(x-c)$ as our factors. We have to combine two of the factors of $2\\cdot 3\\cdot 5\\cdot 67$ , and then sum up the $3$ resulting factors. Since we are minimizing, we choose $2$ and $3$ to combine together. We get $(x-6)(x-5)(x-67)$ which gives us a coefficient of $x^2$ of $-6-5-67=-78$ Therefore $-a=-78$ or $a=\\boxed{78}$", "We want the polynomial $x^3-ax^2+bx-2010$ to have POSITIVE integer roots. That means we want to factor it in to the form $(x-a)(x-b)(x-c).$ We therefore want the prime factorization for $2010$ . The prime factorization of $2010$ is $2 \\cdot 3 \\cdot 5 \\cdot 67$ . We want the smallest difference of the $3$ roots since by Vieta's formulas $a$ is the sum of the $3$ roots.\nWe proceed to factorize it in to $(x-5)(x-6)(x-67)$ . Therefore, our answer is $5+6+67$ $\\boxed{78}$" ]

[ "Let $a^2$ $=$ original population count, $b^2+1$ $=$ the second population count, and $c^2$ $=$ the third population count \nWe first see that $a^2 + 100 = b^2 + 1$ or $99$ $=$ $b^2-a^2$ .\nWe then factor the right side getting $99$ $=$ $(b-a)(b+a)$ . \nSince we can only have an nonnegative integral population, clearly $b+a$ $>$ $b-a$ and both factor $99$ .\nWe factor $99$ into $3^2 \\cdot 11$ $=$ $(b-a)(b+a)$ There are a few cases to look at: $1)$ $b+a$ $=$ $11$ and $b-a$ $=$ $9$ .\nAdding the two equations we get $2b$ $=$ $20$ or $b$ $=$ $10$ , which means $a$ $=$ $1$ .\nBut looking at the restriction that the second population + $100$ $=$ third population... $10^2$ $+$ $1$ $+$ $100$ $=$ $201$ $\\neq$ a perfect square.\n$2)$ $b+a$ $=$ $33$ and $b-a$ $=$ $3$ .\nAdding the two equations we get $2b$ $=$ $36$ or $b$ $=$ $18$ , which means $a$ $=$ $15$ .\nLooking at the same restriction, we get $18^2$ $1$ $100$ $=$ $324$ $101$ $=$ $425$ , which is NOT a perfect square.\nFinally, $b+a$ $=$ $99$ and $b-a$ $=$ $1$ $2b$ $=$ $100$ or $b$ $=$ $50$ , which means $a$ $=$ $49$ .\nLooking at the same restriction, we get $50^2$ $1$ $100$ $=$ $2500$ $101$ $=$ $2601$ $=$ $51^2$ . Thus we find that the original population is $a^2$ $=$ $49^2$ $=$ $7^4$ . Or $a^2$ is a multiple of $\\boxed{7}$", "Let $P$ $=$ original population. Translating the word problem into a system of equations, we got: \\begin{align} P &= x^2 \\\\ P + 100 &= y^2 + 1 \\\\ P + 200 &= z^2 \\end{align} for some positive integers $x$ $y$ and $z$ .\nNow, by subtracting $(2)$ from $(3)$ (i.e. $(3) - (2)$ ), we got: \\begin{align*} 100 &= z^2 - y^2 - 1 \\\\ 101 &= z^2 - y^2 \\\\ 101 &= (z - y)(z + y) \\end{align*} Since y and z are both positive integers and 101 is a prime, by factoring, the only working solution for us is $y = 50$ and $z = 51$ .\nPlugging that back to $(2)$ and simplify, we got $P = 2401 = (49)^2 = x^2$ , a multiple of $7$ .\nTherefore, the answer is $\\boxed{7}$ . -nullptr07" ]

[ "The graph show that the ratio of men to total population is $\\frac{1}{3}$ , so the total number of men is $\\frac{1}{3} \\times 480= \\boxed{160}$" ]

[ "The graph shows $3$ equal squares, each with value $x$ . So $3x = 480$ , so $x = \\boxed{160}$" ]

[ "With about $230$ million people and under $4$ million square miles, there are about $60$ people per square mile. Since a square mile is about $(5000 \\ \\text{ft})^{2} = 25$ million square feet, that gives approximately $\\frac{25}{60}$ of a million square feet per person. $\\frac{25}{60}$ is approximately half, so the answer is approximately half a million, which is $\\boxed{500,000}$" ]

[ "Let the two primes be $a$ and $b$ . We would have $a-b=2$ and $a^{3}-b^{3}=31106$ . Using difference of cubes, we would have $(a-b)(a^{2}+ab+b^{2})=31106$ . Since we know $a-b$ is equal to $2$ $(a-b)(a^{2}+ab+b^{2})$ would become $2(a^{2}+ab+b^{2})=31106$ . Simplifying more, we would get $a^{2}+ab+b^{2}=15553$\nNow let's introduce another variable. Instead of using $a$ and $b$ , we can express the primes as $x+2$ and $x$ where $a$ is $x+2$ and b is $x$ . Plugging $x$ and $x+2$ in, we would have $(x+2)^{2}+x(x+2)+x^{2}$ . When we expand the parenthesis, it would become $x^{2}+4x+4+x^{2}+2x+x^{2}$ . Then we combine like terms to get $3x^{2}+6x+4$ which equals $15553$ . Then we subtract 4 from both sides to get $3x^{2}+6x=15549$ . Since all three numbers are divisible by 3, we can divide by 3 to get $x^{2}+2x=5183$\nNotice how if we add 1 to both sides, the left side would become a perfect square trinomial: $x^{2}+2x+1=5184$ which is $(x+1)^{2}=5184$ . Since $2$ is too small to be a valid number, the two primes must be odd, therefore $x+1$ is the number in the middle of them. Conveniently enough, $5184=72^{2}$ so the two numbers are $71$ and $73$ . The next prime number is $79$ , and $7+9=16$ so the answer is $\\boxed{16}$", "Let the two primes be $a$ and $b$ , with $a$ being the larger prime. We have $a - b = 2$ , and $a^3 - b^3 = 31106$ . Using difference of cubes, we obtain $a^2 + ab + b^2 = 15553$ . Now, we use the equation $a - b = 2$ to obtain $a^2 - 2ab + b^2 = 4$ . Hence, \\[a^2 + ab + b^2 - (a^2 - 2ab + b^2) = 3ab = 15553 - 4 = 15549\\] \\[ab = 5183.\\] Because we have $b = a+2$ $ab = (a+1)^2 - (1)^2$ . Thus, $(a+1)^2 = 5183 + 1 = 5184$ , so $a+1 = 72$ . This implies $a = 71$ $b = 73$ , and thus the next biggest prime is $79$ , so our answer is $7 + 9 = \\boxed{16}$", "Let the two primes be $p$ and $q$ such that $p-q=2$ and $p^{3}-q^{3}=31106$\nBy the difference of cubes formula, $p^{3}-q^{3}=(p-q)(p^{2}+pq+q^{2})$\nPlugging in $p-q=2$ and $p^{3}-q^{3}=31106$\n$31106=2(p^{2}+pq+q^{2})$\nThrough the givens, we can see that $p \\approx q$\nThus, $31106=2(p^{2}+pq+q^{2})\\approx 6p^{2}\\\\p^2\\approx \\tfrac{31106}{6}\\approx 5200$\nRecall that $70^2=4900$ and $80^2=6400$ . It follows that our primes must be only marginally larger than $70$ , where we conveniently find $p=73, q=71$\nThe least prime greater than these two primes is $79 \\implies 7 + 9 \\implies \\boxed{16}$" ]

[ "Let the two primes be $x + 1$ and $x - 1$ . Then, plugging it into the second condition, we get $(x + 1)^3 - (x - 1)^3 = 31106.$ Expanding the left side, \\[6x^2 + 2 = 31106 \\implies x^2 = 5184.\\] Taking the square root of both sides, we get that $x = 72$ and the larger prime is $73$ . The smallest prime larger than $73$ is $79$ , which has a digit sum of $7 + 9 = \\boxed{16}.$" ]

[ "We have that $N^2 - N = N(N - 1)\\equiv 0\\mod{10000}$\nThus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$ . Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$ , which is impossible for a number that is divisible by either $2$ or $5$ . Thus, one of them is divisible by $2^4 = 16$ , and the other is divisible by $5^4 = 625$ . Noting that $625 \\equiv 1\\mod{16}$ , we see that $625$ would work for $N$ , except the thousands digit is $0$ . The other possibility is that $N$ is a multiple of $16$ and $N-1$ is a multiple of $625$ . In order for this to happen, \\[N-1 \\equiv -1 \\pmod {16}.\\] Since $625 \\equiv 1 \\pmod{16}$ , we know that $15 \\cdot 625 = 9375 \\equiv 15 \\equiv -1 \\mod{16}$ . Thus, $N-1 = 9375$ , so $N = 9376$ , and our answer is $\\boxed{937}$", "let $N= 10000t+1000a+100b+10c+d$ for positive integer values $t,a,b,c,d$ .\nWhen we square $N$ we get that \\begin{align*} N^2 &=(10000t+1000a+100b+10c+d)^2\\\\ &=10^8t^2+10^6a^2+10^4b^2+10^2c^2+d^2+2(10^7ta+10^6tb+10^5tc+10^4td+10^5ab+10^4ac+10^3bc+10^ad+10^2bd+10cd) \\end{align*}\nHowever, we don't have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only: \\[2000ad+2000bc+100c^2+200bd+20cd+d^2.\\] Now we need to compare each decimal digit with $1000a+100b+10c+d$ and see whether the digits are congruent in base 10.\nwe first consider the ones digits:\n$d^2\\equiv d \\pmod{10}.$\nThis can happen for only 3 values : 1, 5 and 6.\nWe can try to solve each case:\nConsidering the tenths place, \nwe have that:\n$20cd=20c\\equiv 10c \\pmod {100}$ so $c= 0$\nConsidering the hundreds place we have that\n$200bd+100c^2= 200b \\equiv 100b \\pmod{1000}$ so again $b=0$\nnow considering the thousands place we have that\n$2000ad+2000bc = 2000a \\equiv 1000a \\pmod {10000}$ so we get $a=0$ but $a$ cannot be equal to $0$ so we consider $d=5.$\nconsidering the tenths place\nwe have that:\n$20cd+20=100c+20\\equiv 20 \\equiv 10c \\mod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$ )\nso $c=2$\nconsidering the hundreds place we have that\n$200bd+100c^2+100c= 1000b+600 \\equiv600\\equiv 100b \\pmod{1000}$ ( the extra $100c$ is carried from the tenths place)\nso $b=6$\nnow considering the thousands place we have that\n$2000ad+2000bc +1000b= 10000a+24000+ 6000\\equiv0\\equiv 1000a \\pmod {10000}$ ( the extra $1000b$ is carried from the hundreds place)\nso a is equal 0 again\nconsidering the tenths place\nwe have that:\n$20cd+30=120c+30\\equiv 30+20c \\equiv 10c \\pmod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$ )\nif $c=7$ then we have\n$30+20 \\cdot 7 \\equiv 70\\equiv7 \\cdot 10 \\pmod{100}$\nso $c=7$\nconsidering the hundreds place we have that\n$200bd+100c^2+100c+100= 1200b+4900+800 \\equiv200b+700\\equiv 100b \\pmod{1000}$ ( the extra $100c+100$ is carried from the tenths place)\nif $b=3$ then we have\n$700+200 \\cdot 3 \\equiv 300\\equiv3 \\cdot 100 \\pmod {1000}$\nso $b=3$\nnow considering the thousands place we have that\n$2000ad+2000bc +1000b+5000+1000= 12000a+42000+ 3000+6000\\equiv0\\equiv 2000a+1000\\equiv 1000a \\pmod {10000}$ ( the extra $1000b+6000$ is carried from the hundreds place)\nif $a=9$ then we have\n$2000 \\cdot 9+1000 \\equiv 9000\\equiv9 \\cdot 1000 \\pmod {1000}$\nso $a=9$\nso we have that the last 4 digits of $N$ are $9376$ and $abc$ is equal to $\\boxed{937}$", "By the Chinese Remainder Theorem, the equation $N(N-1)\\equiv 0\\pmod{10000}$ is equivalent to the two equations: \\begin{align*} N(N-1)&\\equiv 0\\pmod{16},\\\\ N(N-1)&\\equiv 0\\pmod{625}. \\end{align*} Since $N$ and $N-1$ are coprime, the only solutions are when $(N\\mod{16},N\\mod{625})\\in\\{(0,0),(0,1),(1,0),(1,1)\\}$\nLet \\[\\varphi:\\mathbb Z/10000\\mathbb Z\\to\\mathbb Z/16\\mathbb Z\\times\\mathbb Z/625\\mathbb Z,\\] \\[x\\mapsto (x\\mod{16},x\\mod{625}).\\] The statement of the Chinese Remainder theorem is that $\\varphi$ is an isomorphism between the two rings. In this language, the solutions are $\\varphi^{-1}(0,0)$ $\\varphi^{-1}(0,1)$ $\\varphi^{-1}(1,0)$ , and $\\varphi^{-1}(1,1)$ . Now we easily see that \\[\\varphi^{-1}(0,0)=0\\] and \\[\\varphi^{-1}(1,1)=1.\\] Noting that $625\\equiv 1\\pmod{16}$ , it follows that \\[\\varphi^{-1}(1,0)=625.\\] To compute $\\varphi^{-1}(0,1)$ , note that \\[(0,1)=15(1,0)+(1,1)\\] in \\[\\mathbb Z/16\\mathbb Z\\times\\mathbb Z/625\\mathbb Z,\\] so since $\\varphi^{-1}$ is linear in its arguments (by virtue of being an isomorphism), \\[\\varphi^{-1}(0,1)=15\\varphi^{-1}(1,0)+\\varphi^{-1}(1,1)=15\\times 625+1=9376.\\]\nThe four candidate digit strings $abcd$ are then $0000,0001,0625,9376$ . Of those, only $9376$ has nonzero first digit, and therefore the answer is $\\boxed{937}$", "WLOG, we can assume that $N$ is a 4-digit integer $\\overline{abcd}$ . Note that the only $d$ that will satisfy $N$ will also satisfy $d^2\\equiv d\\pmod{10}$ , as the units digit of $\\overline{abcd}^2$ is affected only by $d$ , regardless of $a$ $b$ , or $c$\nBy checking the numbers 0-9, we see that the only possible values of $d$ are $d=0, 1, 5, 6$\nNow, we seek to find $c$ . Note that the only $\\overline{cd}$ that will satisfy $N$ will also satisfy $\\overline{cd}^2 \\equiv \\overline{cd}\\pmod{100}$ , by the same reasoning as above - the last two digits of $\\overline{abcd}^2$ are only affected by $c$ and $d$ . As we already have narrowed choices for $d$ , we start reasoning out.\nFirst, we note that if $d=0$ , then $c=0$ , as a number ending in 0, and therefore divisible by 10, is squared, the result is divisible by 100, meaning it ends in two 0's. However, if $N$ ends in $00$ , then recursively, $a$ and $b$ must be $0$ , as a number divisible by 100 squared ends in four zeros. As $a$ cannot be 0, we throw out this possibility, as the only solution in this case is $0$\nNow, let's assume that $d=1$ $\\overline{cd}$ is equal to $10c + d = 10c + 1$ . Squaring this gives $100c^2 + 20c + 1$ , and when modulo 100 is taken, it must equal $10c + 1$ . As $c$ is an integer, $100c^2$ must be divisible by 100, so $100c^2+20c+1 \\equiv 20c + 1\\pmod{100}$ , which must be equivalent to $10c + 1$ . Note that this is really $\\overline{(2c)1}$ and $\\overline{c1}$ , and comparing the 10's digits. So really, we're just looking for when the units digit of $2c$ and $c$ are equal, and a quick check reveals that this is only true when $c=0$ .However, if we extend this process to find $b$ and $a$ , we'd find that they are also 0. The only solution in this case is $1$ , and since $a=0$ here, this is not our solution. Therefore, there are no valid solutions in this case.\nLet's assume that $d=5$ . Note that $(10c + 5)^2 = 100c^2 + 100c + 25$ , and when modulo $100$ is taken, $25$ is the remainder. So all cases here have squares that end in 25, so $\\overline{cd}=25$ is our only case here. A quick check reveals that $25^2=625$ , which works for now.\nNow, let $d=6$ . Note that $(10c + 6)^2 = 100c^2 + 120c + 36$ . Taking modulo 100, this reduces to $20c+36$ , which must be equivalent to $10c+6$ . Again, this is similar to $\\overline{(2c+3)6}$ and $\\overline{c6}$ , so we see when the units digits of $2c+3$ and $c$ are equal. To make checking faster, note that $2c$ is necessarily even, so $2c+3$ is necessarily odd, so $c$ must be odd. Checking all the odds reveals that only $c=3$ works, so this case gives $76$ . Checking quickly $76^2 = 5776$ , which works for now.\nNow, we find $b$ , given two possibilities for $\\overline{cd}$\nStart with $\\overline{cd} = 25$ $\\overline{bcd} = 100b + \\overline{cd} = 100b + 25.$ Note that if we square this, we get $10000b^2 + 5000b + 625$ , which should be equivalent to $100b + 25$ modulo 1000. Note that, since $b$ is an integer, $10000b^2 + 5000 + 625$ simplifies modulo 1000 to $625$ . Therefore, the only $\\overline{bcd}$ that works here is $625$ $625^2 = 390625$\nNow, assume that $\\overline{cd}=76$ . We have $100b + 76$ , and when squared, becomes $10000b^2 + 15200b + 5776$ , which, modulo 1000, should be equivalent to $100b+76$ . Reducing $10000b^2 + 15200b + 5776$ modulo $1000$ gives $200b + 776$ . Using the same technique as before, we must equate the hundreds digit of $\\overline{(2b+7)76}$ to $\\overline{b76}$ , or equate the units digit of $2b+7$ and $b$ . Since $2b+7$ is necessarily odd, any possible $b$ 's must be odd. A quick check reveals that $b=3$ is the only solution, so we get a solution of $376$ $376^2 = 141376$\nFinally, we solve for $a$ . Start with $\\overline{bcd}=625$ . We have $1000a + 625$ , which, squared, gives \\[1000000a^2 + 1250000a + 390625,\\] and reducing modulo 10000 gives simply 625. So $\\overline{abcd}=625$ . However, that makes $a=0$ . Therefore, no solutions exist in this case.\nWe turn to our last case, $\\overline{bcd}=376$ . We have \\[1000a + 376^2 = 1000000a^2 + 752000a + 141376,\\] and reducing modulo $10000$ gives $2000a + 1376$ , which must be equivalent to $1000a + 376$ . So we must have $\\overline{(2a+1)376}$ being equivalent to $\\overline{a376}$ modulo 1000. So, the units digit of $2a+1$ must be equal to $a$ . Since $2a+1$ is odd, $a$ must be odd. Lo and behold, the only possibility for $a$ is $a=3$ . Therefore, $\\overline{abcd}=9376$ , so our answer is $\\boxed{937}$" ]

[ "Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater).\nBreak up the problem into two cases: an even number of sides $2n$ , or an odd number of sides $2n-1$ . For polygons with $2n$ sides, the circumdiameter has endpoints on $2$ vertices. There are $n-1$ points on one side of a diameter, plus $1$ of the endpoints of the diameter for a total of $n$ points. For polygons with $2n - 1$ points, the circumdiameter has $1$ endpoint on a vertex and $1$ endpoint on the midpoint of the opposite side. There are also $n - 1$ points on one side of the diameter, plus the vertex for a total of $n$ points on one side of the diameter.\nCase 1: $2n$ -sided polygon. There are clearly $\\binom{2n}{3}$ different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously $2n$ choices for this point. From there, the other two points must be within the $n-1$ points remaining on the same side of the diameter. So our desired probability is $\\frac{2n\\binom{n-1}{2}}{\\binom{2n}{3}}$ $=\\frac{n(n-1)(n-2)}{\\frac{2n(2n-1)(2n-2)}{6}}$ $=\\frac{6n(n-1)(n-2)}{2n(2n-1)(2n-2)}$ $=\\frac{3(n-2)}{2(2n-1)}$\nso $\\frac{93}{125}=\\frac{3(n-2)}{2(2n-1)}$\n$186(2n-1)=375(n-2)$\n$372n-186=375n-750$\n$3n=564$\n$n=188$ and so the polygon has $376$ sides.\nCase 2: $2n-1$ -sided polygon. Similarly, $\\binom{2n-1}{3}$ total triangles. Again choose the leftmost point, with $2n-1$ choices. For the other two points, there are again $\\binom{n-1}{2}$ possibilities.\nThe probability is $\\frac{(2n-1)\\binom{n-1}{2}}{\\binom{2n-1}{3}}$\n$=\\frac{3(2n-1)(n-1)(n-2)}{(2n-1)(2n-2)(2n-3)}$\n$=\\frac{3(n-2)}{2(2n-3)}$\nso $\\frac{93}{125}=\\frac{3(n-2)}{2(2n-3)}$\n$186(2n-3)=375(n-2)$\n$375n-750=372n-558$\n$3n=192$\n$n=64$ and our polygon has $127$ sides.\nAdding, $127+376=\\boxed{503}$", "We use casework on the locations of the vertices, if we choose the locations of vertices $v_a, v_b, v_c$ on the n-gon (where the vertices of the n-gon are $v_0, v_1, v_2, ... v_{n-1},$ in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that $a<b<c$\nBy symmetry, we can assume W/O LOG that the location of vertex A is vertex $v_0$\nNow, vertex B can be any of $v_1, v_2, ... v_{n-2}$ . We start in on casework.\nCase 1: vertex B is at one of the locations $v_{n-2}, v_{n-3}, ... v_{\\lfloor n/2 \\rfloor +1}$ . (The ceiling function is necessary for the cases in which n is odd.)\nNow, since the clockwise arc from A to B measures more than 180 degrees; for every location of vertex C we can choose in the above restrictions, angle C will be an obtuse angle.\nThere are $\\lceil n/2 \\rceil - 2$ choices for vertex B now (again, the ceiling function is necessary to satisfy both odd and even cases of n). If vertex B is placed at $v_m$ , there are $n - m - 1$ possible places for vertex C.\nSumming over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is $\\frac{(n- \\lceil n/2 \\rceil - 2)(n - \\lceil n/2 \\rceil) - 1}{2}$\nCase 2: vertex B is at one of the locations not covered in the first case.\nNote that this will result in the same number of obtuse triangles as case 1, but multiplied by 2. This is because fixing vertex B in $v_0$ , then counting up the cases for vertices C, and again for vertices C and A, respectively, is combinatorially equivalent to fixing vertex A at $v_0$ , then counting cases for vertex B, as every triangle obtained in this way can be rotated in the n-gon to place vertex A at $v_0$ , and will not be congruent to any obtuse triangle obtained in case 1, as there will be a different side opposite the obtuse angle in this case.\nTherefore, there are $\\frac{3(n- \\lceil n/2 \\rceil - 2)(n - \\lceil n/2 \\rceil - 1)}{2}$ total obtuse triangles obtainable.\nThe total number of triangles obtainable is $1+2+3+...+(n-2) = \\frac{(n-2)(n-1)}{2}$\nThe ratio of obtuse triangles obtainable to all triangles obtainable is therefore\n$\\frac{\\frac{3(n- \\lceil n/2 \\rceil - 2)(n - \\lceil n/2 \\rceil - 1)}{2}}{\\frac{(n-2)(n-1)}{2}} = \\frac{3(n- \\lceil n/2 \\rceil - 2)(n - \\lceil n/2 \\rceil - 1)}{(n-2)(n-1)} = \\frac{93}{125}$\nSo, $\\frac{(n- \\lceil n/2 \\rceil - 2)(n - \\lceil n/2 \\rceil - 1)}{(n-2)(n-1)} = \\frac{31}{125}$\nNow, we have that $(n-2)(n-1)$ is divisible by $125 = 5^3$ . It is now much easier to perform trial-and-error on possible values of n, because we see that $n \\equiv 1,2 \\pmod{125}$\nWe find that $n = 127$ and $n = 376$ both work, so the final answer is $127 + 376 = \\boxed{503}$" ]

[ "Approximating $40.37$ instead of $1.8$ is more effective because larger numbers are less affected by absolute changes (e.g $1001$ is much closer relatively to $1000$ than $2$ is to $1$ ). $74$ is the closest to $72$ , so the answer is $\\boxed{74}$" ]

[ "Converting the decimals to fractions , this is \\begin{align*} 8\\times \\frac{1}{4} \\times 2\\times \\frac{1}{8} &= \\frac{8\\times 2}{4\\times 8} \\\\ &= \\frac{16}{32} \\\\ &= \\frac{1}{2} \\rightarrow \\boxed{12}" ]

[ "Let the three integers be $a, b, c$ $N = abc = 6(a + b + c)$ and $c = a + b$ . Then $N = ab(a + b) = 6(a + b + a + b) = 12(a + b)$ . Since $a$ and $b$ are positive, $ab = 12$ so $\\{a, b\\}$ is one of $\\{1, 12\\}, \\{2, 6\\}, \\{3, 4\\}$ so $a + b$ is one of $13, 8, 7$ so the sum of all possible values of $N$ is $12 \\cdot (13 + 8 + 7) = 12(28) = \\boxed{336}$" ]

[ "Let the lengths of the two congruent sides of the triangle be $x$ , then the product desired is $x^2$\nNotice that the product of the base and twice the height is $4$ times the area of the triangle.\nSet the vertex angle to be $a$ , we derive the equation:\n$x^2=4\\left(\\frac{1}{2}x^2\\sin(a)\\right)$\n$\\sin(a)=\\frac{1}{2}$\nAs the triangle is obtuse, $a=150^\\circ$ only. We get $\\boxed{150}.$", "Denote by $a$ the length of each congruent side. Denote by $\\theta$ the degree measure of each acute angle.\nDenote by $\\phi$ the degree measure of the obtuse angle.\nHence, this problem tells us the following relationship: \\[ a^2 = 2 a \\cos \\theta \\cdot 2 a \\sin \\theta . \\]\nHence, \\begin{align*} 1 & = 2 \\cdot 2 \\sin \\theta \\cos \\theta \\\\ & = 2 \\sin 2 \\theta \\\\ & = 2 \\sin \\left( 180^\\circ - 2 \\theta \\right) \\\\ & = 2 \\sin \\phi . \\end{align*}\nHence, $\\phi = 150^\\circ$\nTherefore, the answer is $\\boxed{150}$" ]

[ "We can first make a small example to find out $A$ and $B$ . So,\n$303\\times505=153015$\nThe ones digit plus thousands digit is $5+3=8$\nNote that the ones and thousands digits are, added together, $8$ . (and so on...) So the answer is $\\boxed{8}$ This is a direct multiplication way." ]

[ "Let the three consecutive positive integers be $a-1$ $a$ , and $a+1$ . Since the mean is $a$ , the sum of the integers is $3a$ . So $8$ times the sum is just $24a$ . With this, we now know that $a(a-1)(a+1)=24a\\Rightarrow(a-1)(a+1)=24$ $24=4\\times6$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=\\boxed{77}$" ]

[ "Let the three consecutive positive integers be $a-1$ $a$ , and $a+1$ . Since the mean is $a$ , the sum of the integers is $3a$ . So $8$ times the sum is just $24a$ . With this, we now know that $a(a-1)(a+1)=24a\\Rightarrow(a-1)(a+1)=24$ $24=4\\times6$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=\\boxed{77}$" ]