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zwhe99

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simplerl-OlympiadBench

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[ "We have that $x^{2}+p x+q=0$ and $x^{2}+p x-q=0$ both have integer solutions.\n\nFor $x^{2}+p x+q=0$, its roots are $\\frac{-p \\pm \\sqrt{p^{2}-4 q}}{2}$.\n\nIn order that these roots be integers, $p^{2}-4 q$ must be a perfect square.\n\nTherefore, $p^{2}-4 q=m^{2}$ for some positive integer $m$.\n\nSimilarly for $x^{2}+p x-q=0$, it has roots $\\frac{-p \\pm \\sqrt{p^{2}+4 q}}{2}$ and in order that these roots be integers $p^{2}+4 q$ must be a perfect square.\n\nThus $p^{2}+4 q=n^{2}$ for some positive integer $n$.\n\nAdding gives $2 p^{2}=m^{2}+n^{2}$ (with $n \\geq m$ since $n^{2}=p^{2}+4 q$\n\n$$\n\\left.\\geq p^{2}-4 q=m^{2}\\right)\n$$\n\nAnd so $p^{2}=\\frac{1}{2} m^{2}+\\frac{1}{2} n^{2}=\\left(\\frac{n+m}{2}\\right)^{2}+\\left(\\frac{n-m}{2}\\right)^{2}$.\n\nWe note that $m$ and $n$ have the same parity since $m^{2}=p^{2}-4 q \\equiv p^{2}(\\bmod 2)$ and $n^{2} \\equiv p^{2}+4 q \\equiv p^{2}(\\bmod 2)$.\n\nSince $\\frac{n+m}{2}$ and $\\frac{n-m}{2}$ are positive integers then $p^{2}=a^{2}+b^{2}$ where $a=\\frac{n+m}{2}$ and $b=\\frac{n-m}{2}$.\n\nFrom above, $a=\\frac{n+m}{2}$ and $b=\\frac{n-m}{2}$ or $n=a+b$ and $m=a-b$.\n\nFrom before, $p^{2}+4 q=n^{2}$\n\n$$\n\\begin{aligned}\n4 q^{2} & =n^{2}-p^{2} \\\\\n& =(a+b)^{2}-\\left(a^{2}+b^{2}\\right) \\\\\n4 q & =2 a b\n\\end{aligned}\n$$\n\nTherefore, $q=\\frac{a b}{2}$." ]

[ "$\\frac{a b}{2}$" ]

[ "For the parabola to have its vertex on the $x$-axis, the equation\n\n$$\ny=k x^{2}+(5 k+3) x+(6 k+5)=0\n$$\n\nmust have two equal real roots.\n\nThat is, its discriminant must equal 0 , and so\n\n$$\n\\begin{aligned}\n(5 k+3)^{2}-4 k(6 k+5) & =0 \\\\\n25 k^{2}+30 k+9-24 k^{2}-20 k & =0 \\\\\nk^{2}+10 k+9 & =0 \\\\\n(k+1)(k+9) & =0\n\\end{aligned}\n$$\n\nTherefore, $k=-1$ or $k=-9$." ]

[ "$-1,-9$" ]

[ "Since $f(x)=f(x-1)+f(x+1)$, then $f(x+1)=f(x)-f(x-1)$, and so\n\n$$\n\\begin{aligned}\n& f(1)=1 \\\\\n& f(2)=3 \\\\\n& f(3)=f(2)-f(1)=3-1=2 \\\\\n& f(4)=f(3)-f(2)=2-3=-1 \\\\\n& f(5)=f(4)-f(3)=-1-2=-3 \\\\\n& f(6)=f(5)-f(4)=-3-(-1)=-2 \\\\\n& f(7)=f(6)-f(5)=-2-(-3)=1=f(1) \\\\\n& f(8)=f(7)-f(6)=1-(-2)=3=f(2)\n\\end{aligned}\n$$\n\nSince the value of $f$ at an integer depends only on the values of $f$ at the two previous integers, then the fact that the first several values form a cycle with $f(7)=f(1)$ and $f(8)=f(2)$ tells us that the values of $f$ will always repeat in sets of 6 .\n\nSince 2008 is 4 more than a multiple of 6 (as $2008=4+2004=4+6(334)$ ), then $f(2008)=f(2008-6(334))=f(4)=-1$." ]

[ "-1" ]

[ "Since $a, b, c$ form an arithmetic sequence, then we can write $a=b-d$ and $c=b+d$ for some real number $d$.\n\nSince $a+b+c=60$, then $(b-d)+b+(b+d)=60$ or $3 b=60$ or $b=20$.\n\nTherefore, we can write $a, b, c$ as $20-d, 20,20+d$.\n\n(We could have written $a, b, c$ instead as $a, a+d, a+2 d$ and arrived at the same result.) Thus, $a-2=20-d-2=18-d$ and $c+3=20+d+3=23+d$, so we can write $a-2, b, c+3$ as $18-d, 20,23+d$.\n\n\n\nSince these three numbers form a geometric sequence, then\n\n$$\n\\begin{aligned}\n\\frac{20}{18-d} & =\\frac{23+d}{20} \\\\\n20^{2} & =(23+d)(18-d) \\\\\n400 & =-d^{2}-5 d+414 \\\\\nd^{2}+5 d-14 & =0 \\\\\n(d+7)(d-2) & =0\n\\end{aligned}\n$$\n\nTherefore, $d=-7$ or $d=2$.\n\nIf $d=-7$, then $a=27, b=20$ and $c=13$.\n\nIf $d=2$, then $a=18, b=20$ and $c=22$.\n\n(We can check that, in each case, $a-2, b, c+3$ is a geometric sequence.)", "Since $a, b, c$ form an arithmetic sequence, then $c-b=b-a$ or $a+c=2 b$.\n\nSince $a+b+c=60$, then $2 b+b=60$ or $3 b=60$ or $b=20$.\n\nThus, $a+c=40$, so $a=40-c$.\n\nTherefore, we can write $a, b, c$ as $40-c, 20, c$.\n\nAlso, $a-2=40-c-2=38-c$, so we can write $a-2, b, c+3$ as $38-c, 20, c+3$.\n\nSince these three numbers form a geometric sequence, then\n\n$$\n\\begin{aligned}\n\\frac{20}{38-c} & =\\frac{c+3}{20} \\\\\n20^{2} & =(38-c)(c+3) \\\\\n400 & =-c^{2}+35 c+114 \\\\\nc^{2}-35 d+286 & =0 \\\\\n(c-13)(c-22) & =0\n\\end{aligned}\n$$\n\nTherefore, $c=13$ or $c=22$.\n\nIf $c=13$, then $a=27$, so $a=27, b=20$ and $c=13$.\n\nIf $c=22$, then $a=18$, so $a=18, b=20$ and $c=22$.\n\n(We can check that, in each case, $a-2, b, c+3$ is a geometric sequence.)" ]

[ "$(27,20,13), (18,20,22)$" ]

[ "Since the average of three consecutive multiples of 3 is $a$, then $a$ is the middle of these three integers, so the integers are $a-3, a, a+3$.\n\nSince the average of four consecutive multiples of 4 is $a+27$, then $a+27$ is halfway in between the second and third of these multiples (which differ by 4), so the second and third of the multiples are $(a+27)-2=a+25$ and $(a+27)+2=a+29$, so the four integers are $a+21, a+25, a+29, a+33$.\n\n(We have used in these two statements the fact that if a list contains an odd number of integers, then there is a middle integer in the list, and if the list contains an even number\n\n\n\nof integers, then the \"middle\" integer is between two integers from the list.)\n\nThe smallest of these seven integers is $a-3$ and the largest is $a+33$.\n\nThe average of these two integers is $\\frac{1}{2}(a-3+a+33)=\\frac{1}{2}(2 a+30)=a+15$.\n\nSince $a+15=42$, then $a=27$." ]

[ "27" ]

[ "Suppose that Billy removes the ball numbered $x$ from his bag and that Crystal removes the ball numbered $y$ from her bag.\n\nThen $b=1+2+3+4+5+6+7+8+9-x=45-x$.\n\nAlso, $c=1+2+3+4+5+6+7+8+9-y=45-y$.\n\nHence, $b-c=(45-x)-(45-y)=y-x$.\n\nSince $1 \\leq x \\leq 9$ and $1 \\leq y \\leq 9$, then $-8 \\leq y-x \\leq 8$.\n\n(This is because $y-x$ is maximized when $y$ is largest (that is, $y=9$ ) and $x$ is smallest (that is, $x=1$ ), so $y-x \\leq 9-1=8$. Similarly, $y-x \\geq-8$.)\n\nSince $b-c=y-x$ is between -8 and 8 , then for it to be a multiple of $4, b-c=y-x$ can be $-8,-4,0,4$, or 8 .\n\nSince each of Billy and Crystal chooses 1 ball from 9 balls and each ball is equally likely to be chosen, then the probability of any specific ball being chosen from one of their bags is $\\frac{1}{9}$. Thus, the probability of any specific pair of balls being chosen (one from each bag) is $\\frac{1}{9} \\times \\frac{1}{9}=\\frac{1}{81}$.\n\nTherefore, to compute the desired probability, we must count the number of pairs $(x, y)$ where $y-x$ is $-8,-4,0,4,8$, and multiply this result by $\\frac{1}{81}$.\n\nMethod 1 \n\nIf $y-x=-8$, then $(x, y)$ must be $(9,1)$.\n\nIf $y-x=8$, then $(x, y)$ must be $(1,9)$.\n\nIf $y-x=-4$, then $(x, y)$ can be $(5,1),(6,2),(7,3),(8,4),(9,5)$.\n\nIf $y-x=4$, then $(x, y)$ can be $(1,5),(2,6),(3,7),(4,8),(5,9)$.\n\nIf $y-x=0$, then $(x, y)$ can be $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9)$.\n\nThere are thus 21 pairs $(x, y)$ that work, so the desired probability is $\\frac{21}{81}=\\frac{7}{27}$.\n\nMethod 2\n\nIf $x=9$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .\n\nIf $x=8$, then for $y-x$ to be a multiple of $4, y$ could be 8 or 4 .\n\nIf $x=7$, then for $y-x$ to be a multiple of $4, y$ could be 7 or 3 .\n\nIf $x=6$, then for $y-x$ to be a multiple of $4, y$ could be 6 or 2 .\n\nIf $x=5$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .\n\nIf $x=4$, then for $y-x$ to be a multiple of $4, y$ could be 8 or 4 .\n\nIf $x=3$, then for $y-x$ to be a multiple of $4, y$ could be 7 or 3 .\n\nIf $x=2$, then for $y-x$ to be a multiple of $4, y$ could be 6 or 2 .\n\nIf $x=1$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .\n\n\n\nThere are thus 21 pairs $(x, y)$ that work, so the desired probability is $\\frac{21}{81}=\\frac{7}{27}$." ]

[ "$\\frac{7}{27}$" ]

[ "Rewriting the equation, we obtain\n\n$$\n\\begin{aligned}\n2^{x+2} 5^{6-x} & =2^{x^{2}} 5^{x^{2}} \\\\\n1 & =2^{x^{2}} 2^{-2-x} 5^{x^{2}} 5^{x-6} \\\\\n1 & =2^{x^{2}-x-2} 5^{x^{2}+x-6} \\\\\n0 & =\\left(x^{2}-x-2\\right) \\log _{10} 2+\\left(x^{2}+x-6\\right) \\log _{10} 5 \\\\\n0 & =(x-2)(x+1) \\log _{10} 2+(x-2)(x+3) \\log _{10} 5 \\\\\n0 & =(x-2)\\left[(x+1) \\log _{10} 2+(x+3) \\log _{10} 5\\right] \\\\\n0 & =(x-2)\\left[\\left(\\log _{10} 2+\\log _{10} 5\\right) x+\\left(\\log _{10} 2+3 \\log 105\\right)\\right] \\\\\n0 & =(x-2)\\left[\\left(\\log _{10} 10\\right) x+\\log _{10}\\left(2 \\cdot 5^{3}\\right)\\right] \\\\\n0 & =(x-2)\\left(x+\\log _{10} 250\\right)\n\\end{aligned}\n$$\n\nTherefore, $x=2$ or $x=-\\log _{10} 250$.", "We take base 10 logarithms of both sides:\n\n$$\n\\begin{aligned}\n\\log _{10}\\left(2^{x+2} 5^{6-x}\\right) & =\\log _{10}\\left(10^{x^{2}}\\right) \\\\\n\\log _{10}\\left(2^{x+2}\\right)+\\log _{10}\\left(5^{6-x}\\right) & =x^{2} \\\\\n(x+2) \\log _{10} 2+(6-x) \\log _{10} 5 & =x^{2} \\\\\nx\\left(\\log _{10} 2-\\log _{10} 5\\right)+\\left(2 \\log _{10} 2+6 \\log _{10} 5\\right) & =x^{2} \\\\\nx^{2}-x\\left(\\log _{10} 2-\\log _{10} 5\\right)-\\left(2 \\log _{10} 2+6 \\log _{10} 5\\right) & =0\n\\end{aligned}\n$$\n\nNow, $\\log _{10} 2+\\log _{10} 5=\\log _{10} 10=1$ so $\\log _{10} 5=1-\\log _{10} 2$, so we can simplify the equation to\n\n$$\nx^{2}-x\\left(2 \\log _{10} 2-1\\right)-\\left(6-4 \\log _{10} 2\\right)=0\n$$\n\nThis is a quadratic equation in $x$, so should have at most 2 real solutions.\n\nBy the quadratic formula,\n\n$$\n\\begin{aligned}\nx & =\\frac{\\left(2 \\log _{10} 2-1\\right) \\pm \\sqrt{\\left(2 \\log _{10} 2-1\\right)^{2}-4(1)\\left(-\\left(6-4 \\log _{10} 2\\right)\\right)}}{2(1)} \\\\\n& =\\frac{\\left(2 \\log _{10} 2-1\\right) \\pm \\sqrt{4\\left(\\log _{10} 2\\right)^{2}-4\\left(\\log _{10} 2\\right)+1+24-16 \\log _{10} 2}}{2} \\\\\n& =\\frac{\\left(2 \\log _{10} 2-1\\right) \\pm \\sqrt{4\\left(\\log _{10} 2\\right)^{2}-20\\left(\\log _{10} 2\\right)+25}}{2} \\\\\n& =\\frac{\\left(2 \\log _{10} 2-1\\right) \\pm \\sqrt{\\left(2 \\log _{10} 2-5\\right)^{2}}}{2} \\\\\n& =\\frac{\\left(2 \\log _{10} 2-1\\right) \\pm\\left(5-2 \\log _{10} 2\\right)}{2}\n\\end{aligned}\n$$\n\nsince $5-2 \\log _{10} 2>0$.\n\nTherefore,\n\n$$\nx=\\frac{\\left(2 \\log _{10} 2-1\\right)+\\left(5-2 \\log _{10} 2\\right)}{2}=\\frac{4}{2}=2\n$$\n\nor\n\n$$\nx=\\frac{\\left(2 \\log _{10} 2-1\\right)-\\left(5-2 \\log _{10} 2\\right)}{2}=\\frac{4 \\log _{10} 2-6}{2}=2 \\log _{10} 2-3\n$$\n\n(Note that at any point, we could have used a calculator to convert to decimal approximations and solve.)" ]

[ "$2,-\\log _{10} 250$" ]

[ "First, we rewrite the system as\n\n$$\n\\begin{aligned}\n& x+\\log _{10} x=y-1 \\\\\n& (y-1)+\\log _{10}(y-1)=z-2 \\\\\n& (z-2)+\\log _{10}(z-2)=x\n\\end{aligned}\n$$\n\nSecond, we make the substitution $a=x, b=y-1$ and $c=z-2$, allowing us to rewrite\n\n\n\nthe system as\n\n$$\n\\begin{aligned}\na+\\log _{10} a & =b \\\\\nb+\\log _{10} b & =c \\\\\nc+\\log _{10} c & =a\n\\end{aligned}\n$$\n\nThird, we observe that $(a, b, c)=(1,1,1)$ is a solution, since $1+\\log _{10} 1=1+0=1$.\n\nNext, if $a>1$, then $\\log _{10} a>0$, so from (1),\n\n$$\nb=a+\\log _{10} a>a+0=a>1\n$$\n\nso $\\log _{10} b>0$, so from $(2)$,\n\n$$\nc=b+\\log _{10} b>b+0=b>a>1\n$$\n\nso $\\log _{10} c>0$, so from (3),\n\n$$\na=c+\\log _{10} c>c+0=c>b>a>1\n$$\n\nBut this says that $a>c>b>a$, which is a contradiction.\n\nTherefore, $a$ cannot be larger than 1 .\n\nLastly, if $0<a<1$ ( $a$ cannot be negative), then $\\log _{10} a<0$, so from (1),\n\n$$\nb=a+\\log _{10} a<a+0=a<1\n$$\n\nso $\\log _{10} b<0$, so from $(2)$,\n\n$$\nc=b+\\log _{10} b<b+0=b<a<1\n$$\n\nso $\\log _{10} c<0$, so from (3),\n\n$$\na=c+\\log _{10} c>c+0=c<b<a<1\n$$\n\nBut this says that $a<c<b<a$, which is a contradiction.\n\nTherefore, $a$ cannot be smaller than 1 either.\n\nThus, $a$ must equal 1.\n\nIf $a=1$, then $b=a+\\log _{10} a=1+\\log _{10} 1=1+0=1$ from (1), which will similarly give $c=1$ from (2).\n\nThus, the only solution to the system is $(a, b, c)=(1,1,1)=(x, y-1, z-2)$ since $a$ cannot be either larger than or smaller than 1 , so $(x, y, z)=(1,2,3)$." ]

[ "$1,2,3$" ]

[ "Since $80=2^{4} \\cdot 5$, its positive divisors are $1,2,4,5,8,10,16,20,40,80$.\n\nFor an integer $n$ to share exactly two positive common divisors with 80, these divisors must be either 1 and 2 or 1 and 5 . ( 1 is a common divisor of any two integers. The second common divisor must be a prime number since any composite divisor will cause there to be at least one more common divisor which is prime.)\n\nSince $1 \\leq n \\leq 30$ and $n$ is a multiple of 2 or of 5 , then the possible values of $n$ come from the list\n\n$$\n2,4,5,6,8,10,12,14,15,16,18,20,22,24,25,26,28,30\n$$\n\nWe remove the multiples of 4 from this list (since they would share at least the divisors $1,2,4$ with 80 ) and the multiples of 10 from this list (since they would share at least the divisors $1,2,5,10$ with 80 ).\n\nThis leaves the list\n\n$$\n2,5,6,14,15,18,22,25,26\n$$\n\nThe common divisors of any number from this list and 80 are either 1 and 2 or 1 and 5 . There are 9 such integers." ]

[ "9" ]

[ "We start with $f(50)$ and apply the given rules for the function until we reach $f(1)$ :\n\n$$\n\\begin{aligned}\nf(50) & =f(25) \\\\\n& =f(24)+1 \\\\\n& =f(12)+1 \\\\\n& =f(6)+1 \\\\\n& =f(3)+1 \\\\\n& =(f(2)+1)+1 \\\\\n& =f(1)+1+1 \\\\\n& =1+1+1 \\\\\n& =3\n\\end{aligned}\n$$\n\n(since 50 is even and $\\frac{1}{2}(50)=25$ )\n\n(since 25 is odd and $25-1=24$ )\n\n$$\n\\left(\\frac{1}{2}(24)=12\\right)\n$$\n\n$$\n\\begin{aligned}\n\\left(\\frac{1}{2}(12)\\right. & =6) \\\\\n\\left(\\frac{1}{2}(6)\\right. & =3) \\\\\n(3-1 & =2) \\\\\n\\left(\\frac{1}{2}(2)\\right. & =1) \\\\\n(f(1) & =1)\n\\end{aligned}\n$$\n\nTherefore, $f(50)=3$." ]

[ "3" ]

[ "Since the hexagon has perimeter 12 and has 6 sides, then each side has length 2 .\n\nSince equilateral $\\triangle P Q R$ has perimeter 12 , then its side length is 4 .\n\nConsider equilateral triangles with side length 2.\n\nSix of these triangles can be combined to form a regular hexagon with side length 2 and four of these can be combined to form an equilateral triangle with side length 4 .\n<img_3579>\n\nNote that the six equilateral triangles around the centre of the hexagon give a total central angle of $6 \\cdot 60^{\\circ}=360^{\\circ}$ (a complete circle) and the three equilateral triangles along each side of the large equilateral triangle make a straight angle of $180^{\\circ}\\left(\\right.$ since $3 \\cdot 60^{\\circ}=180^{\\circ}$ ). Also, the length of each side of the hexagon is 2 and the measure of each internal angle is $120^{\\circ}$, which means that the hexagon is regular. Similarly, the triangle is equilateral.\n\nSince the triangle is made from four identical smaller triangles and the hexagon is made from six of these smaller triangles, the ratio of the area of the triangle to the hexagon is $4: 6$ which is equivalent to $2: 3$." ]

[ "$\\frac{2}{3}$" ]

[ "Let $\\theta=10 k^{\\circ}$.\n\nThe given inequalities become $0^{\\circ}<\\theta<180^{\\circ}$ and $\\frac{5 \\sin \\theta-2}{\\sin ^{2} \\theta} \\geq 2$.\n\nWhen $0^{\\circ}<\\theta<180^{\\circ}, \\sin \\theta \\neq 0$.\n\nThis means that we can can multiply both sides by $\\sin ^{2} \\theta>0$ and obtain the equivalent inequalities:\n\n$$\n\\begin{aligned}\n\\frac{5 \\sin \\theta-2}{\\sin ^{2} \\theta} & \\geq 2 \\\\\n5 \\sin \\theta-2 & \\geq 2 \\sin ^{2} \\theta \\\\\n0 & \\geq 2 \\sin ^{2} \\theta-5 \\sin \\theta+2 \\\\\n0 & \\geq(2 \\sin \\theta-1)(\\sin \\theta-2)\n\\end{aligned}\n$$\n\nSince $\\sin \\theta \\leq 1$, then $\\sin \\theta-2 \\leq-1<0$ for all $\\theta$.\n\nTherefore, $(2 \\sin \\theta-1)(\\sin \\theta-2) \\leq 0$ exactly when $2 \\sin \\theta-1 \\geq 0$.\n\nNote that $2 \\sin \\theta-1 \\geq 0$ exactly when $\\sin \\theta \\geq \\frac{1}{2}$.\n\nTherefore, the original inequality is true exactly when $\\frac{1}{2} \\leq \\sin \\theta \\leq 1$.\n\nNote that $\\sin 30^{\\circ}=\\sin 150^{\\circ}=\\frac{1}{2}$ and $0^{\\circ}<\\theta<180^{\\circ}$.\n\nWhen $\\theta=0^{\\circ}, \\sin \\theta=0$.\n\nFrom $\\theta=0^{\\circ}$ to $\\theta=30^{\\circ}, \\sin \\theta$ increases from 0 to $\\frac{1}{2}$.\n\nFrom $\\theta=30^{\\circ}$ to $\\theta=150^{\\circ}, \\sin \\theta$ increases from $\\frac{1}{2}$ to 1 and then decreases to $\\frac{1}{2}$.\n\nFrom $\\theta=150^{\\circ}$ to $\\theta=180^{\\circ}, \\sin \\theta$ decreases from $\\frac{1}{2}$ to 0 .\n\nTherefore, the original inequality is true exactly when $30^{\\circ} \\leq \\theta \\leq 150^{\\circ}$ which is equivalent to $30^{\\circ} \\leq 10 k^{\\circ} \\leq 150^{\\circ}$ and to $3 \\leq k \\leq 15$.\n\nThe integers $k$ in this range are $k=3,4,5,6, \\ldots, 12,13,14,15$, of which there are 13 ." ]

[ "13" ]

[ "Among a group of $n$ people, there are $\\frac{n(n-1)}{2}$ ways of choosing a pair of these people:\n\nThere are $n$ people that can be chosen first.\n\nFor each of these $n$ people, there are $n-1$ people that can be chosen second.\n\nThis gives $n(n-1)$ orderings of two people.\n\nEach pair is counted twice (given two people A and B, we have counted both the\n\npair $\\mathrm{AB}$ and the pair $\\mathrm{BA})$, so the total number of pairs is $\\frac{n(n-1)}{2}$.\n\nWe label the four canoes W, X, Y, and Z.\n\nFirst, we determine the total number of ways to put the 8 people in the 4 canoes.\n\nWe choose 2 people to put in W. There are $\\frac{8 \\cdot 7}{2}$ pairs. This leaves 6 people for the remaining 3 canoes.\n\nNext, we choose 2 people to put in X. There are $\\frac{6 \\cdot 5}{2}$ pairs. This leaves 4 people for the remaining 2 canoes.\n\nNext, we choose 2 people to put in Y. There are $\\frac{4 \\cdot 3}{2}$ pairs. This leaves 2 people for the remaining canoe.\n\nThere is now 1 way to put the remaining people in $\\mathrm{Z}$.\n\nTherefore, there are\n\n$$\n\\frac{8 \\cdot 7}{2} \\cdot \\frac{6 \\cdot 5}{2} \\cdot \\frac{4 \\cdot 3}{2}=\\frac{8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3}{2^{3}}=7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3\n$$\n\nways to put the 8 people in the 4 canoes.\n\nNow, we determine the number of ways in which no two of Barry, Carrie and Mary will be in the same canoe.\n\nThere are 4 possible canoes in which Barry can go.\n\nThere are then 3 possible canoes in which Carrie can go, because she cannot go in the same canoe as Barry.\n\nThere are then 2 possible canoes in which Mary can go, because she cannot go in the same canoe as Barry or Carrie.\n\nThis leaves 5 people left to put in the canoes.\n\nThere are 5 choices of the person that can go with Barry, and then 4 choices of the person that can go with Carrie, and then 3 choices of the person that can go with Mary.\n\nThe remaining 2 people are put in the remaining empty canoe.\n\nThis means that there are $4 \\cdot 3 \\cdot 2 \\cdot 5 \\cdot 4 \\cdot 3$ ways in which the 8 people can be put in 4 canoes so that no two of Barry, Carrie and Mary are in the same canoe.\n\nTherefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is $\\frac{4 \\cdot 3 \\cdot 2 \\cdot 5 \\cdot 4 \\cdot 3}{7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3}=\\frac{4 \\cdot 3 \\cdot 2}{7 \\cdot 6}=\\frac{24}{42}=\\frac{4}{7}$.", "Let $p$ be the probability that two of Barry, Carrie and Mary are in the same canoe.\n\nThe answer to the original problem will be $1-p$.\n\nLet $q$ be the probability that Barry and Carrie are in the same canoe.\n\nBy symmetry, the probability that Barry and Mary are in the same canoe also equals $q$ as does the probability that Carrie and Mary are in the same canoe.\n\nThis means that $p=3 q$.\n\nSo we calculate $q$.\n\nTo do this, we put Barry in a canoe. Since there are 7 possible people who can go in the canoe with him, then the probability that Carrie is in the canoe with him equals $\\frac{1}{7}$. The other 6 people can be put in the canoes in any way.\n\nThis means that the probability that Barry and Carrie are in the same canoe is $q=\\frac{1}{7}$.\n\nTherefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is $1-3 \\cdot \\frac{1}{7}$ or $\\frac{4}{7}$." ]

[ "$\\frac{4}{7}$" ]

[ "Suppose that $W Y$ makes an angle of $\\theta$ with the horizontal.\n\n<img_3532>\n\nSince the slope of $W Y$ is 2 , then $\\tan \\theta=2$, since the tangent of an angle equals the slope of a line that makes this angle with the horizontal.\n\nSince $\\tan \\theta=2>1=\\tan 45^{\\circ}$, then $\\theta>45^{\\circ}$.\n\nNow $W Y$ bisects $\\angle Z W X$, which is a right-angle.\n\nTherefore, $\\angle Z W Y=\\angle Y W X=45^{\\circ}$.\n\nTherefore, $W X$ makes an angle of $\\theta+45^{\\circ}$ with the horizontal and $W Z$ makes an angle of $\\theta-45^{\\circ}$ with the horizontal. Since $\\theta>45^{\\circ}$, then $\\theta-45^{\\circ}>0$ and $\\theta+45^{\\circ}>90^{\\circ}$.\n\nWe note that since $W Z$ and $X Y$ are parallel, then the slope of $X Y$ equals the slope of $W Z$.\n\nTo calculate the slopes of $W X$ and $W Z$, we can calculate $\\tan \\left(\\theta+45^{\\circ}\\right)$ and $\\tan \\left(\\theta-45^{\\circ}\\right)$.\n\nUsing the facts that $\\tan (A+B)=\\frac{\\tan A+\\tan B}{1-\\tan A \\tan B}$ and $\\tan (A-B)=\\frac{\\tan A-\\tan B}{1+\\tan A \\tan B}$, we obtain:\n\n$$\n\\begin{aligned}\n& \\tan \\left(\\theta+45^{\\circ}\\right)=\\frac{\\tan \\theta+\\tan 45^{\\circ}}{1-\\tan \\theta \\tan 45^{\\circ}}=\\frac{2+1}{1-(2)(1)}=-3 \\\\\n& \\tan \\left(\\theta-45^{\\circ}\\right)=\\frac{\\tan \\theta-\\tan 45^{\\circ}}{1-\\tan \\theta \\tan 45^{\\circ}}=\\frac{2-1}{1+(2)(1)}=\\frac{1}{3}\n\\end{aligned}\n$$\n\nTherefore, the sum of the slopes of $W X$ and $X Y$ is $-3+\\frac{1}{3}=-\\frac{8}{3}$.", "Consider a square $W X Y Z$ whose diagonal $W Y$ has slope 2 .\n\nTranslate this square so that $W$ is at the origin $(0,0)$. Translating a shape in the plane does not affect the slopes of any line segments.\n\nLet the coordinates of $Y$ be $(2 a, 2 b)$ for some non-zero numbers $a$ and $b$.\n\nSince the slope of $W Y$ is 2 , then $\\frac{2 b-0}{2 a-0}=2$ and so $2 b=4 a$ or $b=2 a$.\n\nThus, the coordinates of $Y$ can be written as $(2 a, 4 a)$.\n\nLet $C$ be the centre of square $W X Y Z$.\n\nThen $C$ is the midpoint of $W Y$, so $C$ has coordinates $(a, 2 a)$.\n\nWe find the slopes of $W X$ and $X Y$ by finding the coordinates of $X$.\n\nConsider the segment $X C$.\n\nSince the diagonals of a square are perpendicular, then $X C$ is perpendicular to $W C$.\n\nSince the slope of $W C$ is 2 , then the slopes of $X C$ and $Z C$ are $-\\frac{1}{2}$.\n\nSince the diagonals of a square are equal in length and $C$ is the midpoint of both diagonals, then $X C=W C$.\n\nSince $W C$ and $X C$ are perpendicular and equal in length, then the \"rise/run triangle\" above $X C$ will be a $90^{\\circ}$ rotation of the \"rise/run triangle\" below $W C$.\n\n<img_3997>\n\nThis is because these triangles are congruent (each is right-angled, their hypotenuses are of equal length, and their remaining angles are equal) and their hypotenuses are perpendicular.\n\nIn this diagram, we have assumed that $X$ is to the left of $W$ and $Z$ is to the right of $W$. Since the slopes of parallel sides are equal, it does not matter which vertex is labelled $X$ and which is labelled $Z$. We would obtain the same two slopes, but in a different order. To get from $W(0,0)$ to $C(a, 2 a)$, we go up $2 a$ and right $a$.\n\nThus, to get from $C(a, 2 a)$ to $X$, we go left $2 a$ and up $a$.\n\nTherefore, the coordinates of $X$ are $(a-2 a, 2 a+a)$ or $(-a, 3 a)$.\n\nThus, the slope of $W X$ is $\\frac{3 a-0}{-a-0}=-3$.\n\nSince $X Y$ is perpendicular to $W X$, then its slope is the negative reciprocal of -3 , which is $\\frac{1}{3}$.\n\nThe sum of the slopes of $W X$ and $X Y$ is $-3+\\frac{1}{3}=-\\frac{8}{3}$." ]

[ "$-\\frac{8}{3}$" ]

[ "Since the base of a logarithm must be positive and cannot equal 1 , then $x>0$ and $x \\neq \\frac{1}{2}$ and $x \\neq \\frac{1}{3}$.\n\nThis tells us that $\\log 2 x$ and $\\log 3 x$ exist and do not equal 0 , which we will need shortly when we apply the change of base formula.\n\nWe note further that $48=2^{4} \\cdot 3$ and $162=3^{4} \\cdot 2$ and $\\sqrt[3]{3}=3^{1 / 3}$ and $\\sqrt[3]{2}=2^{1 / 3}$. Using logarithm rules, the following equations are equivalent:\n\n$$\n\\begin{aligned}\n\\log _{2 x}(48 \\sqrt[3]{3}) & =\\log _{3 x}(162 \\sqrt[3]{2}) \\\\\n\\frac{\\log \\left(2^{4} \\cdot 3 \\cdot 3^{1 / 3}\\right)}{\\log 2 x} & =\\frac{\\log \\left(3^{4} \\cdot 2 \\cdot 2^{1 / 3}\\right)}{\\log 3 x} \\quad \\text { (change of base formula) } \\\\\n\\frac{\\log \\left(2^{4} \\cdot 3^{4 / 3}\\right)}{\\log 2+\\log x} & =\\frac{\\log \\left(3^{4} \\cdot 2^{4 / 3}\\right)}{\\log 3+\\log x} \\quad(\\log a b=\\log a+\\log b) \\\\\n\\frac{\\log \\left(2^{4}\\right)+\\log \\left(3^{4 / 3}\\right)}{\\log 2+\\log x} & =\\frac{\\log \\left(3^{4}\\right)+\\log \\left(2^{4 / 3}\\right)}{\\log 3+\\log x} \\quad(\\log a b=\\log a+\\log b) \\\\\n\\frac{4 \\log 2+\\frac{4}{3} \\log 3}{\\log 2+\\log x} & =\\frac{4 \\log 3+\\frac{4}{3} \\log 2}{\\log 3+\\log x} \\quad\\left(\\log \\left(a^{c}\\right)=c \\log a\\right)\n\\end{aligned}\n$$\n\nCross-multiplying, we obtain\n\n$$\n\\left(4 \\log 2+\\frac{4}{3} \\log 3\\right)(\\log 3+\\log x)=\\left(4 \\log 3+\\frac{4}{3} \\log 2\\right)(\\log 2+\\log x)\n$$\n\nExpanding the left side, we obtain\n\n$$\n4 \\log 2 \\log 3+\\frac{4}{3}(\\log 3)^{2}+\\left(4 \\log 2+\\frac{4}{3} \\log 3\\right) \\log x\n$$\n\nExpanding the right side, we obtain\n\n$$\n4 \\log 3 \\log 2+\\frac{4}{3}(\\log 2)^{2}+\\left(4 \\log 3+\\frac{4}{3} \\log 2\\right) \\log x\n$$\n\nSimplifying and factoring, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n\\frac{4}{3}(\\log 3)^{2}-\\frac{4}{3}(\\log 2)^{2} & =\\log x\\left(4 \\log 3+\\frac{4}{3} \\log 2-4 \\log 2-\\frac{4}{3} \\log 3\\right) \\\\\n\\frac{4}{3}(\\log 3)^{2}-\\frac{4}{3}(\\log 2)^{2} & =\\log x\\left(\\frac{8}{3} \\log 3-\\frac{8}{3} \\log 2\\right) \\\\\n(\\log 3)^{2}-(\\log 2)^{2} & =2 \\log x(\\log 3-\\log 2) \\\\\n\\log x & =\\frac{(\\log 3)^{2}-(\\log 2)^{2}}{2(\\log 3-\\log 2)} \\\\\n\\log x & =\\frac{(\\log 3-\\log 2)(\\log 3+\\log 2)}{2(\\log 3-\\log 2)} \\\\\n\\log x & =\\frac{\\log 3+\\log 2}{2} \\\\\n\\log x & =\\frac{1}{2} \\log 6 \\\\\n\\log x & =\\log (\\sqrt{6})\n\\end{aligned}\n$$\n\nand so $x=\\sqrt{6}$." ]

[ "$\\sqrt{6}$" ]

[ "We draw part of the array using the information that $A_{0}=A_{1}=A_{2}=0$ and $A_{3}=1$ :\n\n$$\n\\begin{array}{l|l|l|l|l|l|l|lll|c|c|c|c|c|c}\n\\cdots & A_{0} & A_{1} & A_{2} & A_{3} & A_{4} & A_{5} & \\cdots & \\cdots & 0 & 0 & 0 & 1 & A_{4} & A_{5} & \\cdots \\\\\n\\hline \\cdots & B_{0} & B_{1} & B_{2} & B_{3} & B_{4} & B_{5} & \\cdots & \\cdots & B_{0} & B_{1} & B_{2} & B_{3} & B_{4} & B_{5} & \\cdots\n\\end{array}\n$$\n\nSince $A_{1}$ is the average of $A_{0}, B_{1}$ and $A_{2}$, then $A_{1}=\\frac{A_{0}+B_{1}+A_{2}}{3}$ or $3 A_{1}=A_{0}+B_{1}+A_{2}$. Thus, $3(0)=0+B_{1}+0$ and so $B_{1}=0$.\n\nSince $A_{2}$ is the average of $A_{1}, B_{2}$ and $A_{3}$, then $3 A_{2}=A_{1}+B_{2}+A_{3}$ and so $3(0)=0+B_{2}+1$ which gives $B_{2}=-1$.\n\nSince $B_{2}$ is the average of $B_{1}, A_{2}$ and $B_{3}$, then $3 B_{2}=B_{1}+A_{2}+B_{3}$ and so $3(-1)=0+0+B_{3}$ which gives $B_{3}=-3$.\n\nSo far, this gives\n\n$$\n\\begin{array}{l|c|c|c|c|c|c|l}\n\\cdots & 0 & 0 & 0 & 1 & A_{4} & A_{5} & \\cdots \\\\\n\\hline \\cdots & B_{0} & 0 & -1 & -3 & B_{4} & B_{5} & \\cdots\n\\end{array}\n$$\n\nSince $A_{3}$ is the average of $A_{2}, B_{3}$ and $A_{4}$, then $3 A_{3}=A_{2}+B_{3}+A_{4}$ and so $3(1)=$ $0+(-3)+A_{4}$ which gives $A_{4}=6$." ]

[ "6" ]

[ "If $P$ is the original population of Alphaville and Betaville,\n\n$$\n\\begin{aligned}\nP(.971)(1.089)(1.069) & =P\\left(1+\\frac{r}{100}\\right)^{3} \\\\\n1.1303 & =\\left(1+\\frac{r}{100}\\right)^{3}\n\\end{aligned}\n$$\n\nFrom here,\n\nPossibility 1\n\n$$\n\\begin{aligned}\n1+\\frac{r}{100} & =(1.1303)^{\\frac{1}{3}} \\\\\n1+\\frac{r}{100} & =1.0416 \\\\\nr & \\doteq 4.2 \\%\n\\end{aligned}\n$$\n\nOr, Possibility 2\n\n$$\n\\begin{aligned}\n3 \\log \\left(1+\\frac{r}{100}\\right) & =\\log 1.1303 \\\\\n\\log \\left(1+\\frac{r}{100}\\right) & =.01773 \\\\\n1+\\frac{r}{100} & =1.0416 \\\\\nr & \\doteq 4.2 \\%\n\\end{aligned}\n$$" ]

[ "4.2" ]

[ "The intersection takes place where,\n\n$$\n\\begin{aligned}\n& \\log _{10}(x-2)=1-\\log _{10}(x+1) \\\\\n& \\log _{10}(x-2)+\\log _{10}(x+1)=1 \\\\\n& \\log _{10}\\left(x^{2}-x-2\\right)=1\n\\end{aligned}\n$$\n\n\n\n$$\n\\begin{aligned}\n& x^{2}-x-2=10 \\\\\n& x^{2}-x-12=0 \\\\\n& (x-4)(x+3)=0 \\\\\n& x=4 \\text { or }-3\n\\end{aligned}\n$$\n\nFor $x=-3, y$ is not defined.\n\nFor $x=4, y=\\log _{10} 2 \\doteq 0.3$.\n\nThe graphs therefore intersect at $\\left(4, \\log _{10} 2\\right)$." ]

[ "$(4, \\log _{10} 2)$" ]

[ "Let $N$ be the number formed by the rightmost two digits of the year in which Charlie was born. Then his current age is $100-N+14=114-N$. Setting this equal to $2 N$ and solving yields $N=38$, hence the answer is 1938 .", "Let $N$ be the number formed by the rightmost two digits of the year in which Charlie was born. The number of years from 1900 to 2014 can be thought of as the number of years before Charlie was born plus the number of years since he was born, or $N$ plus Charlie's age. Thus $N+2 N=114$, which leads to $N=38$, so the answer is 1938 ." ]

[ "1938" ]

[ "If Pat and Chris get the same answer, then $A+(B \\cdot C)=(A+B) \\cdot C$, or $A+B C=A C+B C$, or $A=A C$. This equation is true if $A=0$ or $C=1$; the equation places no restrictions on $B$. There are 25 triples $(A, B, C)$ where $A=0,25$ triples where $C=1$, and 5 triples where $A=0$ and $C=1$. As all triples are equally likely, the answer is $\\frac{25+25-5}{5^{3}}=\\frac{45}{125}=\\frac{\\mathbf{9}}{\\mathbf{2 5}}$." ]

[ "$\\frac{9}{25}$" ]

[ "There are 6 people, so there are $6 !=720$ permutations. However, for each arrangement of the boys, there are $3 !=6$ permutations of the girls, of which only one yields an acceptable lineup. The same logic holds for the boys. Thus the total number of permutations must be divided by $3 ! \\cdot 3 !=36$, yielding $6 ! /(3 ! \\cdot 3 !)=\\mathbf{2 0}$ acceptable lineups.", "Once the positions of Greg, Peter, and Bobby are determined, the entire lineup is determined, because there is only one acceptable ordering of the three girls. Because the boys occupy three of the six positions, there are $\\left(\\begin{array}{l}6 \\\\ 3\\end{array}\\right)=\\mathbf{2 0}$ acceptable lineups." ]

[ "20" ]

[ "Using the Law of Cosines, $a^{2}+b^{2}-2 a b \\cos C=c^{2}$ implies\n\n$$\nb \\cos C=\\frac{a^{2}+b^{2}-c^{2}}{2 a}\n$$\n\nSimilarly,\n\n$$\nc \\cos B=\\frac{a^{2}-b^{2}+c^{2}}{2 a}\n$$\n\nThus\n\n$$\n\\begin{aligned}\nb \\cos C-c \\cos B & =\\frac{a^{2}+b^{2}-c^{2}}{2 a}-\\frac{a^{2}-b^{2}+c^{2}}{2 a} \\\\\n& =\\frac{2 b^{2}-2 c^{2}}{2 a} \\\\\n& =\\frac{b^{2}-c^{2}}{a} .\n\\end{aligned}\n$$\n\n\n\nWith the given values, the result is $\\left(17^{2}-13^{2}\\right) / 12=120 / 12=\\mathbf{1 0}$.", "Let $H$ be the foot of the altitude from $A$ to $\\overline{B C}$; let $B H=x$, $C H=y$, and $A H=h$. Then $b \\cos C=y, c \\cos B=x$, and the desired quantity is $Q=y-x$. However, $y+x=a$, so $y^{2}-x^{2}=a Q$. By the Pythagorean Theorem, $y^{2}=b^{2}-h^{2}$ and $x^{2}=c^{2}-h^{2}$, so $y^{2}-x^{2}=\\left(b^{2}-h^{2}\\right)-\\left(c^{2}-h^{2}\\right)=b^{2}-c^{2}$. Thus $a Q=b^{2}-c^{2}$, and $Q=\\frac{b^{2}-c^{2}}{a}$\n\nWith the given values, the result is $\\left(17^{2}-13^{2}\\right) / 12=120 / 12=\\mathbf{1 0}$." ]

[ "10" ]

[ "Let $P$ denote a palindromic word, let $Q$ denote any word, and let $\\bar{R}$ denote the reverse of word $R$. Note that if two consecutive terms of the sequence are $a_{n}=P, a_{n+1}=Q$, then $a_{n+2}=Q \\bar{P}=Q P$ and $a_{n+3}=Q P \\bar{Q}$. Thus if $a_{n}$ is a palindrome, so is $a_{n+3}$. Because $a_{1}$ and $a_{2}$ are both palindromes, then so must be all terms in the subsequences $a_{4}, a_{7}, a_{10}, \\ldots$ and $a_{5}, a_{8}, a_{11}, \\ldots$\n\nTo show that the other terms are not palindromes, note that if $P^{\\prime}$ is not a palindrome, then $Q P^{\\prime} \\bar{Q}$ is also not a palindrome. Thus if $a_{n}$ is not a palindrome, then $a_{n+3}$ is not a palindrome either. Because $a_{3}=O X$ is not a palindrome, neither is any term of the subsequence $a_{6}, a_{9}, a_{12}, \\ldots$ (Alternatively, counting the number of $X$ 's in each word $a_{i}$ shows that the number of $X$ 's in $a_{3 k}$ is odd. So if $a_{3 k}$ were to be a palindrome, it would have to have an odd number of letters, with an $X$ in the middle. However, it can be shown that the length of $a_{3 k}$ is even. Thus $a_{3 k}$ cannot be a palindrome.)\n\nIn total there are $1000-333=\\mathbf{6 6 7}$ palindromes among the first 1000 terms." ]

[ "667" ]

[ "Let $D(n)$ be the number of divisors of the integer $n$. Note that if $D(214 n)=D(2014 n)$ and if some $p$ divides $n$ and is relatively prime to both 214 and 2014 , then $D\\left(\\frac{214 n}{p}\\right)=D\\left(\\frac{2014 n}{p}\\right)$. Thus any prime divisor of the smallest possible positive $n$ will be a divisor of $214=2 \\cdot 107$ or $2014=2 \\cdot 19 \\cdot 53$. For the sake of convenience, write $n=2^{a-1} 19^{b-1} 53^{c-1} 107^{d-1}$, where $a, b, c, d \\geq 1$. Then $D(214 n)=(a+1) b c(d+1)$ and $D(2014 n)=(a+1)(b+1)(c+1) d$. Divide both sides by $a+1$ and expand to get $b c d+b c=b c d+b d+c d+d$, or $b c-b d-c d-d=0$.\n\nBecause the goal is to minimize $n$, try $d=1$ : $b c-b-c-1=0 \\Rightarrow(b-1)(c-1)=2$, which has solutions $(b, c)=(2,3)$ and $(3,2)$. The latter gives the smaller value for $n$, namely $19^{2} \\cdot 53=$ 19133. The only quadruples $(a, b, c, d)$ that satisfy $2^{a-1} 19^{b-1} 53^{c-1} 107^{d-1}<19133$ and $d>1$ are $(1,1,2,2),(1,2,1,2)$, and $(1,1,1,3)$. None of these quadruples satisfies $b c-b d-c d-d=0$, so the minimum value is $n=\\mathbf{1 9 1 3 3}$." ]

[ "19133" ]

[ "Because $N$ is greater than 20, the base-20 and base-14 representations of $N$ must be at least two digits long. The smallest possible case is that $N$ is a two-digit palindrome in both bases. Then $N=20 a+a=21 a$, where $1 \\leq a \\leq 19$. Similarly, in order to be a two-digit palindrome in base $14, N=14 b+b=15 b$, with $1 \\leq b \\leq 13$. So $N$ would have to be a multiple of both 21 and 15 . The least common multiple of 21 and 15 is 105 , which has the base 20 representation of $105=55_{20}$ and the base-14 representation of $105=77_{14}$, both of which are palindromes. Thus the answer is 105." ]

[ "105" ]

[ "The ratio of binomial coefficients $\\left(\\begin{array}{c}1000 \\\\ k\\end{array}\\right) /\\left(\\begin{array}{c}1000 \\\\ k+1\\end{array}\\right)=\\frac{k+1}{1000-k}$. Because 1000 is 1 less than a multiple of 7 , namely $1001=7 \\cdot 11 \\cdot 13$, either $1000-k$ and $k+1$ are both multiples of 7 or neither is. Hence whenever the numerator is divisible by 7, the denominator is also. Thus for the largest value of $k$ such that $\\left(\\begin{array}{c}1000 \\\\ k\\end{array}\\right)$ is a multiple of $7, \\frac{k+1}{1000-k}$ must equal $7 \\cdot \\frac{p}{q}$, where $p$ and $q$ are relatively prime integers and $7 \\nmid q$. The only way this can happen is when $k+1$ is a multiple of 49 , the greatest of which less than 1000 is 980 . Therefore the greatest value of $k$ satisfying the given conditions is $980-1=\\mathbf{9 7 9}$.", "Rewrite 1000 in base 7: $1000=2626_{7}$. Let $k=\\underline{a} \\underline{b} \\underline{c}_{7}$. By Lucas's Theorem, $\\left(\\begin{array}{c}1000 \\\\ k\\end{array}\\right) \\equiv\\left(\\begin{array}{l}2 \\\\ a\\end{array}\\right)\\left(\\begin{array}{l}6 \\\\ b\\end{array}\\right)\\left(\\begin{array}{l}2 \\\\ c\\end{array}\\right)\\left(\\begin{array}{l}6 \\\\ d\\end{array}\\right) \\bmod 7$. The binomial coefficient $\\left(\\begin{array}{l}p \\\\ q\\end{array}\\right) \\stackrel{a}{=} 0$ only when $q>p$. Base 7 digits cannot exceed 6 , and $k \\leq 1000$, thus the greatest value of $k$ that works is $2566_{7}=\\mathbf{9 7 9}$. (Alternatively, the least value of $k$ that works is $30_{7}=21$; because $\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)=\\left(\\begin{array}{c}n \\\\ n-k\\end{array}\\right)$, the greatest such $k$ is $1000-21=979$.)" ]

[ "979" ]

[ "For a tenuous function $g$, let $S_{g}=g(1)+g(2)+\\cdots+g(20)$. Then:\n\n$$\n\\begin{aligned}\nS_{g} & =(g(1)+g(20))+(g(2)+g(19))+\\cdots+(g(10)+g(11)) \\\\\n& \\geq\\left(20^{2}+1\\right)+\\left(19^{2}+1\\right)+\\cdots+\\left(11^{2}+1\\right) \\\\\n& =10+\\sum_{k=11}^{20} k^{2} \\\\\n& =2495 .\n\\end{aligned}\n$$\n\n\n\nThe following argument shows that if a tenuous function $g$ attains this sum, then $g(1)=$ $g(2)=\\cdots=g(10)$. First, if the sum equals 2495 , then $g(1)+g(20)=20^{2}+1, g(2)+g(19)=$ $19^{2}+1, \\ldots, g(10)+g(11)=11^{2}+1$. If $g(1)<g(2)$, then $g(1)+g(19)<19^{2}+1$, which contradicts the tenuousness of $g$. Similarly, if $g(2)>g(1)$, then $g(2)+g(20)<20^{2}+1$. Therefore $g(1)=g(2)$. Analogously, comparing $g(1)$ and $g(3), g(1)$ and $g(4)$, etc. shows that $g(1)=g(2)=g(3)=\\cdots=g(10)$.\n\nNow consider all functions $g$ for which $g(1)=g(2)=\\cdots=g(10)=a$ for some integer $a$. Then $g(n)=n^{2}+1-a$ for $n \\geq 11$. Because $g(11)+g(11)>11^{2}=121$, it is the case that $g(11) \\geq 61$. Thus $11^{2}+1-a \\geq 61 \\Rightarrow a \\leq 61$. Thus the smallest possible value for $g(14)$ is $14^{2}+1-61=\\mathbf{1 3 6}$." ]

[ "136" ]

[ "Pentagon $T N Y W R$ fits inside square $T A Y B$, where $A=(6,0)$ and $B=(0,6)$. The region of $T A Y B$ not in $T N Y W R$ consists of triangles $\\triangle N A Y$ and $\\triangle W B R$, as shown below.\n\n<img_3654>\n\nThus\n\n$$\n\\begin{aligned}\n{[T N Y W R] } & =[T A Y B]-[N A Y]-[W B R] \\\\\n& =6^{2}-\\frac{1}{2} \\cdot 4 \\cdot 6-\\frac{1}{2} \\cdot 2 \\cdot 4 \\\\\n& =\\mathbf{2 0} .\n\\end{aligned}\n$$" ]

[ "20" ]

[ "Let $r$ and $s$ denote the zeros of the polynomial $x^{2}-3 T x+T^{2}$. The rectangle's diagonal has length $\\sqrt{r^{2}+s^{2}}=\\sqrt{(r+s)^{2}-2 r s}$. Recall that for a quadratic polynomial $a x^{2}+b x+c$, the sum of its zeros is $-b / a$, and the product of its zeros is $c / a$. In this particular instance, $r+s=3 T$ and $r s=T^{2}$. Thus the length of the rectangle's diagonal is $\\sqrt{9 T^{2}-2 T^{2}}=T \\cdot \\sqrt{7}$. With $T=20$, the rectangle's diagonal is $\\mathbf{2 0} \\sqrt{\\mathbf{7}}$." ]

[ "$20 \\sqrt{7}$" ]

[ "Write $w=k+\\alpha$, where $k$ is an integer, and $0 \\leq \\alpha<1$. Then\n\n$$\nT=1^{2}+2^{2}+\\cdots+k^{2}+(k+1)^{2} \\cdot \\alpha .\n$$\n\nComputing $\\lceil 2 w\\rceil$ requires computing $w$ to the nearest half-integer. First obtain the integer $k$. As $\\sqrt{7}>2$, with $T=20 \\sqrt{7}$, one obtains $T>40$. As $1^{2}+2^{2}+3^{2}+4^{2}=30$, it follows that $k \\geq 4$. To obtain an upper bound for $k$, note that $700<729$, so $10 \\sqrt{7}<27$, and $T=20 \\sqrt{7}<54$. As $1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=55$, it follows that $4<w<5$, and hence $k=4$.\n\nIt now suffices to determine whether or not $\\alpha>0.5$. To this end, one must determine whether $T>1^{2}+2^{2}+3^{2}+4^{2}+5^{2} / 2=42.5$. Indeed, note that $2.5^{2}=6.25<7$, so $T>(20)(2.5)=50$. It follows that $\\alpha>0.5$, so $4.5<w<5$. Thus $9<2 w<10$, and $\\lceil 2 w\\rceil=\\mathbf{1 0}$.", "Once it has been determined that $4<w<5$, the formula for $T$ yields $1+4+9+16+25 \\cdot \\alpha=20 \\sqrt{7}$, hence $\\alpha=\\frac{4 \\sqrt{7}-6}{5}$. Thus $2 \\alpha=\\frac{8 \\sqrt{7}-12}{5}=\\frac{\\sqrt{448}-12}{5}>\\frac{21-12}{5}=1.8$. Because $2 w=2 k+2 \\alpha$, it follows that $\\lceil 2 w\\rceil=\\lceil 8+2 \\alpha\\rceil=\\mathbf{1 0}$, because $1.8<2 \\alpha<2$." ]

[ "10" ]

[ "Note that if $p$ is prime, then $\\operatorname{gcd}\\left(p^{3}, p !\\right)=p$. A good strategy is to look for values of $n$ with several (not necessarily distinct) prime factors so that $n^{3}$ and $n$ ! will have many factors in common. For example, if $n=6, n^{3}=216=2^{3} \\cdot 3^{3}$ and $n !=720=2^{4} \\cdot 3^{2} \\cdot 5$, so $\\operatorname{gcd}(216,720)=2^{3} \\cdot 3^{2}=72$. Because 7 is prime, try $n=8$. Notice that $8^{3}=2^{9}$ while $8 !=2^{7} \\cdot 3^{2} \\cdot 5 \\cdot 7$. Thus $\\operatorname{gcd}(512,8 !)=2^{7}=128>100$, hence the smallest value of $n$ is $\\mathbf{8}$." ]

[ "8" ]

[ "If there were $n$ people at the party, including Ed, and if Ed had not left early, there would have been $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)$ handshakes. Because Ed left early, the number of handshakes is strictly less than that, but greater than $\\left(\\begin{array}{c}n-1 \\\\ 2\\end{array}\\right)$ (everyone besides Ed shook everyone else's hand). So find the least number $n$ such that $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right) \\geq 160$. The least such $n$ is 19 , because $\\left(\\begin{array}{c}18 \\\\ 2\\end{array}\\right)=153$ and $\\left(\\begin{array}{c}19 \\\\ 2\\end{array}\\right)=171$. Therefore there were 19 people at the party. However, $171-160=11$ handshakes never took place. Therefore the number of people who shook hands with Ed is $19-11-1=7$." ]

[ "7" ]

[ "By the recursive definition, notice that $u_{6}=89=3 u_{5}-u_{4}$ and $u_{5}=3 u_{4}-u_{3}=3 u_{4}-5$. This is a linear system of equations. Write $3 u_{5}-u_{4}=89$ and $-3 u_{5}+9 u_{4}=15$ and add to obtain $u_{4}=13$. Now apply the recursive definition to obtain $u_{5}=34$ and $u_{7}=\\mathbf{2 3 3}$.", "Notice that the given values are both Fibonacci numbers, and that in the Fibonacci sequence, $f_{1}=f_{2}=1, f_{5}=5$, and $f_{11}=89$. That is, 5 and 89 are six terms apart in the Fibonacci sequence, and only three terms apart in the given sequence. This relationship is not a coincidence: alternating terms in the Fibonacci sequence satisfy the given recurrence relation for the sequence $\\left\\{u_{n}\\right\\}$, that is, $f_{n+4}=3 f_{n+2}-f_{n}$. Proof: if $f_{n}=a$ and $f_{n+1}=b$, then $f_{n+2}=a+b, f_{n+3}=a+2 b$, and $f_{n+4}=2 a+3 b=3(a+b)-b=3 f_{n+2}-f_{n}$. To compute the final result, continue out the Fibonacci sequence to obtain $f_{12}=144$ and $u_{7}=f_{13}=233$." ]

[ "233" ]

[ "There are $\\left(\\begin{array}{c}17 \\\\ 2\\end{array}\\right)=136$ possible pairs of dishes, so $\\mathcal{F}_{17}$ must have 136 people." ]

[ "136" ]

[ "With $d$ dishes there are $\\left(\\begin{array}{l}d \\\\ 2\\end{array}\\right)=\\frac{d^{2}-d}{2}$ possible pairs, so $n=\\frac{d^{2}-d}{2}$. Then $2 n=d^{2}-d$, or $d^{2}-d-2 n=0$. Using the quadratic formula yields $d=\\frac{1+\\sqrt{1+8 n}}{2}$ (ignoring the negative value)." ]

[ "$d=\\frac{1+\\sqrt{1+8 n}}{2}$" ]

[ "Because the town is full, each pair of dishes is cooked by exactly one resident, so it is simplest to identify residents by the pairs of dishes they cook. Suppose the first resident cooks $\\left(d_{1}, d_{2}\\right)$, the second resident $\\left(d_{2}, d_{3}\\right)$, the third resident $\\left(d_{3}, d_{4}\\right)$, and so on, until the sixth resident, who cooks $\\left(d_{6}, d_{1}\\right)$. Then there are 8 choices for $d_{1}$ and 7 choices for $d_{2}$. There are only 6 choices for $d_{3}$, because $d_{3} \\neq d_{1}$ (otherwise two residents would cook the same pair of dishes). For $k>3$, the requirement that no two intermediate residents cook the same dishes implies that $d_{k+1}$ cannot equal any of $d_{1}, \\ldots, d_{k-1}$, and of course $d_{k}$ and $d_{k+1}$ must be distinct dishes. Hence there are $8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3=20,160$ six-person resident cycles, not accounting for different starting points in the cycle and the two different directions to go around the cycle. Taking these into account, there are $20,160 /(6 \\cdot 2)=1,680$ distinguishable resident cycles." ]

[ "1680" ]

[ "First, we compute the number of distinguishable resident cycles of length 6 in $\\mathcal{F}_{8}$.\n\nBecause the town is full, each pair of dishes is cooked by exactly one resident, so it is simplest to identify residents by the pairs of dishes they cook. Suppose the first resident cooks $\\left(d_{1}, d_{2}\\right)$, the second resident $\\left(d_{2}, d_{3}\\right)$, the third resident $\\left(d_{3}, d_{4}\\right)$, and so on, until the sixth resident, who cooks $\\left(d_{6}, d_{1}\\right)$. Then there are 8 choices for $d_{1}$ and 7 choices for $d_{2}$. There are only 6 choices for $d_{3}$, because $d_{3} \\neq d_{1}$ (otherwise two residents would cook the same pair of dishes). For $k>3$, the requirement that no two intermediate residents cook the same dishes implies that $d_{k+1}$ cannot equal any of $d_{1}, \\ldots, d_{k-1}$, and of course $d_{k}$ and $d_{k+1}$ must be distinct dishes. Hence there are $8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3=20,160$ six-person resident cycles, not accounting for different starting points in the cycle and the two different directions to go around the cycle. Taking these into account, there are $20,160 /(6 \\cdot 2)=1,680$ distinguishable resident cycles.\n\nUsing the logic above, there are $d(d-1) \\cdots(d-k+1)$ choices for $d_{1}, d_{2}, \\ldots, d_{k}$. To account for indistinguishable cycles, divide by $k$ possible starting points and 2 possible directions, yielding $\\frac{d(d-1) \\cdots(d-k+1)}{2 k}$ or $\\frac{d !}{2 k(d-k) !}$ distinguishable resident cycles." ]

[ "$\\frac{d !}{2 k(d-k) !}$" ]

[ "Because the given repetend has ten digits, the original had four digits. If $\\frac{1}{N}=.0 \\underline{A} \\underline{B} \\underline{C} \\underline{D}=$ $\\frac{\\underline{A} \\underline{B} \\underline{C} \\underline{D}}{99990}$, then the numerator must divide $99990=10 \\cdot 99 \\cdot 101=2 \\cdot 3^{2} \\cdot 5 \\cdot 11 \\cdot 101$.\n\nNote that all 3- and 4-digit multiples of 101 contain at least one digit which appears twice. Because the 10-digit string under the vinculum (i.e., 0231846597) contains no repeated digits, $\\underline{A} \\underline{B} \\underline{C} \\underline{D}$ cannot be a multiple of 101 . So $\\underline{A} \\underline{B} \\underline{C} \\underline{D}$ divides $2 \\cdot 3^{2} \\cdot 5 \\cdot 11=990$. The only divisor of 990 that can be formed from four of the given digits (taken in order) is 0165, that is, 165. Hence $\\frac{1}{N}=\\frac{165}{99990}=\\frac{1}{606} \\Rightarrow N=\\mathbf{6 0 6}$." ]

[ "606" ]

[ "Because $\\sqrt{n}$ is a multiple of $3, n$ must be a multiple of 9 . Therefore the sum of the digits of $n$ is a multiple of 9 . Thus $\\sqrt{n}$ must be a multiple of 27 , which implies that $n$ is a multiple of $27^{2}$. The only candidates to consider are $54^{2}(=2916)$ and $81^{2}(=6561)$, and only 2916 satisfies the desired conditions." ]

[ "2916" ]

[ "The map $n \\mapsto 24 / n$ establishes a one-to-one correspondence among the positive integer divisors of 24 . Thus\n\n$$\n\\begin{aligned}\n\\sum_{\\substack{n \\mid 24 \\\\\nn>0}} \\frac{1}{n} & =\\sum_{\\substack{n \\mid 24 \\\\\nn>0}} \\frac{1}{24 / n} \\\\\n& =\\frac{1}{24} \\sum_{\\substack{n \\mid 24 \\\\\nn>0}} n\n\\end{aligned}\n$$\n\nBecause $24=2^{3} \\cdot 3$, the sum of the positive divisors of 24 is $\\left(1+2+2^{2}+2^{3}\\right)(1+3)=15 \\cdot 4=60$. Hence the sum is $60 / 24=\\mathbf{5} / \\mathbf{2}$.", "Because $24=2^{3} \\cdot 3$, any positive divisor of 24 is of the form $2^{a} 3^{b}$ where $a=0,1,2$, or 3 , and $b=0$ or 1 . So the sum of the positive divisors of 24 can be represented as the product $(1+2+4+8)(1+3)$. Similarly, the sum of their reciprocals can be represented as the product $\\left(\\frac{1}{1}+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}\\right)\\left(\\frac{1}{1}+\\frac{1}{3}\\right)$. The first sum is $\\frac{15}{8}$ and the second is $\\frac{4}{3}$, so the product is $\\mathbf{5 / 2}$." ]

[ "$\\frac{5}{2}$" ]

[ "Consider the ordered pairs of digits $(X, Y)$ for which $\\underline{1} \\underline{2} \\underline{3} \\underline{X} \\underline{5} \\underline{Y} \\underline{7}$ is a multiple of 11 . Recall that a number is a multiple of 11 if and only if the alternating sum of the digits is a multiple of 11 . Because $1+3+5+7=16$, the sum of the remaining digits, namely $2+X+Y$, must equal 5 or 16 . Thus $X+Y$ must be either 3 or 14 , making $X=3-Y$ (if $Y=0,1,2$, or 3 ) or $14-Y$ (if $Y=5,6,7,8$, or 9 ). Thus a solution $(X, Y)$ exists unless $Y=4$." ]

[ "4" ]

[ "Label the triangle as $\\triangle A B C$, with $A B=2 \\sqrt{3}$ and $B C=4$. Let $D$ and $E$ lie on $\\overline{A B}$ such that $D B=A E=2$. Let $F$ be the midpoint of $\\overline{B C}$, so that $B F=F C=2$. Let $G$ and $H$ lie on $\\overline{A C}$, with $A G=H C=2$. Now draw the arcs of radius 2 between $E$ and $G, D$ and $F$, and $F$ and $H$. Let the intersection of arc $D F$ and $\\operatorname{arc} E G$ be $J$. Finally, let $M$ be the midpoint of $\\overline{A B}$. The completed diagram is shown below.\n\n<img_3875>\n\nThe region $R$ consisting of all points within $\\triangle A B C$ that lie within 2 units of any vertex is the union of the three sectors $E A G, D B F$, and $F C H$. The angles of these sectors, being the angles $\\angle A, \\angle B$, and $\\angle C$, sum to $180^{\\circ}$, so the sum of their areas is $2 \\pi$. Computing the area of $R$ requires subtracting the areas of all intersections of the three sectors that make up $R$.\n\nThe only sectors that intersect are $E A G$ and $D B F$. Half this area of intersection, the part above $\\overline{M J}$, equals the difference between the areas of sector $D B J$ and of $\\triangle M B J$. Triangle $M B J$ is a $1: \\sqrt{3}: 2$ right triangle because $B M=\\sqrt{3}$ and $B J=2$, so the area of $\\triangle M B J$ is $\\frac{\\sqrt{3}}{2}$. Sector $D B J$ has area $\\frac{1}{12}(4 \\pi)=\\frac{\\pi}{3}$, because $\\mathrm{m} \\angle D B J=30^{\\circ}$. Therefore the area of intersection of the sectors is $2\\left(\\frac{\\pi}{3}-\\frac{\\sqrt{3}}{2}\\right)=\\frac{2 \\pi}{3}-\\sqrt{3}$. Hence the total area of $R$ is $2 \\pi-\\left(\\frac{2 \\pi}{3}-\\sqrt{3}\\right)=\\frac{4 \\pi}{3}+\\sqrt{3}$. The total area of $\\triangle A B C$ is $4 \\sqrt{3}$, therefore the desired probability is $\\frac{\\frac{4 \\pi}{3}+\\sqrt{3}}{4 \\sqrt{3}}=\\frac{\\pi}{3 \\sqrt{3}}+\\frac{1}{4}$. Then $a=\\frac{1}{4}$ and $b=\\left(\\frac{1}{3 \\sqrt{3}}\\right)^{2}=\\frac{1}{27}$, hence the answer is $\\left(\\frac{1}{4}, \\frac{1}{27}\\right)$." ]

[ "$(\\frac{1}{4}, \\frac{1}{27})$" ]

[ "Consider repeatedly reflecting square $A R M L$ over its sides so that the entire plane is covered by copies of $A R M L$. A path starting at $(2 / 7,3 / 7)$ that touches one or more sides and returns to $(2 / 7,3 / 7)$ corresponds to a straight line starting at $(2 / 7,3 / 7)$ and ending at the image of $(2 / 7,3 / 7)$ in one of the copies of $A R M L$. To touch three sides, the path must cross three lines, at least one of which must be vertical and at least one of which must be horizontal.\n\n<img_3221>\n\nIf the path crosses two horizontal lines and the line $x=0$, it will have traveled a distance of 2 units vertically and $4 / 7$ units vertically for a total distance of $\\sqrt{2^{2}+(4 / 7)^{2}}$ units. Similarly, the total distance traveled when crossing two horizontal lines and $x=1$ is $\\sqrt{2^{2}+(10 / 7)^{2}}$, the total distance traveled when crossing two vertical lines and $y=0$ is $\\sqrt{2^{2}+(6 / 7)^{2}}$, and the total distance traveled when crossing two vertical lines and $y=1$ is $\\sqrt{2^{2}+(8 / 7)^{2}}$. The least of these is\n\n$$\n\\sqrt{2^{2}+(4 / 7)^{2}}=\\frac{2}{\\mathbf{7}} \\sqrt{\\mathbf{5 3}}\n$$" ]

[ "$\\frac{2}{7} \\sqrt{53}$" ]

[ "If 306 is an element of $S_{k}$, then there exists an integer $m \\geq 0$ such that $306=k+m k^{2}$. Thus $k \\mid 306$ and $k^{2} \\mid 306-k$. The second relation can be rewritten as $k \\mid 306 / k-1$, which implies that $k \\leq \\sqrt{306}$ unless $k=306$. The prime factorization of 306 is $2 \\cdot 3^{2} \\cdot 17$, so the set of factors of 306 less than $\\sqrt{306}$ is $\\{1,2,3,6,9,17\\}$. Check each in turn:\n\n$$\n\\begin{aligned}\n306-1 & =305, & & 1^{2} \\mid 305 \\\\\n306-2 & =304, & & 2^{2} \\mid 304 \\\\\n306-3 & =303, & & 3^{2} \\nmid 303 \\\\\n306-6 & =300, & & 6^{2} \\nmid 300 \\\\\n306-9 & =297, & & 9^{2} \\nmid 297 \\\\\n306-17 & =289, & & 17^{2} \\mid 289 .\n\\end{aligned}\n$$\n\nThus the set of possible $k$ is $\\{1,2,17,306\\}$, and the sum is $1+2+17+306=\\mathbf{3 2 6}$." ]

[ "326" ]

[ "Let $\\log _{x} y=a$. Then the first equation is equivalent to $2 a+\\frac{5}{a}=2 k-1$, and the second equation is equivalent to $\\frac{5 a}{2}-\\frac{3}{2 a}=k-3$. Solving this system by eliminating $k$ yields the quadratic equation $3 a^{2}+5 a-8=0$, hence $a=1$ or $a=-\\frac{8}{3}$. Substituting each of these values\n\n\n\nof $a$ into either of the original equations and solving for $k$ yields $(a, k)=(1,4)$ or $\\left(-\\frac{8}{3},-\\frac{149}{48}\\right)$. Adding the values of $k$ yields the answer of $43 / 48$.", "In terms of $a=\\log _{x} y$, the two equations become $2 a+\\frac{5}{a}=2 k-1$ and $\\frac{5 a}{2}-\\frac{3}{2 a}=k-3$. Eliminate $\\frac{1}{a}$ to obtain $31 a=16 k-33$; substitute this into either of the original equations and clear denominators to get $96 k^{2}-86 k-1192=0$. The sum of the two roots is $86 / 96=\\mathbf{4 3} / \\mathbf{4 8}$." ]

[ "$\\frac{43}{48}$" ]

[ "Define a fault line to be a side of a tile other than its base. Any tiling of $W A S H$ can be represented as a sequence of tiles $t_{1}, t_{2}, \\ldots, t_{14}$, where $t_{1}$ has a fault line of $\\overline{W H}, t_{14}$ has a fault line of $\\overline{A S}$, and where $t_{k}$ and $t_{k+1}$ share a fault line for $1 \\leq k \\leq 13$. Also note that to determine the position of tile $t_{k+1}$, it is necessary and sufficient to know the fault line that $t_{k+1}$ shares with $t_{k}$, as well as whether the base of $t_{k+1}$ lies on $\\overline{W A}$ (abbreviated \" $\\mathrm{B}$ \" for \"bottom\") or on $\\overline{S H}$ (abbreviated \"T\" for \"top\"). Because rectangle $W A S H$ has width 7 , precisely 7 of the 14 tiles must have their bases on $\\overline{W A}$. Thus any permutation of 7 B's and 7 T's determines a unique tiling $t_{1}, t_{2}, \\ldots, t_{14}$, and conversely, any tiling $t_{1}, t_{2}, \\ldots, t_{14}$ corresponds to a unique permutation of 7 B's and 7 T's. Thus the answer is $\\left(\\begin{array}{c}14 \\\\ 7\\end{array}\\right)=\\mathbf{3 4 3 2}$.", "Let $T(a, b)$ denote the number of ways to triangulate the polygon with vertices at $(0,0),(b, 0),(a, 1),(0,1)$, where each triangle has area $1 / 2$ and vertices at lattice points. The problem is to compute $T(7,7)$. It is easy to see that $T(a, 0)=T(0, b)=1$ for all $a$ and $b$. If $a$ and $b$ are both positive, then either one of the triangles includes the edge from $(a-1,1)$ to $(b, 0)$ or one of the triangles includes the edge from $(a, 1)$ to $(b-1,0)$, but not both. (In fact, as soon as there is an edge from $(a, 1)$ to $(x, 0)$ with $x<b$, there must be edges from $(a, 1)$ to $\\left(x^{\\prime}, 0\\right)$ for all $x \\leq x^{\\prime}<b$.) If there is an edge from $(a-1,1)$ to $(b, 0)$, then the number of ways to complete the triangulation is $T(a-1, b)$; if there is an edge from $(a, 1)$ to $(b-1,0)$, then the number of ways to complete the triangulation is $T(a, b-1)$; thus $T(a, b)=T(a-1, b)+T(a, b-1)$. The recursion and the initial conditions describe Pascal's triangle, so $T(a, b)=\\left(\\begin{array}{c}a+b \\\\ a\\end{array}\\right)$. In particular, $T(7,7)=\\left(\\begin{array}{c}14 \\\\ 7\\end{array}\\right)=3432$." ]

[ "3432" ]

[ "Because $\\cos 2 x=1-2 \\sin ^{2} x, \\sin ^{2} x=\\frac{1-\\cos 2 x}{2}$. Thus the desired sum can be rewritten as\n\n$$\n\\frac{1-\\cos 8^{\\circ}}{2}+\\frac{1-\\cos 16^{\\circ}}{2}+\\cdots+\\frac{1-\\cos 352^{\\circ}}{2}=\\frac{44}{2}-\\frac{1}{2}\\left(\\cos 8^{\\circ}+\\cos 16^{\\circ}+\\cdots+\\cos 352^{\\circ}\\right) .\n$$\n\nIf $\\alpha=\\cos 8^{\\circ}+i \\sin 8^{\\circ}$, then $\\alpha$ is a primitive $45^{\\text {th }}$ root of unity, and $1+\\alpha+\\alpha^{2}+\\alpha^{3}+\\cdots+\\alpha^{44}=0$. Hence $\\alpha+\\alpha^{2}+\\cdots+\\alpha^{44}=-1$, and because the real part of $\\alpha^{n}$ is simply $\\cos 8 n^{\\circ}$,\n\n$$\n\\cos 8^{\\circ}+\\cos 16^{\\circ}+\\cdots+\\cos 352^{\\circ}=-1 .\n$$\n\nThus the desired sum is $22-(1 / 2)(-1)=\\mathbf{4 5} / \\mathbf{2}$.", "The problem asks to simplify the sum\n\n$$\n\\sin ^{2} a+\\sin ^{2} 2 a+\\sin ^{2} 3 a+\\cdots+\\sin ^{2} n a\n$$\n\n\n\nwhere $a=4^{\\circ}$ and $n=44$. Because $\\cos 2 x=1-2 \\sin ^{2} x, \\sin ^{2} x=\\frac{1-\\cos 2 x}{2}$. Thus the desired sum can be rewritten as\n\n$$\n\\frac{1-\\cos 2 a}{2}+\\frac{1-\\cos 4 a}{2}+\\cdots+\\frac{1-\\cos 2 n a}{2}=\\frac{n}{2}-\\frac{1}{2}(\\cos 2 a+\\cos 4 a+\\cdots+\\cos 2 n a) .\n$$\n\nLet $Q=\\cos 2 a+\\cos 4 a+\\cdots+\\cos 2 n a$. By the sum-to-product identity,\n\n$$\n\\begin{aligned}\n\\sin 3 a-\\sin a & =2 \\cos 2 a \\sin a \\\\\n\\sin 5 a-\\sin 3 a & =2 \\cos 4 a \\sin a \\\\\n& \\vdots \\\\\n\\sin (2 n+1) a-\\sin (2 n-1) a & =2 \\cos 2 n a \\sin a\n\\end{aligned}\n$$\n\nThus\n\n$$\n\\begin{aligned}\nQ \\cdot 2 \\sin a & =(\\sin 3 a-\\sin a)+(\\sin 5 a-\\sin 3 a)+\\cdots+(\\sin (2 n+1) a-\\sin (2 n-1) a) \\\\\n& =\\sin (2 n+1) a-\\sin a\n\\end{aligned}\n$$\n\nWith $a=4^{\\circ}$ and $n=44$, the difference on the right side becomes $\\sin 356^{\\circ}-\\sin 4^{\\circ}$; note that the terms in this difference are opposites, because of the symmetry of the unit circle. Hence\n\n$$\n\\begin{aligned}\nQ \\cdot 2 \\sin 4^{\\circ} & =-2 \\sin 4^{\\circ}, \\text { and } \\\\\nQ & =-1 .\n\\end{aligned}\n$$\n\nThus the original sum becomes $44 / 2-(1 / 2)(-1)=\\mathbf{4 5} / \\mathbf{2}$." ]

[ "$\\frac{45}{2}$" ]

[ "Call the region $R$, and let $R_{q}$ be the portion of $R$ in the $q^{\\text {th }}$ quadrant. Noting that the point $(x, y)$ is in $R$ if and only if $( \\pm x, \\pm y)$ is in $R$, it follows that $\\left[R_{1}\\right]=\\left[R_{2}\\right]=\\left[R_{3}\\right]=\\left[R_{4}\\right]$, and so $[R]=4\\left[R_{1}\\right]$. So it suffices to determine $\\left[R_{1}\\right]$.\n\nIn the first quadrant, the boundary equation is just $x^{2}+y^{2}=x+y \\Rightarrow\\left(x-\\frac{1}{2}\\right)^{2}+\\left(y-\\frac{1}{2}\\right)^{2}=\\frac{1}{2}$. This equation describes a circle of radius $\\frac{\\sqrt{2}}{2}$ centered at $\\left(\\frac{1}{2}, \\frac{1}{2}\\right)$. The portion of the circle's interior which is inside the first quadrant can be decomposed into a right isosceles triangle with side length 1 and half a circle of radius $\\frac{\\sqrt{2}}{2}$. Thus $\\left[R_{1}\\right]=\\frac{1}{2}+\\frac{\\pi}{4}$, hence $[R]=\\mathbf{2}+\\boldsymbol{\\pi}$." ]

[ "$2+\\pi$" ]

[ "Write $a_{n}=a_{1}+r(n-1)$ and $b_{n}=b_{1}+s(n-1)$. Then $a_{20}+b_{14}=a_{1}+b_{1}+19 r+13 s$, while $b_{20}+a_{14}=a_{1}+b_{1}+13 r+19 s=a_{20}+b_{14}+6(s-r)$. Because both sequences consist only of integers, $r$ and $s$ must be integers, so $b_{20}+a_{14} \\equiv a_{20}+b_{14} \\bmod 6$. Thus the least possible value of $b_{20}+a_{14}$ is 4 . If $b_{20}=3$ and $a_{14}=1$, then $\\left\\{a_{n}\\right\\}$ must be a decreasing sequence (else $a_{13}$ would not be positive) and $a_{20} \\leq-5$, which is impossible. The case $b_{20}=a_{14}=2$ violates the requirement that the terms be distinct, and by reasoning analogous to the first case, $b_{20}=1, a_{14}=3$ is also impossible. Hence the sum $b_{20}+a_{14}$ is at least 10 . To show that 10 is attainable, make $\\left\\{a_{n}\\right\\}$ decreasing and $b_{20}$ as small as possible: set $b_{20}=1, a_{14}=9$, and $a_{n}=23-n$. Then $a_{20}=3$, yielding $b_{14}=997$. Hence $s=\\frac{997-1}{14-20}=\\frac{996}{-6}=-166$ and\n\n\n\n$b_{1}=997-(13)(-166)=3155$, yielding $b_{n}=3155-166(n-1)$. Because $b_{20}=1 \\leq a_{20}$ and $b_{19}=167 \\geq a_{1}$, the sequences $\\left\\{b_{n}\\right\\}$ and $\\left\\{a_{n}\\right\\}$ are distinct for $1 \\leq n \\leq 20$, completing the proof. Hence the minimum possible value of $b_{20}+a_{14}$ is 10." ]

[ "10" ]

[ "First, eliminate $x: y\\left(y^{2} z-x\\right)+\\left(x y-z^{2}\\right)=14(y+1) \\Rightarrow z^{2}-y^{3} z+14(y+1)=0$. Viewed as a quadratic in $z$, this equation implies $z=\\frac{y^{3} \\pm \\sqrt{y^{6}-56(y+1)}}{2}$. In order for $z$ to be an integer, the discriminant must be a perfect square. Because $y^{6}=\\left(y^{3}\\right)^{2}$ and $\\left(y^{3}-1\\right)^{2}=y^{6}-2 y^{3}+1$, it follows that $|56(y+1)| \\geq 2\\left|y^{3}\\right|-1$. This inequality only holds for $|y| \\leq 5$. Within that range, the only values of $y$ for which $y^{6}-56 y-56$ is a perfect square are -1 and -3 . If $y=-1$, then $z=-1$ or $z=0$. If $y=-3$, then $z=1$ or $z=-28$. After solving for the respective values of $x$ in the various cases, the four lattice points satisfying the system are $(-15,-1,-1),(-14,-1,0),(-5,-3,1)$, and $(-266,-3,-28)$. The farthest solution point from the origin is therefore $(-\\mathbf{2 6 6}, \\mathbf{- 3 , - 2 8})$." ]

[ "$(-266,-3,-28)$" ]

[ "Let $r$ be the common ratio of the sequence. Then $a_{20}=r^{20-14} \\cdot a_{14}$, hence $8=r^{6} \\cdot 2^{21} \\Rightarrow r^{6}=$ $\\frac{2^{3}}{2^{21}}=2^{-18}$, so $r=2^{-3}=\\frac{1}{8}$. Thus $a_{21}=r \\cdot a_{20}=\\frac{1}{8} \\cdot 8=\\mathbf{1}$." ]

[ "1" ]

[ "Because $\\overline{O E}$ is tangent to circle $L, \\overline{L E} \\perp \\overline{O E}$. Also note that $L O=4 T-T=3 T$. Hence, by the Pythagorean Theorem, $O E=\\sqrt{(3 T)^{2}-T^{2}}=2 T \\sqrt{2}$ (this also follows from the TangentSecant Theorem). With $T=1, O E=\\mathbf{2} \\sqrt{\\mathbf{2}}$." ]

[ "$2 \\sqrt{2}$" ]

[ "Let $c$ be the length of the hypotenuse. Then, by the Pythagorean Theorem, $\\left(T^{2}\\right)^{2}+(c-2)^{2}=$ $c^{2} \\Rightarrow c=\\frac{T^{4}}{4}+1$. With $T=2 \\sqrt{2}, T^{4}=64$, and $c=17$. So the triangle is a $8-15-17$ triangle with perimeter 40 ." ]

[ "40" ]

[ "Multiply each side of the first equation by $T$ to obtain $T x+9 T y=17 T$. Subtract the second equation to yield $9 T y-T y-y=16 T-2 \\Rightarrow y(8 T-1)=2(8 T-1)$. Hence either $T=\\frac{1}{8}$ (in which case, the value of $y$ is not uniquely determined) or $y=2$. Plug $y=2$ into the first equation to obtain $x=-1$. Hence $20 x+14 y=-20+28=\\mathbf{8}$." ]

[ "8" ]

[ "Using Vièta's Formula, write $f(x)=a x^{2}+b x+T a$. Substituting the coordinates of the given points yields the system of equations: $4 a-2 b+T a=20$ and $a+b+T a=14$. Multiply each side of the latter equation by 2 and add the resulting equation to the former equation to eliminate $b$. Simplifying yields $a=\\frac{16}{T+2}$. With $T=8, a=8 / 5$." ]

[ "$\\frac{8}{5}$" ]

[ "Note that $z_{1}-z_{2}=14+(5-K) i$, hence $\\left|z_{1}-z_{2}\\right|=\\sqrt{14^{2}+(5-K)^{2}}$. With $T=8 / 5,15 T=24$, hence $14^{2}+(5-K)^{2} \\geq 24^{2}$. Thus $|5-K| \\geq \\sqrt{24^{2}-14^{2}}=\\sqrt{380}$. Because $K$ is a positive integer, it follows that $K-5 \\geq 20$, hence the desired value of $K$ is $\\mathbf{2 5}$." ]

[ "25" ]

[ "First count the number of arrangements in which Chris stands next to Charlie. This is $(T-1) \\cdot 2 ! \\cdot(T-2) !=2 \\cdot(T-1)$ ! because there are $T-1$ possible leftmost positions for the pair $\\{$ Charlie, Chris $\\}$, there are 2 ! orderings of this pair, and there are $(T-2)$ ! ways to arrange the remaining people. There are equally many arrangements in which Abby stands next to Charlie. However, adding these overcounts the arrangements in which Abby, Charlie, and Chris are standing next to each other, with Charlie in the middle. Using similar reasoning as above, there are $(T-2) \\cdot 2 ! \\cdot(T-3) !=2 \\cdot(T-2)$ ! such arrangements. Hence the desired probability is $\\frac{2 \\cdot 2 \\cdot(T-1) !-2 \\cdot(T-2) !}{T !}=\\frac{2 \\cdot(T-2) !(2 T-2-1)}{T !}=\\frac{2(2 T-3)}{T(T-1)}$. With $T=25$, the fraction simplifies to $\\frac{\\mathbf{4 7}}{\\mathbf{3 0 0}}$." ]

[ "$\\frac{47}{300}$" ]

[ "The given conditions are equivalent to $\\sin \\alpha=A$ and $\\cos \\beta=B$. Using either the sumto-product or the sine of a sum/difference identities, the desired expression is equivalent to $2(\\sin \\alpha)(\\cos \\beta)=2 \\cdot A \\cdot B$. With $A=\\frac{47}{300}$ and $B=\\frac{12}{169}, 2 \\cdot A \\cdot B=\\frac{2 \\cdot 47}{25 \\cdot 169}=\\frac{\\mathbf{9 4}}{\\mathbf{4 2 2 5}}$." ]

[ "$\\frac{94}{4225}$" ]

[ "The cone's volume is $\\frac{1}{3} \\pi r^{2}\\left(T-r^{2}\\right)$. Maximizing this is equivalent to maximizing $x(T-x)$, where $x=r^{2}$. Using the formula for the vertex of a parabola (or the AM-GM inequality), the maximum value occurs when $x=\\frac{T}{2}$. Hence $V=\\frac{1}{3} \\pi \\cdot \\frac{T}{2} \\cdot \\frac{T}{2}=\\frac{\\pi T^{2}}{12}$, and $\\pi / V=12 / T^{2}$. With $T=13, V=\\frac{\\mathbf{1 2}}{\\mathbf{1 6 9}}$." ]

[ "$\\frac{12}{169}$" ]

[ "Write $2=\\log 100$ and use the well-known properties for the sum/difference of two logs to obtain $\\log T=\\log \\left(\\frac{100 k}{2}\\right)$, hence $k=\\frac{T}{50}$. With $T=650, k=13$." ]

[ "13" ]

[ "Let $D$ be the distance in miles traveled by the plane. The given conditions imply that $\\frac{D}{T}-\\frac{D}{T+30}=1.5 \\Rightarrow \\frac{30 D}{T(T+30)}=1.5 \\Rightarrow D=\\frac{T(T+30)}{20}$. With $T=100, D=5 \\cdot 130=\\mathbf{6 5 0}$." ]

[ "650" ]

[ "The given radical equals $\\left(\\left(\\left(10^{T^{2}-T}\\right)^{\\frac{1}{T}}\\right)^{\\frac{1}{2}}\\right)^{\\frac{1}{2}}=10^{(T-1) / 4}$. With $T=9$, this simplifies to $10^{2}=100$" ]

[ "100" ]

[ "Because $\\overline{S U}$ and $\\overline{R E}$ are sides of the hexagon, $S U=R E=\\sqrt{T}$. Let $H$ be the foot of the altitude from $R$ to $\\overline{B E}$ in $\\triangle B R E$ and note that each interior angle of a regular hexagon is $120^{\\circ}$. Thus $B E=B H+H E=2\\left(\\frac{\\sqrt{3}}{2}\\right)(\\sqrt{T})=\\sqrt{3 T}$. Thus $B E \\cdot S U \\cdot R E=\\sqrt{3 T} \\cdot \\sqrt{T} \\cdot \\sqrt{T}=T \\sqrt{3 T}$. With $T=3$, the answer is $\\mathbf{9}$." ]

[ "9" ]

[ "Using the Multiplication Principle, Chef Selma can make $4 \\cdot 3 \\cdot 2 \\cdot K=24 K$ different burrito varieties. With $T=70$, the smallest integral value of $K$ such that $24 K \\geq 70$ is $\\left\\lceil\\frac{70}{24}\\right\\rceil=3$." ]

[ "3" ]

[ "Because $\\operatorname{gcd}(14,20)=2$, the problem is equivalent to computing the smallest positive integer $N$ such that $7 \\mid 10 N$ and $10 \\mid 7 N$. Thus $7 \\mid N$ and $10 \\mid N$, and the desired value of $N$ is $\\operatorname{lcm}(7,10)=\\mathbf{7 0}$." ]

[ "70" ]

[ "The largest fibbish number is 10112369. First, if $\\underline{A_{1}} \\underline{A_{2}} \\cdots \\underline{A_{n}}$ is an $n$-digit fibbish number with $A_{1}$ and $A_{2} \\neq 0$, the number created by prepending the ${\\text { digits }} A_{1}$ and 0 to the number is larger and still fibbish: $\\underline{A_{1}} \\underline{0} \\underline{A_{1}} \\underline{A_{2}} \\cdots \\underline{A_{n}}>\\underline{A_{1}} \\underline{A_{2}} \\cdots \\underline{A_{n}}$. Suppose that $A_{2}=0$ and $A_{3}=A_{1}$, so that the number begins $\\underline{A_{1}} \\underline{0} \\underline{A_{1}} \\underline{A_{4}}$. If the number is to be fibbish, $A_{4} \\geq A_{1}>0$. Then if $A_{1} \\geq 2$ and $A_{4} \\geq 2$, because the number is fibbish, $A_{5} \\geq 4$, and $A_{6} \\geq 6$. In this case there can be no more digits, because $A_{5}+A_{6} \\geq 10$. So the largest possible fibbish number beginning with 20 is 202246. If $A_{1}=2$ and $A_{2}=1$, then $A_{3}$ must be at least 3 , and the largest possible number is 21459; changing $A_{3}$ to 3 does not increase the length. Now consider $A_{1}=1$. If $A_{2}=1$, then $A_{3} \\geq 2, A_{4} \\geq 3, A_{5} \\geq 5$, and $A_{6} \\geq 8$. There can be no seventh digit because that digit would have to be at least 13 . Increasing $A_{3}$ to 3 yields only two additional digits, because $A_{4} \\geq 4, A_{5} \\geq 7$. So $A_{3}=2$ yields a longer (and thus larger) number. Increasing $A_{4}$ to 4 yields only one additional digit, $A_{5} \\geq 6$, because $A_{4}+A_{5} \\geq 10$. But if $A_{4}=3$, increasing $A_{5}$ to 6 still allows $A_{6}=9$, yielding the largest possible number of digits (8) and the largest fibbish number with that many digits." ]

[ "10112369" ]

[ "Note that any portion of side length $m \\geq 4$ will overlap the center square, so consider only portions of side length 3 or less. If there were no hole in the candy bar, the number of portions could be counted by conditioning on the possible location of the upper-left corner of the portion. If the portion is of size $1 \\times 1$, then the corner can occupy any of the $7^{2}$ squares of the bar. If the portion is of size $2 \\times 2$, then the corner can occupy any of the top 6 rows and any of the left 6 columns, for $6^{2}=36$ possible $2 \\times 2$ portions. In general, the upper-left corner of an $m \\times m$ portion can occupy any of the top $8-m$ rows and any of the left $8-m$ columns. So the total number of portions from an intact bar would be $7^{2}+6^{2}+5^{2}$. Now when $m \\leq 3$, the number of $m \\times m$ portions that include the missing square is simply $m^{2}$, because the missing square could be any square of the portion. So the net number of portions is\n\n$$\n\\begin{aligned}\n7^{2}+6^{2}+5^{2}-3^{2}-2^{2}-1^{2} & =(49+36+25)-(9+4+1) \\\\\n& =110-14 \\\\\n& =\\mathbf{9 6}\n\\end{aligned}\n$$", "First ignore the missing square. As in the previous solution, the number of $m \\times m$ portions that can fit in the bar is $(8-m)^{2}$. So the total number of portions of all sizes is simply\n\n$$\n7^{2}+6^{2}+\\cdots+1^{2}=\\frac{7(7+1)(2 \\cdot 7+1)}{6}=140\n$$\n\nTo exclude portions that overlap the missing center square, it is useful to consider the location of the missing square within the portion. If an $m \\times m$ portion includes the missing center\n\n\n\nsquare, and $m \\leq 4$, then the missing square could be any one of the $m^{2}$ squares in the portion. If $m=5$, then the missing square cannot be in the leftmost or rightmost columns of the portion, because then the entire bar would have to extend at least four squares past the hole, and it only extends three. By similar logic, the square cannot be in the top or bottom rows of the portion. So for $m=5$, there are $3 \\cdot 3=9$ possible positions. For $m=6$, the two left and two right columns are excluded, as are the two top and the two bottom rows, for $2 \\cdot 2=4$ possible positions for the portion. And in a $7 \\times 7$ square, the only possible location of the hole is in the center. So the total number of portions overlapping the missing square is\n\n$$\n1^{2}+2^{2}+3^{2}+4^{2}+3^{2}+2^{2}+1^{2}=44 .\n$$\n\nThe difference is thus $140-44=\\mathbf{9 6}$" ]

[ "96" ]

[ "The diagram below shows the hexagons.\n\n<img_3234>\n\nThe area of hexagon $G B C D K L$ can be computed as $[G B C D K L]=[A B C D E F]-[A G L K E F]$, and $[A G L K E F]$ can be computed by dividing concave hexagon $A G L K E F$ into two parallelograms sharing $\\overline{F L}$. If $A B=s$, then the height $A E$ is $s \\sqrt{3}$, so the height of parallelogram $A G L F$ is $\\frac{s \\sqrt{3}}{2}$. Thus $[A G L F]=L F \\cdot \\frac{s \\sqrt{3}}{2}$ and $[A G L K E F]=L F \\cdot s \\sqrt{3}$. On the other hand, the area of a regular hexagon of side length $s$ is $\\frac{3 s^{2} \\sqrt{3}}{2}$. Because $[G B C D K L]=\\frac{1}{2}[A B C D E F]$, it follows that $[A G L K E F]=\\frac{1}{2}[A B C D E F]$, and\n\n$$\nL F \\cdot s \\sqrt{3}=\\frac{1}{2}\\left(\\frac{3 s^{2} \\sqrt{3}}{2}\\right)=\\frac{3 s^{2} \\sqrt{3}}{4}\n$$\n\nwhence $L F=\\frac{3}{4} s$. With $s=24$, the answer is $\\mathbf{1 8}$.", "Compute $[B C D K L G]$ as twice the area of trapezoid $B C L G$. If $A B=s$, then $B G=s-L F$ and $C L=2 s-L F$, while the height of the trapezoid is $\\frac{s \\sqrt{3}}{2}$.[^0]\n\n\n[^0]: ${ }^{1}$ The answer 115 was also accepted for this problem because of an alternate (and unintended) reasonable interpretation of the problem statement. Some students also counted portions that contained the \"hole\", with the hole being strictly inside the portion, and not along its edges.\n\n\n\nThus the area of the trapezoid is:\n\n$$\n\\frac{1}{2}\\left(\\frac{s \\sqrt{3}}{2}\\right)((s-L F)+(2 s-L F))=\\frac{s \\sqrt{3}(3 s-2 L F)}{4}\n$$\n\nDouble that area to obtain\n\n$$\n[B C D K L G]=\\frac{s \\sqrt{3}(3 s-2 L F)}{2}\n$$\n\nOn the other hand, $[A B C D E F]=\\frac{3 s^{2} \\sqrt{3}}{2}$, so\n\n$$\n\\begin{aligned}\n\\frac{s \\sqrt{3}(3 s-2 L F)}{2} & =\\frac{3 s^{2} \\sqrt{3}}{4} \\\\\n3 s-2 L F & =\\frac{3 s}{2} \\\\\nL F & =\\frac{3}{4} s .\n\\end{aligned}\n$$\n\nSubstituting $s=24$ yields $L F=\\mathbf{1 8}$." ]

[ "18" ]

[ "Let $\\underline{A} \\underline{B} \\underline{C} \\underline{D}=N$. Because $7 !=5040$ and $8 !=40,320, N$ must be no greater than $7 !+6 !+6 !=6480$. This value of $N$ does not work, so work through the list of possible sums in decreasing order: $7 !+6 !+5 !, 7 !+6 !+4$ !, etc. The first value that works is $N=5762=7 !+6 !+2 !$.", "Let $\\underline{A} \\underline{B} \\underline{C} \\underline{D}=N$. Because $7 !=5040$ and $8 !=40,320$, to find the maximal value, first consider values of $N$ that include 7 as a digit. Suppose then that $N=5040+X !+Y$ !. To force a 7 to appear in this sum with maximal $N$, let $X=6$, which yields $N=5040+720+Y !=5760+Y$ !. This value of $N$ has a 7 (and a 6 ), so search for values of $Y$ to find ones that satisfy the conditions of the problem. Only $Y=1$ and $Y=2$ will do, giving 5761 and 5762 . Hence $\\mathbf{5 7 6 2}$ is the maximum possible value of $N$." ]

[ "5762" ]

[ "The number of digits of $n$ is $\\lfloor\\log n\\rfloor+1$. Because $100^{1000^{10,000}}=\\left(10^{2}\\right)^{1000^{10,000}}, X=2$. $1000^{10,000}+1$. Similarly, $Y=3 \\cdot 10,000^{100,000}+1$. Using the change-of-base formula,\n\n$$\n\\begin{aligned}\n\\log _{X} Y=\\frac{\\log Y}{\\log X} & \\approx \\frac{\\log 3+\\log 10,000^{100,000}}{\\log 2+\\log 1000^{10,000}} \\\\\n& =\\frac{\\log 3+100,000 \\log 10,000}{\\log 2+10,000 \\log 1000} \\\\\n& =\\frac{\\log 3+100,000 \\cdot 4}{\\log 2+10,000 \\cdot 3} \\\\\n& =\\frac{400,000+\\log 3}{30,000+\\log 2}\n\\end{aligned}\n$$\n\n\n\nBoth $\\log 3$ and $\\log 2$ are tiny compared to the integers to which they are being added. If the quotient 400,000/30,000 were an integer (or extremely close to an integer), the values of these logarithms might matter, but $400,000 / 30,000=40 / 3=13 . \\overline{3}$, so in this case, they are irrelevant. Hence\n\n$$\n\\left\\lfloor\\log _{X} Y\\right\\rfloor=\\left\\lfloor\\frac{400,000}{30,000}\\right\\rfloor=\\left\\lfloor\\frac{40}{3}\\right\\rfloor=13\n$$" ]

[ "13" ]

[ "Let the vertices of the polygon be $A_{0}, A_{1}, \\ldots, A_{n-1}$. Considering the polygon as inscribed in a circle, the angle between diagonals $\\overline{A_{0} A_{i}}$ and $\\overline{A_{0} A_{j}}$ is $\\frac{1}{2} \\cdot\\left(\\frac{360^{\\circ}}{n}\\right) \\cdot|j-i|=\\left(\\frac{180|j-i|}{n}\\right)^{\\circ}$. The diagonal $\\overline{A_{k} A_{k+j}}$ can be considered as the rotation of $\\overline{A_{0} A_{j}}$ through $k / n$ of a circle, or $\\left(\\frac{360 k}{n}\\right)^{\\circ}$. So the diagonals $A_{0} A_{i}$ and $A_{k} A_{k+j}$ intersect at a combined angle of $\\left(\\frac{180|j-i|}{n}\\right)^{\\circ}+\\left(\\frac{360 k}{n}\\right)^{\\circ}$. Without loss of generality, assume $i<j$ (otherwise relabel vertices in the opposite order, with $A_{k}$ becoming $A_{0}$ ). Then the desired number is the least $n$ such that\n\n$$\n\\left(\\frac{180(j-i)}{n}\\right)+\\frac{360 k}{n}=159\n$$\n\nMultiply both sides of the equation by $n$ and factor to obtain $180(j-i+2 k)=159 n$, thus $60(j-i+2 k)=53 n$. Because 53 and 60 are relatively prime and $(j-i+2 k)$ is an integer, it follows that $60 \\mid n$. So the smallest possible value is $n=\\mathbf{6 0}$; one set of values that satisfies the equation is $i=1, j=54, k=0$." ]

[ "60" ]

[ "Because the graph passes through $(0,0)$, conclude that $c=0$. Then\n\n$$\nf(15)=225 \\Rightarrow a(15)^{2}+b(15)=225 a+15 b=225\n$$\n\nfrom which $b=15-15 a$. On the other hand, $f$ can be factored as $f(x)=a x(x+b / a)$, so if the roots are integers, $b / a$ must be an integer. Divide both sides of the equation $b=15-15 a$ by $a$ to obtain $b / a=15 / a-15$. Thus $15 / a$ must be an integer, and $a \\in\\{ \\pm 1, \\pm 3, \\pm 5, \\pm 15\\}$. Because $b=15-15 a$ is linear, each of these values for $a$ yields a unique integer value for $b$, so there are 8 such ordered pairs. The values of $a, b$, and the nonnegative root are given in the table below.\n\n| $a$ | $b$ | Root |\n| ---: | ---: | ---: |\n| 1 | 0 | 0 |\n| 3 | -30 | 10 |\n| 5 | -60 | 12 |\n| 15 | -210 | 14 |\n| -1 | 30 | 30 |\n| -3 | 60 | 20 |\n| -5 | 90 | 18 |\n| -15 | 240 | 16 |" ]

[ "8" ]

[ "The first step is to compute the radius $r$ of one of the marbles. The diagram below shows a cross-section through the centers of two diagonally opposite marbles.\n\n<img_3908>\n\nTriangle $B Q R$ appears to be equilateral, and in fact, it is. Reflect the diagram in the tabletop $\\overline{A C}$ to obtain six mutually tangent congruent circles inside a larger circle:\n\n<img_3444>\n\nBecause the circles are congruent, their centers are equidistant from $B$, and the distances between adjacent centers are equal. So $Q$ can be obtained as the image of $R$ under a rotation of $360^{\\circ} / 6=60^{\\circ}$ counterclockwise around $B$. Then $P Q=r \\Rightarrow B Q=B R=2 r \\Rightarrow B D=$ $3 r$, hence $r=1 / 3$. Notice too that the height of the pyramid is simply the radius $r$ and the diagonal of the square base is twice the altitude of the equilateral triangle $B Q R$, that is, $2 \\cdot \\frac{r \\sqrt{3}}{2}=r \\sqrt{3}$. So the area of the base is $3 r^{2} / 2$. Thus the volume of the pyramid is $(1 / 3)\\left(3 r^{2} / 2\\right)(r)=r^{3} / 2$. Because $r=1 / 3$, the volume is $\\mathbf{1} / \\mathbf{5 4}$." ]

[ "$\\frac{1}{54}$" ]

[ "Because 9 is used as a digit, $b \\geq 10$. The conditions require that $b+6$ be prime and $9 b+7$ be a perfect square. The numbers modulo 9 whose squares are congruent to 7 modulo 9 are 4 and 5. So $9 b+7=(9 k+4)^{2}$ or $(9 k+5)^{2}$ for some integer $k$. Finally, $b$ must be odd (otherwise $b+6$ is even), so $9 b+7$ must be even, which means that for any particular value of $k$, only one of $9 k+4$ and $9 k+5$ is possible. Taking these considerations together, $k=0$ is too small. Using $k=1$ makes $9 k+4$ odd, and while $(9 \\cdot 1+5)^{2}=196=9 \\cdot 21+7$ is even, because $21+6=27$ is composite, $b \\neq 21$. Using $k=2$ makes $9 k+4$ even, yielding $22^{2}=484=9 \\cdot 53+7$, and $53+6=59$ is prime. Thus $b=\\mathbf{5 3}$, and $53+6=59$ is prime while $9 \\cdot 53+7=484=22^{2}$." ]

[ "53" ]

[ "Group values of $n$ according to the number of bits (digits) in their binary representations:\n\n| Bits | $C(n)$ values | Total |\n| :---: | :---: | :---: |\n| 1 | $C\\left(1_{2}\\right)=0$ | 0 |\n| 2 | $C\\left(10_{2}\\right)=0$ <br> $C\\left(11_{2}\\right)=1$ | 1 |\n| 3 | $C\\left(100_{2}\\right)=0$ $C\\left(101_{2}\\right)=0$ <br> $C\\left(110_{2}\\right)=1$ $C\\left(111_{2}\\right)=2$ | 3 |\n| 4 | $C\\left(1000_{2}\\right)=0$ $C\\left(1001_{2}\\right)=0$ $C\\left(1100_{2}\\right)=1$ $C\\left(1101_{2}\\right)=1$ <br> $C\\left(1010_{2}\\right)=0$ $C\\left(1011_{2}\\right)=1$ $C\\left(1110_{2}\\right)=2$ $C\\left(1111_{2}\\right)=3$ | 8 |\n\nLet $B_{n}$ be the set of $n$-bit integers, and let $c_{n}=\\sum_{k \\in B_{n}} C(k)$ be the sum of the $C$-values for all $n$-bit integers. Observe that the integers in $B_{n+1}$ can be obtained by appending a 1 or a 0 to the integers in $B_{n}$. Appending a bit does not change the number of consecutive 1's in the previous (left) bits, but each number in $B_{n}$ generates two different numbers in $B_{n+1}$. Thus $c_{n+1}$ equals twice $2 c_{n}$ plus the number of new 11 pairs. Appending a 1 will create a new pair of consecutive 1's in (and only in) numbers that previously terminated in 1. The number of such numbers is half the number of elements in $B_{n}$. Because there are $2^{n-1}$ numbers in $B_{n}$, there are $2^{n-2}$ additional pairs of consecutive 1's among the elements in $B_{n+1}$. Thus for $n \\geq 2$, the sequence $\\left\\{c_{n}\\right\\}$ satisfies the recurrence relation\n\n$$\nc_{n+1}=2 c_{n}+2^{n-2}\n$$\n\n(Check: the table shows $c_{3}=3$ and $c_{4}=8$, and $8=2 \\cdot 3+2^{3-1}$.) Thus\n\n$$\n\\begin{aligned}\n& c_{5}=2 \\cdot c_{4}+2^{4-2}=2 \\cdot 8+4=20, \\\\\n& c_{6}=2 \\cdot c_{5}+2^{5-2}=2 \\cdot 20+8=48, \\\\\n& c_{7}=2 \\cdot c_{6}+2^{6-2}=2 \\cdot 48+16=112, \\text { and } \\\\\n& c_{8}=2 \\cdot c_{7}+2^{7-2}=2 \\cdot 112+32=256 .\n\\end{aligned}\n$$\n\nBecause $C(256)=0$, the desired sum is $c_{1}+c_{2}+c_{3}+c_{4}+c_{5}+c_{6}+c_{7}+c_{8}$, which equals 448 ." ]

[ "448" ]

[ "Let $S_{L}$ be the set of the least six integers in $S$, let $m$ be the median of $S$, and let $S_{G}$ be the set of the greatest six integers in $S$. In order to maximize the median, the elements of $S_{L}$ should be as small as possible, so start with $S_{L}=\\{1,2,3,4,5,6\\}$. Then the sum of $S_{L}$ 's elements is 21, leaving 99 as the sum of $m$ and the six elements of $S_{G}$. If $m=11$ and $S_{G}=\\{12,13,14,15,16,17\\}$, then the sum of all thirteen elements of $S$ is 119 . It is impossible to increase $m$ any further, because then the smallest set of numbers for $S_{G}$ would be $\\{13,14,15,16,17,18\\}$, and the sum would be at least 126 . To get the sum to be exactly 120, simply increase either 6 to 7 or 17 to 18 . The answer is $\\mathbf{1 1 .}$" ]

[ "11" ]

[ "For any integers $n$ and $b$, define $d(n, b)$ to be the unique nonnegative integer $k$ such that $b^{k} \\mid n$ and $b^{k+1} \\nmid n$; for example, $d(9,3)=2, d(9,4)=0$, and $d(18,6)=1$. So the problem asks for the smallest value of $b$ such that $d(T !, b)=2$. If $p$ is a prime and $p \\mid b$, then $d(T !, b) \\leq d(T !, p)$, so the least value of $b$ such that $d(T !, b)=2$ must be prime. Also, if $b$ is prime, then $d(T !, b)=\\lfloor T / b\\rfloor+\\left\\lfloor T / b^{2}\\right\\rfloor+\\left\\lfloor T / b^{3}\\right\\rfloor+\\cdots$. The only way that $d(T, b)$ can equal 2 is if the first term $\\lfloor T / b\\rfloor$ equals 2 and all other terms equal zero. (If $T \\geq b^{2}$, then $b \\geq 2$ implies $T / b \\geq b \\geq 2$, which would mean the first two terms by themselves would have a sum of at least 3.) Thus $2 b \\leq T<3 b$, hence $b \\leq T / 2$ and $T / 3<b$. For $T=11$, the only such $b$ is 5 ." ]

[ "5" ]

[ "Start by computing the first few terms of the sequence: $a_{1}=1, a_{2}=\\lceil\\sqrt{35}\\rceil=6, a_{3}=$ $\\lceil\\sqrt{70}\\rceil=9$, and $a_{4}=\\lceil\\sqrt{115}\\rceil=11$. Note that when $m \\geq 17,(m+1)^{2}=m^{2}+2 m+1>$ $m^{2}+34$, so if $a_{n} \\geq 17, a_{n+1}=\\left[\\sqrt{a_{n}^{2}+34}\\right\\rceil=a_{n}+1$. So it remains to continue the sequence until $a_{n} \\geq 17: a_{5}=13, a_{6}=15, a_{7}=17$. Then for $n>7, a_{n}=17+(n-7)=n+10$, and $a_{n}>100 T \\Rightarrow n>100 T-10$. With $T=5, n>490$, and the least value of $n$ is 491 ." ]

[ "491" ]

[ "If the polygon is inscribed in a circle, then the arc $\\overparen{A_{1} A_{13}}$ intercepted by $\\angle A_{1} A_{20} A_{13}$ has measure $12\\left(360^{\\circ} / n\\right)$, and thus $\\mathrm{m} \\angle A_{1} A_{20} A_{13}=6\\left(360^{\\circ} / n\\right)$. If $6(360 / n)<60$, then $n>6(360) / 60=$ 36. Thus the smallest value of $n$ is $\\mathbf{3 7}$." ]

[ "37" ]

[ "After the holes have been drilled, each face of the cube has area $T^{2}-1$. The three holes meet in a $1 \\times 1 \\times 1$ cube in the center, forming six holes in the shape of rectangular prisms whose bases are $1 \\times 1$ squares and whose heights are $(T-1) / 2$. Each of these holes thus contributes $4(T-1) / 2=2(T-1)$ to the surface area, for a total of $12(T-1)$. Thus the total area is $6\\left(T^{2}-1\\right)+12(T-1)$, which can be factored as $6(T-1)(T+1+2)=6(T-1)(T+3)$. With $T=37$, the total surface area is $6(36)(40)=\\mathbf{8 6 4 0}$." ]

[ "8640" ]

[ "Let $S=\\log _{4}\\left(1+2+4+\\cdots+2^{T}\\right)$. Because $1+2+4+\\cdots+2^{T}=2^{T+1}-1$, the change-of-base formula yields\n\n$$\nS=\\frac{\\log _{2}\\left(2^{T+1}-1\\right)}{\\log _{2} 4}\n$$\n\n\n\nLet $k=\\log _{2}\\left(2^{T+1}-1\\right)$. Then $T<k<T+1$, so $T / 2<S<(T+1) / 2$. If $T$ is even, then $\\lfloor S\\rfloor=T / 2$; if $T$ is odd, then $\\lfloor S\\rfloor=(T-1) / 2$. With $T=8640$, the answer is 4320 ." ]

[ "4320" ]

[ "Factoring, $6=2 \\cdot 3^{1}, 16=16 \\cdot 3^{0}$, and $72=8 \\cdot 3^{2}$, so $d(6)=1 / 3, d(16)=1$, and $d(72)=1 / 9$." ]

[ "$\\frac{1}{3},1,\\frac{1}{9}$" ]

[ "If $n=3^{k} m$ where $3 \\nmid m$, then $d(n)=1 / 3^{k}$. So the smallest values of $d(n)$ occur when $k$ is largest. The largest power of 3 less than 100 is $3^{4}=81$, so $d(81)=1 / 3^{4}=1 / 81$ is minimal." ]

[ "81" ]

[ "Here, $\\mathcal{N}(n)=\\{m \\mid m=27 k$, where $3 \\nmid k\\}$. The ten smallest elements of $\\mathcal{N}(n)$ are 27, $54,108,135,189,216,270,297,351$, and 378." ]

[ "27,54,108,135,189,216,270,297,351,378" ]

[ "Because $d(17, m)=1 / 81,17-m=81 l$, where $l \\in \\mathbb{Z}$ and $3 \\nmid l$. So $m=17-81 l$ and $16-m=81 l-1$. Hence $3 \\nmid 16-m$, and $d(16, m)=d(16-m)=1$." ]

[ "1" ]

[ "The maximum possible distance $d\\left(34, n_{k}\\right)$ is $1 / 3$. This can be proved by induction on $k: d\\left(n_{1}, 34\\right) \\leq 1 / 3$, and if both $d\\left(n_{k-1}, 34\\right) \\leq 1 / 3$ and $d\\left(n_{k-1}, n_{k}\\right) \\leq 1 / 3$, then $\\max \\left\\{d\\left(n_{k-1}, 34\\right), d\\left(n_{k-1}, n_{k}\\right)\\right\\} \\leq 1 / 3$ so by 5 b, $d\\left(34, n_{k}\\right) \\leq 1 / 3$." ]

[ "$1/3$" ]

[ "$\\frac{1}{3}, 1, 9$" ]

[ "$\\frac{1}{3}, 1, 9$" ]

[ "First, determine the angles of $A R M L$. Let $\\mathrm{m} \\angle M=x$. Then $\\mathrm{m} \\angle L R M=x$ because $\\triangle L R M$ is isosceles, and $\\mathrm{m} \\angle R L M=180^{\\circ}-2 x$. Because $\\overline{A R} \\| \\overline{L M}, \\mathrm{~m} \\angle A R M=180^{\\circ}-x$ and $\\mathrm{m} \\angle A R L=180^{\\circ}-2 x$, as shown in the diagram below.\n\n<img_3627>\n\nHowever, $\\triangle A R L$ is also isosceles (because $A R=A L$ ), so $\\mathrm{m} \\angle A L R=180^{\\circ}-2 x$, yielding $\\mathrm{m} \\angle A L M=360^{\\circ}-4 x$. Because $\\mathrm{m} \\angle R M L=\\mathrm{m} \\angle A L M$, conclude that $360^{\\circ}-4 x=x$, so $x=72^{\\circ}$. Therefore the base angles $L$ and $M$ have measure $72^{\\circ}$ while the other base angles $A$ and $R$ have measure $108^{\\circ}$. Finally, the angle formed by diagonals $\\overline{A M}$ and $\\overline{L R}$ is as follows: $\\mathrm{m} \\angle R D M=180^{\\circ}-\\mathrm{m} \\angle L R M-\\mathrm{m} \\angle A M R=180^{\\circ}-72^{\\circ}-36^{\\circ}=72^{\\circ}$.\n\nNow construct equilateral $\\triangle R O M$ with $O$ on the exterior of the trapezoid, as shown below.\n\n<img_3202>\n\nBecause $A R=R M=R O$, triangle $O A R$ is isosceles with base $\\overline{A O}$. The measure of $\\angle A R O$ is $108^{\\circ}+60^{\\circ}=168^{\\circ}$, so $\\mathrm{m} \\angle R A O=(180-168)^{\\circ} / 2=6^{\\circ}$. Thus $P$ lies on $\\overline{A O}$. Additionally, $\\mathrm{m} \\angle P O M=\\mathrm{m} \\angle A O M=60^{\\circ}-6^{\\circ}=54^{\\circ}$, and $\\mathrm{m} \\angle P M O=60^{\\circ}+12^{\\circ}=72^{\\circ}$ by construction. Thus $\\mathrm{m} \\angle M P O=180^{\\circ}-72^{\\circ}-54^{\\circ}=54^{\\circ}$, hence $\\triangle P M O$ is isosceles with $P M=O M$. But because $O M=R M, \\triangle R M P$ is isosceles with $R M=M P$, and $R M=D M$ implies that $\\triangle P D M$ is also isosceles. But $\\mathrm{m} \\angle R M P=12^{\\circ}$ implies that $\\mathrm{m} \\angle P M D=36^{\\circ}-12^{\\circ}=24^{\\circ}$, so $\\mathrm{m} \\angle D P M=78^{\\circ}$. Thus $\\mathrm{m} \\angle A P D=180^{\\circ}-\\mathrm{m} \\angle O P M-\\mathrm{m} \\angle D P M=180^{\\circ}-54^{\\circ}-78^{\\circ}=48^{\\circ}$." ]

[ "48" ]

[ "There are 6 triangles of side lengths $1,1, \\sqrt{3} ; 2$ equilateral triangles of side length $\\sqrt{3}$; and 12 triangles of side lengths $1, \\sqrt{3}, 2$. One triangle of each type is shown in the diagram below.\n<img_3233>\n\nEach triangle in the first set has area $\\sqrt{3} / 4$; each triangle in the second set has area $3 \\sqrt{3} / 4$; and each triangle in the third set has area $\\sqrt{3} / 2$. The average is\n\n$$\n\\frac{6\\left(\\frac{\\sqrt{3}}{4}\\right)+2\\left(\\frac{3 \\sqrt{3}}{4}\\right)+12\\left(\\frac{\\sqrt{3}}{2}\\right)}{20}=\\frac{\\frac{6 \\sqrt{3}}{4}+\\frac{6 \\sqrt{3}}{4}+\\frac{24 \\sqrt{3}}{4}}{20}=\\frac{\\mathbf{9} \\sqrt{\\mathbf{3}}}{\\mathbf{2 0}} .\n$$" ]

[ "$\\frac{9 \\sqrt{3}}{20}$" ]

[ "The problem does not state the number of mugs Paul intended to buy, but the actual number is irrelevant. Suppose Paul plans to buy $M$ mugs and $20-M$ shirts. The total cost is $10 M+6(20-M)$ However, he puts back $40 \\%$ of the mugs, so he ends up spending $10(0.6 M)+$ $6(20-M)=6 M+120-6 M=\\mathbf{1 2 0}$ dollars." ]

[ "120" ]

[ "In order for $1584 \\cdot x$ to be a perfect cube, all of its prime factors must be raised to powers divisible by 3 . Because $1584=2^{4} \\cdot 3^{2} \\cdot 11$, $x$ must be of the form $2^{3 k+2} \\cdot 3^{3 m+1} \\cdot 11^{3 n+2} \\cdot r^{3}$, for nonnegative integers $k, m, n, r, r>0$. Thus the least positive value of $x$ is $2^{2} \\cdot 3 \\cdot 11^{2}=1452$. But in order for $x y$ to be a positive multiple of $1584, x y$ must be of the form $2^{a} \\cdot 3^{b} \\cdot 11^{c} \\cdot d$, where $a \\geq 4, b \\geq 2, c \\geq 1$, and $d \\geq 1$. Thus $y$ must equal $2^{2} \\cdot 3^{1}=\\mathbf{1 2}$." ]

[ "12" ]

[ "Let $a$ be the price of one apple and $p$ be the price of one peach, in cents. The first transaction shows that $500<5 a+5 p<1000$, hence $100<a+p<200$. The second transaction shows that $1000<2 a+12 p<2000$, so $500<a+6 p<1000$. Subtracting the inequalities yields $300<5 p<900$, so $60<p<180$. Therefore the price of 25 peaches is at least $25 \\cdot 61=\\mathbf{1 5 2 5}$ cents." ]

[ "1525" ]

[ "Extend $\\overline{A B}$ to point $Q$ such that $\\overline{P Q} \\perp \\overline{A Q}$ as shown, and let $M$ be the midpoint of $\\overline{A B}$. (The problem does not specify whether $A$ or $B$ is nearer $P$, but $B$ can be assumed to be nearer $P$ without loss of generality.)\n\n<img_3454>\n\nThen $O P=10, P Q=O M=2$, and $O B=6$. Thus $M B=\\sqrt{6^{2}-2^{2}}=4 \\sqrt{2}$. Because $Q M=O P=10$, it follows that $Q B=10-4 \\sqrt{2}$ and $Q A=10+4 \\sqrt{2}$. So\n\n$$\n\\begin{aligned}\nP A^{2}+P B^{2} & =\\left(Q A^{2}+Q P^{2}\\right)+\\left(Q B^{2}+Q P^{2}\\right) \\\\\n& =(10+4 \\sqrt{2})^{2}+2^{2}+(10-4 \\sqrt{2})^{2}+2^{2} \\\\\n& =\\mathbf{2 7 2}\n\\end{aligned}\n$$" ]

[ "272" ]

[ "If $a+b \\geq 2014$, then at least one of $a, b$ must be greater than 1006 . The palindromes greater than 1006 but less than 2014 are, in descending order, 2002, 1991, 1881, ..., 1111. Let a\n\n\n\nrepresent the larger of the two palindromes. Then for $n=2014, a=2002$ is impossible, because $2014-2002=12$. Any value of $a$ between 1111 and 2000 ends in 1 , so if $a+b=2014$, $b$ ends in 3 , and because $b<1000$, it follows that $303 \\leq b \\leq 393$. Subtracting 303 from 2014 yields 1711, and so $a \\leq 1711$. Thus $a=1661$ and $b=353$. A similar analysis shows the following results:\n\n$$\n\\begin{aligned}\n& 2015=1551+464 ; \\\\\n& 2016=1441+575 ; \\\\\n& 2017=1331+686 ; \\text { and } \\\\\n& 2018=1221+797\n\\end{aligned}\n$$\n\nBut 2019 cannot be expressed as the sum of two palindromes: $b$ would have to end in 8 , so $b=808+10 d$ for some digit $d$. Then $2019-898 \\leq a \\leq 2019-808$, hence $1121 \\leq a \\leq 1211$, and there is no palindrome in that interval." ]

[ "2019" ]

[ "First consider the problem with $x, y, z$ positive real numbers. If $x y+z=160$ and $z$ is constant, then $y=\\frac{160-z}{x}$, yielding $x+y z=x+\\frac{z(160-z)}{x}$. For $a, x>0$, the quantity $x+\\frac{a}{x}$ is minimized when $x=\\sqrt{a}$ (proof: use the Arithmetic-Geometric Mean Inequality $\\frac{A+B}{2} \\geq \\sqrt{A B}$ with $A=x$ and $\\left.B=\\frac{a}{x}\\right)$; in this case, $x+\\frac{a}{x}=2 \\sqrt{a}$. Thus $x+y z \\geq 2 \\sqrt{z(160-z)}$. Considered as a function of $z$, this lower bound is increasing for $z<80$.\n\nThese results suggest the following strategy: begin with small values of $z$, and find a factorization of $160-z$ such that $x$ is close to $\\sqrt{z(160-z)}$. (Equivalently, such that $\\frac{x}{y}$ is close to $z$.) The chart below contains the triples $(x, y, z)$ with the smallest values of $x+y z$, conditional upon $z$.\n\n| $z$ | $(x, y, z)$ | $x+y z$ |\n| :---: | :---: | :---: |\n| 1 | $(53,3,1)$ | 56 |\n| 2 | $(79,2,2)$ | 83 |\n| 3 | $(157,1,3)$ | 160 |\n| 4 | $(26,6,4)$ | 50 |\n| 5 | $(31,5,5)$ | 56 |\n| 6 | $(22,7,6)$ | 64 |\n\nBecause $x+y z \\geq 2 \\sqrt{z(160-z)}$, it follows that $x+y z \\geq 64$ for $6 \\leq z \\leq 80$. And because $x+y z>80$ for $z \\geq 80$, the minimal value of $x+y z$ is $\\mathbf{5 0}$." ]

[ "50" ]

[ "The identity $\\cos 3 \\theta=4 \\cos ^{3} \\theta-3 \\cos \\theta$ can be rewritten into the power-reducing identity\n\n$$\n\\cos ^{3} \\theta=\\frac{1}{4} \\cos 3 \\theta+\\frac{3}{4} \\cos \\theta\n$$\n\n\n\nThus if $D$ is the desired sum,\n\n$$\n\\begin{aligned}\nD & =\\cos ^{3} \\frac{2 \\pi}{7}+\\cos ^{3} \\frac{4 \\pi}{7}+\\cos ^{3} \\frac{8 \\pi}{7} \\\\\n& =\\frac{1}{4}\\left(\\cos \\frac{6 \\pi}{7}+\\cos \\frac{12 \\pi}{7}+\\cos \\frac{24 \\pi}{7}\\right)+\\frac{3}{4}\\left(\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{8 \\pi}{7}\\right) .\n\\end{aligned}\n$$\n\nObserve that $\\cos \\frac{24 \\pi}{7}=\\cos \\frac{10 \\pi}{7}$, so\n\n$$\nD=\\frac{1}{4}\\left(\\cos \\frac{6 \\pi}{7}+\\cos \\frac{12 \\pi}{7}+\\cos \\frac{10 \\pi}{7}\\right)+\\frac{3}{4}\\left(\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{8 \\pi}{7}\\right) .\n$$\n\nNotice also that $\\cos \\theta=\\cos (2 \\pi-\\theta)$ implies $\\cos \\frac{12 \\pi}{7}=\\cos \\frac{2 \\pi}{7}, \\cos \\frac{10 \\pi}{7}=\\cos \\frac{4 \\pi}{7}$, and $\\cos \\frac{8 \\pi}{7}=$ $\\cos \\frac{6 \\pi}{7}$. Rewriting $D$ using the least positive equivalent angles yields\n\n$$\n\\begin{aligned}\nD & =\\frac{1}{4}\\left(\\cos \\frac{6 \\pi}{7}+\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}\\right)+\\frac{3}{4}\\left(\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{6 \\pi}{7}\\right) \\\\\n& =\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{6 \\pi}{7} .\n\\end{aligned}\n$$\n\nTo evaluate this sum, use the identity $\\cos \\theta=\\cos (2 \\pi-\\theta)$ again to write\n\n$$\n2 D=\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{6 \\pi}{7}+\\cos \\frac{8 \\pi}{7}+\\cos \\frac{10 \\pi}{7}+\\cos \\frac{12 \\pi}{7}\n$$\n\nIf $\\alpha=\\cos \\frac{2 \\pi}{7}+i \\sin \\frac{2 \\pi}{7}$, notice that the right side of the equation above is simply the real part of the sum $\\alpha+\\alpha^{2}+\\alpha^{3}+\\alpha^{4}+\\alpha^{5}+\\alpha^{6}$. Because $\\alpha^{n}$ is a solution to the equation $z^{7}=1$ for $n=0,1, \\ldots, 6$, the sum $1+\\alpha+\\alpha^{2}+\\cdots+\\alpha^{6}$ equals 0 . Hence $\\alpha+\\alpha^{2}+\\cdots+\\alpha^{6}=-1$ and $D=-1 / 2$.", "Construct a cubic polynomial in $x$ for which $\\cos \\frac{2 \\pi}{7}, \\cos \\frac{4 \\pi}{7}$, and $\\cos \\frac{8 \\pi}{7}$ are zeros; then the sum of their cubes can be found using techniques from the theory of equations. In particular, suppose the three cosines are zeros of $x^{3}+b x^{2}+c x+d$. Then\n\n$$\n\\begin{aligned}\nb & =-\\left(\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{8 \\pi}{7}\\right) \\\\\nc & =\\cos \\frac{2 \\pi}{7} \\cos \\frac{4 \\pi}{7}+\\cos \\frac{2 \\pi}{7} \\cos \\frac{8 \\pi}{7}+\\cos \\frac{4 \\pi}{7} \\cos \\frac{8 \\pi}{7}, \\text { and } \\\\\nd & =-\\cos \\frac{2 \\pi}{7} \\cos \\frac{4 \\pi}{7} \\cos \\frac{8 \\pi}{7}\n\\end{aligned}\n$$\n\nUse complex seventh roots of unity (as in the previous solution) to find $b=1 / 2$. To find $c$, use the product-to-sum formula $2 \\cos A \\cos B=\\cos (A+B)+\\cos (A-B)$ three times:\n\n$$\n\\begin{aligned}\n2 c & =\\left(\\cos \\frac{6 \\pi}{7}+\\cos \\frac{2 \\pi}{7}\\right)+\\left(\\cos \\frac{10 \\pi}{7}+\\cos \\frac{6 \\pi}{7}\\right)+\\left(\\cos \\frac{4 \\pi}{7}+\\cos \\frac{12 \\pi}{7}\\right) \\\\\n& \\left.=\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{6 \\pi}{7}+\\cos \\frac{8 \\pi}{7}+\\cos \\frac{10 \\pi}{7}+\\cos \\frac{12 \\pi}{7} \\text { [because } \\cos \\theta=\\cos (2 \\pi-\\theta)\\right] \\\\\n& =-1\n\\end{aligned}\n$$\n\n\n\nThus $c=-1 / 2$.\n\nTo compute $d$, multiply both sides by $\\sin \\frac{2 \\pi}{7}$ and use the identity $2 \\sin \\theta \\cos \\theta=\\sin 2 \\theta$ :\n\n$$\n\\begin{aligned}\nd \\sin \\frac{2 \\pi}{7} & =-\\sin \\frac{2 \\pi}{7} \\cos \\frac{2 \\pi}{7} \\cos \\frac{4 \\pi}{7} \\cos \\frac{8 \\pi}{7} \\\\\n& =-\\frac{1}{2} \\sin \\frac{4 \\pi}{7} \\cos \\frac{4 \\pi}{7} \\cos \\frac{8 \\pi}{7} \\\\\n& =-\\frac{1}{4} \\sin \\frac{8 \\pi}{7} \\cos \\frac{8 \\pi}{7} \\\\\n& =-\\frac{1}{8} \\sin \\frac{16 \\pi}{7} .\n\\end{aligned}\n$$\n\nBecause $\\sin \\frac{16 \\pi}{7}=\\sin \\frac{2 \\pi}{7}$, the factors on both sides cancel, leaving\n\n$$\nd=-1 / 8\n$$\n\nThus $\\cos \\frac{2 \\pi}{7}, \\cos \\frac{4 \\pi}{7}$, and $\\cos \\frac{8 \\pi}{7}$ are roots of $x^{3}+\\frac{1}{2} x^{2}-\\frac{1}{2} x-\\frac{1}{8}$; so each value also satisfies the equation $x^{3}=-\\frac{1}{2} x^{2}+\\frac{1}{2} x+\\frac{1}{8}$. Hence the desired sum can be rewritten as\n\n$$\n\\begin{aligned}\n\\cos ^{3} \\frac{2 \\pi}{7}+\\cos ^{3} \\frac{4 \\pi}{7}+\\cos ^{3} \\frac{8 \\pi}{7} & =-\\frac{1}{2}\\left(\\cos ^{2} \\frac{2 \\pi}{7}+\\cos ^{2} \\frac{4 \\pi}{7}+\\cos ^{2} \\frac{8 \\pi}{7}\\right) \\\\\n& +\\frac{1}{2}\\left(\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{8 \\pi}{7}\\right)+\\frac{3}{8}\n\\end{aligned}\n$$\n\nPrevious work has already established that $\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{8 \\pi}{7}=-1 / 2$, so it remains to compute $\\cos ^{2} \\frac{2 \\pi}{7}+\\cos ^{2} \\frac{4 \\pi}{7}+\\cos ^{2} \\frac{8 \\pi}{7}$. The identity $A^{2}+B^{2}+C^{2}=(A+B+C)^{2}-2(A B+B C+A C)$ allows the use of previous results: $\\cos ^{2} \\frac{2 \\pi}{7}+\\cos ^{2} \\frac{4 \\pi}{7}+\\cos ^{2} \\frac{8 \\pi}{7}=(-1 / 2)^{2}-2(-1 / 2)=5 / 4$. Thus\n\n$$\n\\cos ^{3} \\frac{2 \\pi}{7}+\\cos ^{3} \\frac{4 \\pi}{7}+\\cos ^{3} \\frac{8 \\pi}{7}=-\\frac{1}{2}\\left(\\frac{5}{4}\\right)+\\frac{1}{2}\\left(-\\frac{1}{2}\\right)+\\frac{3}{8}=-\\frac{1}{2} .\n$$" ]

[ "$-\\frac{1}{2}$" ]

[ "Note that in right triangle $A B C$ with right angle $C$, the inradius $r$ is equal to $\\frac{a+b-c}{2}$, where $a=B C, b=A C$, and $c=A B$, because the inradius equals the distance from the vertex of the right angle $C$ to (either) point of tangency along $\\overline{A C}$ or $\\overline{B C}$. Thus the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$ is equal to one-half the difference between the sum of the lengths of the legs of these triangles and the sum of the lengths of the hypotenuses of these triangles. Let $t$ be the side length of square $P Q R S$. Then the sum of the lengths of the legs of triangles $A X P, C X S, C Y R$, and $B Y Q$ is\n\n$$\n\\begin{aligned}\n& A P+P X+X S+S C+C R+R Y+Y Q+Q B \\\\\n= & A P+P S+S R+R Q+Q B \\\\\n= & A P+t+t+t+Q B \\\\\n= & A B-P Q+3 t \\\\\n= & c-t+3 t \\\\\n= & c+2 t .\n\\end{aligned}\n$$\n\n\n\nThe sum of the lengths of the hypotenuses of triangles $A X P, C X S, C Y R$, and $B Y Q$ is $A X+X C+C Y+Y B=A C+C B=b+a$. Hence the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$ is $\\frac{c+2 t-(a+b)}{2}=t-r$. Thus the desired sum equals $\\sqrt{576}-10=24-10=\\mathbf{1 4}$." ]

[ "14" ]

[ "Because the quantity on the left side is the difference of two integers, $x / 7$ must be an integer, hence $x$ is an integer (in fact a multiple of 7). Because the denominators on the left side are 2 and 3 , it is convenient to write $x=6 q+r$, where $0 \\leq r \\leq 5$, so that $\\lfloor x / 2\\rfloor=3 q+\\lfloor r / 2\\rfloor$ and $\\lfloor x / 3\\rfloor=2 q+\\lfloor r / 3\\rfloor$. Then for $r=0,1, \\ldots, 5$ these expressions can be simplified as shown in the table below.\n\n| $r$ | 0 | 1 | 2 | 3 | 4 | 5 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\left\\lfloor\\frac{x}{2}\\right\\rfloor$ | $3 q$ | $3 q$ | $3 q+1$ | $3 q+1$ | $3 q+2$ | $3 q+2$ |\n| $\\left\\lfloor\\frac{x}{3}\\right\\rfloor$ | $2 q$ | $2 q$ | $2 q$ | $2 q+1$ | $2 q+1$ | $2 q+1$ |\n| $\\left\\lfloor\\frac{x}{2}\\right\\rfloor-\\left\\lfloor\\frac{x}{3}\\right\\rfloor$ | $q$ | $q$ | $q+1$ | $q$ | $q+1$ | $q+1$ |\n\nNow proceed by cases:\n\n$r=0:$ Then $q=x / 6$. But from the statement of the problem, $q=x / 7$, so $x=0$.\n\n$r=1: \\quad$ Then $q=(x-1) / 6=x / 7 \\Rightarrow x=7$.\n\n$r=2: \\quad$ Then $q=(x-2) / 6$ and $q+1=x / 7$, so $(x+4) / 6=x / 7$, and $x=-28$.\n\n$r=3$ : Then $q=(x-3) / 6$ and $q=x / 7$, so $x=21$.\n\n$r=4: \\quad$ Then $q=(x-4) / 6$ and $q+1=x / 7$, so $(x+2) / 6=x / 7$, and $x=-14$.\n\n$r=5$ : Then $q=(x-5) / 6$ and $q+1=x / 7$, so $(x+1) / 6=x / 7$, and $x=-7$.\n\nThe sum of these values is $0+7+-28+21+-14+-7=\\mathbf{- 2 1}$." ]

[ "-21" ]

[ "If $f$ is simply a permutation of $S$, then $\\left\\{s_{n}\\right\\}$ is periodic. To understand why, consider a smaller set $T=\\{1,2,3,4,5,6,7,8,9,10\\}$. If $f:[1,2,3,4,5,6,7,8,9,10] \\rightarrow[2,3,4,5,1,7,8,6,9,10]$, then $f$ has one cycle of period 5 and one cycle of period 3 , so the period of $f$ is 15 . However,\n\n$$\nf(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)+f(10)=\n$$\n\n\n\n$$\n2+3+4+5+1+7+8+6+9+10=55,\n$$\n\nbecause $f$ just rearranges the order of the summands. So $s_{1}=s_{0}$, and for all $n, s_{n}=s_{n+1}$; in short, the period of $\\left\\{s_{n}\\right\\}$ is just 1 .\n\nIn order for $\\left\\{s_{n}\\right\\}$ to have a period greater than $1, f$ must be many-to-one, so that some values occur more than once (and some values do not occur at all) in the sum $f(1)+f(2)+\\cdots+f(10)$ (or, in the original problem, $f(1)+f(2)+\\cdots+f(20)$ ). For example, consider the function $f_{2}$ below:\n\n$$\nf_{2}:[1,2,3,4,5,6,7,8,9,10] \\rightarrow[2,3,4,5,1,10,9,10,7,3]\n$$\n\nNote that $s_{1}=2+3+4+5+1+10+9+10+7+3 \\neq 55$, so $\\left\\{s_{n}\\right\\}$ is not immediately periodic. But $\\left\\{s_{n}\\right\\}$ is eventually periodic, as the following argument shows. The function $f_{2}$ has two cycles: $1 \\rightarrow 2 \\rightarrow 3 \\rightarrow 4 \\rightarrow 5 \\rightarrow 1$, and $7 \\rightarrow 9 \\rightarrow 7$. There are also two paths that meet up with the first cycle: $6 \\rightarrow 10 \\rightarrow 3 \\rightarrow \\cdots$ and $8 \\rightarrow 10 \\rightarrow 3 \\rightarrow \\cdots$. Thus for all $k$ in $T, f_{2}\\left(f_{2}(k)\\right)$ is an element of one of these two extended cycles. Thus $\\left\\{s_{n}\\right\\}$ eventually becomes periodic.\n\nThe criterion that the function be many-to-one is necessary, but not sufficient, for $\\left\\{s_{n}\\right\\}$ to have period greater than 1 . To see why, consider the function $g:[1,2,3,4,5,6,7,8,9,10] \\rightarrow$ $[2,3,4,5,6,1,8,7,8,7]$. This function is many-to-one, and contains two cycles, $1 \\rightarrow 2 \\rightarrow$ $3 \\rightarrow 4 \\rightarrow 5 \\rightarrow 6 \\rightarrow 1$ and $7 \\rightarrow 8 \\rightarrow 7$. But because $g(9)=8$ and $g(10)=7$, the sum $s_{1}=2+3+4+5+6+1+8+7+8+7$, while $s_{2}=3+4+5+6+1+2+7+8+7+8$. In fact, for $n>1, s_{n+1}=s_{n}$, because applying $f$ only permutes the 6 -cycle and switches the two 7 's and two 8's. That is, in the list $\\underbrace{(g \\circ \\cdots \\circ g)}_{n}(1), \\ldots, \\underbrace{(g \\circ \\cdots \\circ g)}_{n}(10)$, the values 7 and 8 both show up exactly twice. This cycle is balanced: each of its elements shows up the same number of times for all $n$ in the list $\\underbrace{(g \\circ \\cdots \\circ g)}_{n}(1), \\ldots, \\underbrace{(g \\circ \\cdots \\circ g)}_{n}(10)$, for all $n$ after a certain point. The conclusion is that not all many-to-one functions produce unbalanced cycles.\n\nThere are two ways a function $g$ can produce balanced cycles. First, the cycles can be selfcontained, so no element outside of the cycle is ever absorbed into the cycle, as happens with the 6-cycle in the example above. Alternatively, the outside elements that are absorbed into a cycle can all arrive at different points of the cycle, so that each element of the cycle occurs equally often in each iteration of $g$. In the example above, the values $g(9)=7$ and $g(10)=8$ balance the $7 \\rightarrow 8 \\rightarrow 7$ cycle. On the other hand, in the function $f_{2}$ above, $f(f(6))=f(f(8))=f(f(1))=3$, making the large cycle unbalanced: in $s_{2}$, the value 3 appears three times in $s_{2}$, but the value 2 only appears once in $s_{2}$.\n\nThe foregoing shows that only unbalanced cycles can affect the periodicity of $\\left\\{s_{n}\\right\\}$. Because each element of a balanced cycle occurs equally often in each iteration, the period of that component of the sum $s_{n}$ attributed to the cycle is simply 1. (The case where $f$ is a permutation of $S$ is simply a special case of this result.) In the above example, the large cycle is\n\n\n\nunbalanced. Note the following results under $f_{2}$.\n\n| $n$ | $\\overbrace{\\left(f_{2} \\circ \\cdots \\circ f_{2}\\right)}^{n}(T)$ | $s_{n}$ |\n| :---: | :---: | :---: |\n| 1 | $[2,3,4,5,1,10,9,10,7,3]$ | 54 |\n| 2 | $[3,4,5,1,2,3,7,3,9,4]$ | 41 |\n| 3 | $[4,5,1,2,3,4,9,4,7,5]$ | 40 |\n| 4 | $[5,1,2,3,4,5,7,5,9,1]$ | 42 |\n| 5 | $[1,2,3,4,5,1,9,1,7,2]$ | 35 |\n| 6 | $[2,3,4,5,1,2,7,2,9,3]$ | 38 |\n| 7 | $[3,4,5,1,2,3,9,3,7,4]$ | 41 |\n| 8 | $[4,5,1,2,3,4,7,4,9,5]$ | 40 |\n| 9 | $[5,1,2,3,4,5,9,5,7,1]$ | 42 |\n\nThe period of $\\left\\{s_{n}\\right\\}$ for $f_{2}$ is 5 , the period of the unbalanced cycle.\n\nThe interested reader may inquire whether all unbalanced cycles affect the periodicity of $\\left\\{s_{n}\\right\\}$; we encourage those readers to explore the matter independently. For the purposes of solving this problem, it is sufficient to note that unbalanced cycles can affect $\\left\\{s_{n}\\right\\}$ 's periodicity.\n\nFinally, note that an unbalanced $k$-cycle actually requires at least $k+1$ elements: $k$ to form the cycle, plus at least 1 to be absorbed into the cycle and cause the imbalance. For the original set $S$, one way to create such an imbalance would be to have $f(20)=f(1)=$ $2, f(2)=3, f(3)=4, \\ldots, f(19)=1$. This arrangement creates an unbalanced cycle of length 19. But breaking up into smaller unbalanced cycles makes it possible to increase the period of $\\left\\{s_{n}\\right\\}$ even more, because then in most cases the period is the least common multiple of the periods of the unbalanced cycles. For example, $f:[1,2,3, \\ldots, 20]=$ $[2,3,4,5,6,7,8,9,1,1,12,13,14,15,16,17,18,11,11,11]$ has an unbalanced cycle of length 9 and an unbalanced cycle of length 8 , giving $\\left\\{s_{n}\\right\\}$ a period of 72 .\n\nSo the goal is to maximize $\\operatorname{lcm}\\left\\{k_{1}, k_{2}, \\ldots, k_{m}\\right\\}$ such that $k_{1}+k_{2}+\\cdots+k_{m}+m \\leq 20$. With $m=2$, the maximal period is 72 , achieved with $k_{1}=9$ and $k_{2}=8$. With $m=3$, $k_{1}+k_{2}+k_{3} \\leq 17$, but $\\operatorname{lcm}\\{7,6,4\\}=84<\\operatorname{lcm}\\{7,5,4\\}=140$. This last result can be obtained with unbalanced cycles of length 4,5 , and 7 , with the remaining four points entering the three cycles (or with one point forming a balanced cycle of length 1, i.e., a fixed point). Choosing larger values of $m$ decreases the values of $k$ so far that they no longer form long cycles: when $m=4, k_{1}+k_{2}+k_{3}+k_{4} \\leq 16$, and even if $k_{4}=2, k_{3}=3$, and $k_{2}=5$, for a period of 30 , the largest possible value of $k_{1}=6$, which does not alter the period. (Even $k_{1}=7, k_{2}=5$, and $k_{3}=k_{4}=2$ only yields a period of 70 .) Thus the maximum period of $s_{n}$ is $\\mathbf{1 4 0}$. One such function $f$ is given below.\n\n$$\n\\begin{array}{c|cccccccccccccccccccc}\nn & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\\\\n\\hline f(n) & 2 & 3 & 4 & 1 & 1 & 7 & 8 & 9 & 10 & 6 & 6 & 13 & 14 & 15 & 16 & 17 & 18 & 12 & 12 & 20\n\\end{array}\n$$" ]

[ "140" ]

[ "Note that $n^{2}+n^{0}+n^{1}+n^{3}=n^{2}+1+n+n^{3}=\\left(n^{2}+1\\right)(1+n)$. Because 13 is prime, 13 must be a divisor of one of these factors. The smallest positive integer $n$ such that $13 \\mid 1+n$ is $n=12$, whereas the smallest positive integer $n$ such that $13 \\mid n^{2}+1$ is $n=\\mathbf{5}$." ]

[ "5" ]