id int64 | question string | solution list | final_answer list | context string | image_1 image | image_2 image | image_3 image | image_4 image | image_5 image | modality string | difficulty string | is_multiple_answer bool | unit string | answer_type string | error string | question_type string | subfield string | subject string | language string |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
2,649 | Let $T=1$. Circles $L$ and $O$ are internally tangent and have radii $T$ and $4 T$, respectively. Point $E$ lies on circle $L$ such that $\overline{O E}$ is tangent to circle $L$. Compute $O E$. | [
"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 $T=1, O E=\\mathbf{2} \\sqrt{\\mathbf{2}}$."
] | [
"$2 \\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,650 | Let $T=2 \sqrt{2}$. In a right triangle, one leg has length $T^{2}$ and the other leg is 2 less than the hypotenuse. Compute the triangle's perimeter. | [
"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"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,651 | $\quad$ Let $T=40$. If $x+9 y=17$ and $T x+(T+1) y=T+2$, compute $20 x+14 y$. | [
"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... | [
"8"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,652 | Let $T=8$. Let $f(x)=a x^{2}+b x+c$. The product of the roots of $f$ is $T$. If $(-2,20)$ and $(1,14)$ lie on the graph of $f$, compute $a$. | [
"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 $a=\\fra... | [
"$\\frac{8}{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,653 | Let $T=\frac{8}{5}$. Let $z_{1}=15+5 i$ and $z_{2}=1+K i$. Compute the smallest positive integral value of $K$ such that $\left|z_{1}-z_{2}\right| \geq 15 T$. | [
"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 ... | [
"25"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,654 | Let $T=25$. Suppose that $T$ people are standing in a line, including three people named Charlie, Chris, and Abby. If the people are assigned their positions in line at random, compute the probability that Charlie is standing next to at least one of Chris or Abby. | [
"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 remai... | [
"$\\frac{47}{300}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,655 | Let $A$ be the number you will receive from position 7 and let $B$ be the number you will receive from position 9. Let $\alpha=\sin ^{-1} A$ and let $\beta=\cos ^{-1} B$. Compute $\sin (\alpha+\beta)+\sin (\alpha-\beta)$. | [
"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 $B=\\frac{12}{169}, 2 \\cdot A \\cdot ... | [
"$\\frac{94}{4225}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,656 | Let $T=13$. If $r$ is the radius of a right circular cone and the cone's height is $T-r^{2}$, let $V$ be the maximum possible volume of the cone. Compute $\pi / V$. | [
"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 $V=\\frac{1}{3} \\pi \\cdot \\frac{T}{2} \\cdo... | [
"$\\frac{12}{169}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,657 | Let $T=650$. If $\log T=2-\log 2+\log k$, compute the value of $k$. | [
"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"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,658 | Let $T=100$. Nellie has a flight from Rome to Athens that is scheduled to last for $T+30$ minutes. However, owing to a tailwind, her flight only lasts for $T$ minutes. The plane's speed is 1.5 miles per minute faster than what it would have been for the originally scheduled flight. Compute the distance (in miles) that the plane travels. | [
"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"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,659 | Let $T=9$. Compute $\sqrt{\sqrt{\sqrt[T]{10^{T^{2}-T}}}}$. | [
"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"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,660 | Let $T=3$. Regular hexagon $S U P E R B$ has side length $\sqrt{T}$. Compute the value of $B E \cdot S U \cdot R E$. | [
"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)(\\s... | [
"9"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,661 | Let $T=70$. Chef Selma is preparing a burrito menu. A burrito consists of: (1) a choice of chicken, beef, turkey, or no meat, (2) exactly one of three types of beans, (3) exactly one of two types of rice, and (4) exactly one of $K$ types of cheese. Compute the smallest value of $K$ such that Chef Selma can make at least $T$ different burrito varieties. | [
"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"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,662 | Compute the smallest positive integer $N$ such that $20 N$ is a multiple of 14 and $14 N$ is a multiple of 20 . | [
"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"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,663 | Call a positive integer fibbish if each digit, after the leftmost two, is at least the sum of the previous two digits. Compute the greatest fibbish number. | [
"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}}... | [
"10112369"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,664 | An ARMLbar is a $7 \times 7$ grid of unit squares with the center unit square removed. A portion of an ARMLbar is a square section of the bar, cut along the gridlines of the original bar. Compute the number of different ways there are to cut a single portion from an ARMLbar. | [
"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... | [
"96"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,665 | Regular hexagon $A B C D E F$ and regular hexagon $G H I J K L$ both have side length 24 . The hexagons overlap, so that $G$ is on $\overline{A B}, B$ is on $\overline{G H}, K$ is on $\overline{D E}$, and $D$ is on $\overline{J K}$. If $[G B C D K L]=\frac{1}{2}[A B C D E F]$, compute $L F$. | [
"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 $A E$... | [
"18"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,666 | Compute the largest base-10 integer $\underline{A} \underline{B} \underline{C} \underline{D}$, with $A>0$, such that $\underline{A} \underline{B} \underline{C} \underline{D}=B !+C !+D !$. | [
"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=5... | [
"5762"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,667 | Let $X$ be the number of digits in the decimal expansion of $100^{1000^{10,000}}$, and let $Y$ be the number of digits in the decimal expansion of $1000^{10,000^{100,000}}$. Compute $\left\lfloor\log _{X} Y\right\rfloor$. | [
"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$$\n\\begin{aligned}\n\\log _{X} Y=\\frac{\\log Y}{\\log X} & \\approx \\frac{\\l... | [
"13"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,668 | Compute the smallest possible value of $n$ such that two diagonals of a regular $n$-gon intersect at an angle of 159 degrees. | [
"Let the vertices of the polygon be $A_{0}, A_{1}, \\ldots, A_{n-1}$. Considering the polygon as inscribed in a circle, the angle between diagonals $\\overline{A_{0} A_{i}}$ and $\\overline{A_{0} A_{j}}$ is $\\frac{1}{2} \\cdot\\left(\\frac{360^{\\circ}}{n}\\right) \\cdot|j-i|=\\left(\\frac{180|j-i|}{n}\\right)^{\\... | [
"60"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,669 | Compute the number of quadratic functions $f(x)=a x^{2}+b x+c$ with integer roots and integer coefficients whose graphs pass through the points $(0,0)$ and $(15,225)$. | [
"Because the graph passes through $(0,0)$, conclude that $c=0$. Then\n\n$$\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 ... | [
"8"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,670 | A bubble in the shape of a hemisphere of radius 1 is on a tabletop. Inside the bubble are five congruent spherical marbles, four of which are sitting on the table and one which rests atop the others. All marbles are tangent to the bubble, and their centers can be connected to form a pyramid with volume $V$ and with a square base. Compute $V$. | [
"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 mut... | [
"$\\frac{1}{54}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,671 | Compute the smallest positive integer base $b$ for which $16_{b}$ is prime and $97_{b}$ is a perfect square. | [
"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... | [
"53"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,672 | For a positive integer $n$, let $C(n)$ equal the number of pairs of consecutive 1's in the binary representation of $n$. For example, $C(183)=C\left(10110111_{2}\right)=3$. Compute $C(1)+C(2)+$ $C(3)+\cdots+C(256)$. | [
"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... | [
"448"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,673 | A set $S$ contains thirteen distinct positive integers whose sum is 120 . Compute the largest possible value for the median of $S$. | [
"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 element... | [
"11"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,674 | Let $T=11$. Compute the least positive integer $b$ such that, when expressed in base $b$, the number $T$ ! ends in exactly two zeroes. | [
"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)... | [
"5"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,675 | Let $T=5$. Suppose that $a_{1}=1$, and that for all positive integers $n, a_{n+1}=$ $\left\lceil\sqrt{a_{n}^{2}+34}\right\rceil$. Compute the least value of $n$ such that $a_{n}>100 T$. | [
"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... | [
"491"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,676 | Compute the smallest $n$ such that in the regular $n$-gon $A_{1} A_{2} A_{3} \cdots A_{n}, \mathrm{~m} \angle A_{1} A_{20} A_{13}<60^{\circ}$. | [
"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... | [
"37"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,677 | Let $T=37$. A cube has edges of length $T$. Square holes of side length 1 are drilled from the center of each face of the cube through the cube's center and across to the opposite face; the edges of each hole are parallel to the edges of the cube. Compute the surface area of the resulting solid. | [
"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-... | [
"8640"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,678 | Let $T=8640$. Compute $\left\lfloor\log _{4}\left(1+2+4+\cdots+2^{T}\right)\right\rfloor$. | [
"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 $\\l... | [
"4320"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,679 | In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum "location, location, location," this Power Question will refer to "houses" and "house numbers" interchangeably.
Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$.
Compute $d(6), d(16)$, and $d(72)$. | [
"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}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,680 | In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum "location, location, location," this Power Question will refer to "houses" and "house numbers" interchangeably.
Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$.
Of the houses with positive numbers less than 100, find, with proof, the house or houses which is (are) closest to City Hall. | [
"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"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,684 | In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum "location, location, location," this Power Question will refer to "houses" and "house numbers" interchangeably.
Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$.
The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$.
Suppose that $n$ is a house with $d(n)=1 / 27$. Determine the ten smallest positive integers $m$ (in the standard ordering of the integers) such that $m \in \mathcal{N}(n)$. | [
"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"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,687 | In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum "location, location, location," this Power Question will refer to "houses" and "house numbers" interchangeably.
Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$.
The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$.
Suppose that $d(17, m)=1 / 81$. Determine the possible values of $d(16, m)$. | [
"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"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,701 | In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum "location, location, location," this Power Question will refer to "houses" and "house numbers" interchangeably.
Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$.
The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$.
Unfortunately for new development, ARMLopolis is full: every nonnegative integer corresponds to (exactly one) house (or City Hall, in the case of 0). However, eighteen families arrive and are looking to move in. After much debate, the connotations of using negative house numbers are deemed unacceptable, and the city decides on an alternative plan. On July 17, Shewad Movers arrive and relocate every family from house $n$ to house $n+18$, for all positive $n$ (so that City Hall does not move). For example, the family in house number 17 moves to house number 35.
Ross takes a walk starting at his house, which is number 34 . He first visits house $n_{1}$, such that $d\left(n_{1}, 34\right)=1 / 3$. He then goes to another house, $n_{2}$, such that $d\left(n_{1}, n_{2}\right)=1 / 3$. Continuing in that way, he visits houses $n_{3}, n_{4}, \ldots$, and each time, $d\left(n_{i}, n_{i+1}\right)=1 / 3$. At the end of the day, what is his maximum possible distance from his original house? Justify your answer. | [
"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... | [
"$1/3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,702 | In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum "location, location, location," this Power Question will refer to "houses" and "house numbers" interchangeably.
Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$.
The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$.
Later, ARMLopolis finally decides on a drastic expansion plan: now house numbers will be rational numbers. To define $d(p / q)$, with $p$ and $q$ integers such that $p q \neq 0$, write $p / q=3^{k} p^{\prime} / q^{\prime}$, where neither $p^{\prime}$ nor $q^{\prime}$ is divisible by 3 and $k$ is an integer (not necessarily positive); then $d(p / q)=3^{-k}$.
Compute $d(3 / 5), d(5 / 8)$, and $d(7 / 18)$. | [
"$\\frac{1}{3}, 1, 9$"
] | [
"$\\frac{1}{3}, 1, 9$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,705 | Let $A R M L$ be a trapezoid with bases $\overline{A R}$ and $\overline{M L}$, such that $M R=R A=A L$ and $L R=$ $A M=M L$. Point $P$ lies inside the trapezoid such that $\angle R M P=12^{\circ}$ and $\angle R A P=6^{\circ}$. Diagonals $A M$ and $R L$ intersect at $D$. Compute the measure, in degrees, of angle $A P D$. | [
"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 ... | [
"48"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,706 | A regular hexagon has side length 1. Compute the average of the areas of the 20 triangles whose vertices are vertices of the hexagon. | [
"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 seco... | [
"$\\frac{9 \\sqrt{3}}{20}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,707 | Paul was planning to buy 20 items from the ARML shop. He wanted some mugs, which cost $\$ 10$ each, and some shirts, which cost $\$ 6$ each. After checking his wallet he decided to put $40 \%$ of the mugs back. Compute the number of dollars he spent on the remaining items. | [
"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}$ dolla... | [
"120"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,708 | Let $x$ be the smallest positive integer such that $1584 \cdot x$ is a perfect cube, and let $y$ be the smallest positive integer such that $x y$ is a multiple of 1584 . Compute $y$. | [
"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... | [
"12"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,709 | Emma goes to the store to buy apples and peaches. She buys five of each, hands the shopkeeper one $\$ 5$ bill, but then has to give the shopkeeper another; she gets back some change. Jonah goes to the same store, buys 2 apples and 12 peaches, and tries to pay with a single $\$ 10$ bill. But that's not enough, so Jonah has to give the shopkeeper another $\$ 10$ bill, and also gets some change. Finally, Helen goes to the same store to buy 25 peaches. Assuming that the price in cents of each fruit is an integer, compute the least amount of money, in cents, that Helen can expect to pay. | [
"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... | [
"1525"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,710 | Circle $O$ has radius 6. Point $P$ lies outside circle $O$, and the shortest distance from $P$ to circle $O$ is 4. Chord $\overline{A B}$ is parallel to $\overleftrightarrow{O P}$, and the distance between $\overline{A B}$ and $\overleftrightarrow{O P}$ is 2 . Compute $P A^{2}+P B^{2}$. | [
"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 ... | [
"272"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,711 | A palindrome is a positive integer, not ending in 0 , that reads the same forwards and backwards. For example, 35253,171,44, and 2 are all palindromes, but 17 and 1210 are not. Compute the least positive integer greater than 2013 that cannot be written as the sum of two palindromes. | [
"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 val... | [
"2019"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,712 | Positive integers $x, y, z$ satisfy $x y+z=160$. Compute the smallest possible value of $x+y z$. | [
"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} \\... | [
"50"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,713 | Compute $\cos ^{3} \frac{2 \pi}{7}+\cos ^{3} \frac{4 \pi}{7}+\cos ^{3} \frac{8 \pi}{7}$. | [
"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} \\... | [
"$-\\frac{1}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,714 | In right triangle $A B C$ with right angle $C$, line $\ell$ is drawn through $C$ and is parallel to $\overline{A B}$. Points $P$ and $Q$ lie on $\overline{A B}$ with $P$ between $A$ and $Q$, and points $R$ and $S$ lie on $\ell$ with $C$ between $R$ and $S$ such that $P Q R S$ is a square. Let $\overline{P S}$ intersect $\overline{A C}$ in $X$, and let $\overline{Q R}$ intersect $\overline{B C}$ in $Y$. The inradius of triangle $A B C$ is 10 , and the area of square $P Q R S$ is 576 . Compute the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$. | [
"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 t... | [
"14"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,715 | Compute the sum of all real numbers $x$ such that
$$
\left\lfloor\frac{x}{2}\right\rfloor-\left\lfloor\frac{x}{3}\right\rfloor=\frac{x}{7}
$$ | [
"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+\\lfloo... | [
"-21"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,717 | Let $S=\{1,2, \ldots, 20\}$, and let $f$ be a function from $S$ to $S$; that is, for all $s \in S, f(s) \in S$. Define the sequence $s_{1}, s_{2}, s_{3}, \ldots$ by setting $s_{n}=\sum_{k=1}^{20} \underbrace{(f \circ \cdots \circ f)}_{n}(k)$. That is, $s_{1}=f(1)+$ $\cdots+f(20), s_{2}=f(f(1))+\cdots+f(f(20)), s_{3}=f(f(f(1)))+f(f(f(2)))+\cdots+f(f(f(20)))$, etc. Compute the smallest integer $p$ such that the following statement is true: The sequence $s_{1}, s_{2}, s_{3}, \ldots$ must be periodic after a certain point, and its period is at most $p$. (If the sequence is never periodic, then write $\infty$ as your answer.) | [
"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 .... | [
"140"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,718 | Compute the smallest positive integer $n$ such that $n^{2}+n^{0}+n^{1}+n^{3}$ is a multiple of 13 . | [
"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"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,720 | Let T be any number you want. Compute $2^{\log _{T} 8}-8^{\log _{T} 2}$. | [
"Let $\\log _{T} 8=x$. Then $T^{x}=8$. Thus the given expression equals $2^{x}-\\left(T^{x}\\right)^{\\log _{T} 2}=2^{x}-T^{x \\log _{T} 2}=$ $2^{x}-T^{\\log _{T} 2^{x}}=2^{x}-2^{x}=\\mathbf{0}$ (independent of $T$ )."
] | [
"0"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,721 | Let $T=0$. At some point during a given week, a law enforcement officer had issued $T+2$ traffic warnings, 20 tickets, and had made $T+5$ arrests. How many more tickets must the officer issue in order for the combined number of tickets and arrests to be 20 times the number of warnings issued that week? | [
"The problem requests the value of $k$ such that $20+k+T+5=20(T+2)$, thus $k=19 T+15$. With $T=0$, it follows that $k=\\mathbf{1 5}$."
] | [
"15"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,722 | Let $T=15$. In parallelogram $A R M L$, points $P$ and $Q$ trisect $\overline{A R}$ and points $W, X, Y, Z$ divide $\overline{M L}$ into fifths (where $W$ is closest to $M$, and points $X$ and $Y$ are both between $W$ and $Z$ ). If $[A R M L]=T$, compute $[P Q W Z]$. | [
"Let $h$ be the distance between $\\overline{A R}$ and $\\overline{M L}$, and for simplicity, let $A R=M L=15 n$. Then $[A R M L]=15 n h$, and $[P Q W Z]=(1 / 2)(P Q+W Z) h$. Note that $P Q=15 n / 3=5 n$ and $W Z=15 n-3 n-3 n=9 n$. Thus $[P Q W Z]=7 n h=(7 / 15) \\cdot[A R M L]=7 T / 15$. With $T=15$, the answer is... | [
"7"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,723 | Let $T=7$. Compute the number of positive perfect cubes that are divisors of $(T+10) !$. | [
"Let $N=T+10$. In order for $k^{3}(k \\in \\mathbb{N})$ to be a divisor of $N$ !, the largest odd prime factor of $k$ (call it $p$ ) must be less than or equal to $N / 3$ so that there are at least three multiples of $p$ among the product of the first $N$ positive integers. If $p=3$, then the smallest possible valu... | [
"36"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,724 | Let $T=36$. The graph of $y=x^{2}+2 x-T$ intersects the $x$-axis at points $A$ and $M$, which are diagonally opposite vertices of square $A R M L$. Compute $[A R M L]$. | [
"Note that the $x$-coordinates of $A$ and $M$ correspond to the two roots $r_{1}, r_{2}$ of $x^{2}+2 x-T$. If $s$ is the side length of square $A R M L$, then $A M=s \\sqrt{2}=\\left|r_{1}-r_{2}\\right|=\\sqrt{\\left(r_{1}-r_{2}\\right)^{2}}=$ $\\sqrt{\\left(r_{1}+r_{2}\\right)^{2}-4 r_{1} r_{2}}=\\sqrt{(-2)^{2}-4(... | [
"74"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,725 | Let $S$ be the set of prime factors of the numbers you receive from positions 7 and 9 , and let $p$ and $q$ be the two least distinct elements of $S$, with $p<q$. Hexagon HEXAGO is inscribed in circle $\omega$, and every angle of $H E X A G O$ is $120^{\circ}$. If $H E=X A=G O=p$ and $E X=A G=O H=q$, compute the area of circle $\omega$. | [
"The given information implies that triangles $H E X, X A G$, and $G O H$ are congruent, hence triangle $H X G$ is equilateral. If $H X=s$, then the radius of the circle circumscribing $\\triangle H X G$ is $s / \\sqrt{3}$ so that the circle's area is $\\pi s^{2} / 3$. It remains to compute $s$. With $\\mathrm{m} \... | [
"$\\frac{67 \\pi}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,726 | Let $T=184$. A group of $n$ friends goes camping; two of them are selected to set up the campsite when they arrive and two others are selected to take down the campsite the next day. Compute the smallest possible value of $n$ such that there are at least $T$ ways of selecting the four helpers. | [
"There are $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)$ ways of choosing the two people to set up and $\\left(\\begin{array}{c}n-2 \\\\ 2\\end{array}\\right)$ ways of choosing the two people to take down the campsite, so there are $\\frac{n(n-1)}{2} \\cdot \\frac{(n-2)(n-3)}{2}$ ways of choosing the four ... | [
"7"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,727 | Let $T=8$. The parabola $y=x^{2}+T x$ is tangent to the parabola $y=-(x-2 T)^{2}+b$. Compute $b$. | [
"In this case, the two parabolas are tangent exactly when the system of equations has a unique solution. (Query: Is this the case for every pair of equations representing parabolas?) So set the right sides equal to each other: $x^{2}+T x=-(x-2 T)^{2}+b$. Then $x^{2}+T x=$ $-x^{2}+4 T x-4 T^{2}+b$, or equivalently, ... | [
"184"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,728 | Let $T=19$. The first two terms of a sequence are $a_{1}=3 / 5$ and $a_{2}=4 / 5$. For $n>2$, if $n$ is odd, then $a_{n}=a_{n-1}^{2}-a_{n-2}^{2}$, while if $n$ is even, then $a_{n}=2 a_{n-2} a_{n-3}$. Compute the sum of the squares of the first $T-3$ terms of the sequence. | [
"Using the identity $\\left(x^{2}-y^{2}\\right)^{2}+(2 x y)^{2}=\\left(x^{2}+y^{2}\\right)^{2}$, notice that $a_{2 n+1}^{2}+a_{2 n+2}^{2}=\\left(a_{2 n}^{2}-a_{2 n-1}^{2}\\right)^{2}+$ $\\left(2 a_{2 n} a_{2 n-1}\\right)^{2}=\\left(a_{2 n}^{2}+a_{2 n-1}^{2}\\right)^{2}$. So surprisingly, for all $n \\in \\mathbb{N}... | [
"8"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,729 | Let $T=17$. A regular $n$-gon has exactly $T$ more diagonals than a regular $(n-1)$-gon. Compute the value of $n$. | [
"Using the formula $D(n)=\\frac{n(n-3)}{2}$ twice yields $D(n)-D(n-1)=\\frac{n^{2}-3 n}{2}-\\frac{n^{2}-5 n+4}{2}=\\frac{2 n-4}{2}=n-2$. So $T=n-2$, thus $n=T+2$, and with $T=17, n=19$."
] | [
"19"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,730 | Let $T=27/2$. The sequence $a_{1}, a_{2}, a_{3}, \ldots$, is arithmetic with $a_{16}=13$ and $a_{30}=20$. Compute the value of $k$ for which $a_{k}=T$. | [
"If $d$ is the common difference of the sequence, then the $n^{\\text {th }}$ term of the sequence is $a_{n}=$ $a_{16}+d(n-16)$. The values $a_{16}=13$ and $a_{30}=20$ yield $d=(20-13) /(30-16)=1 / 2$, hence $a_{n}=13+(1 / 2)(n-16)$. If $a_{n}=T$, then $n=2(T-13)+16=2 T-10$. With $T=27 / 2$, it follows that $n=\\ma... | [
"17"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,731 | Let $T=114$. A rectangular prism has a length of 1 , a width of 3 , a height of $h$, and has a total surface area of $T$. Compute the value of $h$. | [
"The surface area is given by the expression $2 \\cdot 1 \\cdot 3+2 \\cdot 1 \\cdot h+2 \\cdot 3 \\cdot h=6+8 h$. Because $6+8 h=T, h=\\frac{T-6}{8}$. With $T=114, h=108 / 8=\\mathbf{2 7} / \\mathbf{2}$."
] | [
"$\\frac{27}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,732 | The zeros of $x^{2}+b x+93$ are $r$ and $s$. If the zeros of $x^{2}-22 x+c$ are $r+1$ and $s+1$, compute $c$. | [
"Use sums and products of roots formulas: the desired quantity $c=(r+1)(s+1)=r s+r+s+1$. From the first equation, $r s=93$, while from the second equation, $(r+1)+(s+1)=r+s+2=$ 22. So $r s+r+s+1=93+22-1=\\mathbf{1 1 4}$."
] | [
"114"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,733 | Let $N=888,888 \times 9,999,999$. Compute the sum of the digits of $N$. | [
"Write $N$ as\n\n$$\n\\begin{aligned}\n& (10,000,000-1) \\cdot 888,888 \\\\\n= & 8,888,880,000,000-888,888 \\\\\n= & 8,888,879,111,112 .\n\\end{aligned}\n$$\n\nThe sum of the digits of $N$ is 63 ."
] | [
"63"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,734 | Five equilateral triangles are drawn in the plane so that no two sides of any of the triangles are parallel. Compute the maximum number of points of intersection among all five triangles. | [
"Any two of the triangles intersect in at most six points, because each side of one triangle can intersect the other triangle in at most two points. To count the total number of intersections among the five triangles, note that there are $\\left(\\begin{array}{l}5 \\\\ 2\\end{array}\\right)=10$ ways to select a pai... | [
"60"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,735 | $\quad$ Let $S$ be the set of four-digit positive integers for which the sum of the squares of their digits is 17 . For example, $2023 \in S$ because $2^{2}+0^{2}+2^{2}+3^{2}=17$. Compute the median of $S$. | [
"In order for the sums of the squares of four digits to be 17 , the digits must be either $0,2,2$, and 3 , or $0,0,1$, and 4 , in some order. If the leading digit is 2 , there are $3 !=6$ possible four-digit numbers. If the leading digit is 1,3 , or 4 , there are $\\frac{3 !}{2 !}=3$ possible four-digit numbers. In... | [
"2302"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,736 | Let $E U C L I D$ be a hexagon inscribed in a circle of radius 5 . Given that $E U=U C=L I=I D=6$, and $C L=D E$, compute $C L$. | [
"Let $C L=x$. Because the quadrilaterals $E U C L$ and $L I D E$ are congruent, $\\overline{E L}$ is a diameter of the circle in which the hexagon is inscribed, so $E L=10$. Furthermore, because $\\overline{E L}$ is a diameter of the circle, it follows that the inscribed $\\angle E U L$ is a right angle, hence $U L... | [
"$\\frac{14}{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,737 | The ARMLLexicon consists of 10 letters: $\{A, R, M, L, e, x, i, c, o, n\}$. A palindrome is an ordered list of letters that read the same backwards and forwards; for example, MALAM, n, oncecno, and MoM are palindromes. Compute the number of 15-letter palindromes that can be spelled using letters in the ARMLLexicon, among which there are four consecutive letters that spell out $A R M L$. | [
"Any 15-letter palindrome is determined completely by its first 8 letters, because the last 7 letters must be the first 7 in reverse. Such a palindrome contains the string $A R M L$ if and only if its first 8 letters contain either $A R M L$ or $L M R A$. (The string $A R M L$ cannot cross the middle of the palindr... | [
"99956"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,738 | Let $10^{y}$ be the product of all real numbers $x$ such that $\log x=\frac{3+\left\lfloor(\log x)^{2}\right\rfloor}{4}$. Compute $y$. | [
"First, note that\n\n$$\n\\left\\lfloor(\\log x)^{2}\\right\\rfloor \\leq(\\log x)^{2} \\Longrightarrow \\frac{3+\\left\\lfloor(\\log x)^{2}\\right\\rfloor}{4} \\leq \\frac{3+(\\log x)^{2}}{4}\n$$\n\nTherefore\n\n$$\n\\log x \\leq \\frac{(\\log x)^{2}+3}{4} \\Longrightarrow 0 \\leq(\\log x)^{2}-4 \\log x+3=(\\log x... | [
"8"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,739 | The solutions to the equation $x^{2}-180 x+8=0$ are $r_{1}$ and $r_{2}$. Compute
$$
\frac{r_{1}}{\sqrt[3]{r_{2}}}+\frac{r_{2}}{\sqrt[3]{r_{1}}}
$$ | [
"First note that the solutions of the given equation are real because the equation's discriminant is positive. By Vieta's Formulas, $r_{1}+r_{2}=180(*)$ and $r_{1} r_{2}=8(* *)$. The expression to be computed can be written with a common denominator as\n\n$$\n\\frac{\\sqrt[3]{r_{1}^{4}}+\\sqrt[3]{r_{2}^{4}}}{\\sqrt... | [
"508"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,740 | Circle $\omega$ is tangent to parallel lines $\ell_{1}$ and $\ell_{2}$ at $A$ and $B$ respectively. Circle $\omega_{1}$ is tangent to $\ell_{1}$ at $C$ and to $\omega$ externally at $P$. Circle $\omega_{2}$ is tangent to $\ell_{2}$ at $D$ and to $\omega$ externally at $Q$. Circles $\omega_{1}$ and $\omega_{2}$ are also externally tangent to each other. Given that $A Q=12$ and $D Q=8$, compute $C D$. | [
"Let $O, O_{1}$ and $O_{2}$ be the centers, and let $r, r_{1}$ and $r_{2}$ be the radii of the circles $\\omega, \\omega_{1}$, and $\\omega_{2}$, respectively. Let $R$ be the point of tangency between $\\omega_{1}$ and $\\omega_{2}$.\n\nLet $H_{1}$ and $H_{2}$ be the projections of $O_{1}$ and $O_{2}$ onto $\\overl... | [
"$5 \\sqrt{10}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,741 | Given quadrilateral $A R M L$ with $A R=20, R M=23, M L=25$, and $A M=32$, compute the number of different integers that could be the perimeter of $A R M L$. | [
"Notice that $\\triangle A R M$ is fixed, so the number of integers that could be the perimeter of $A R M L$ is the same as the number of integers that could be the length $A L$ in $\\triangle A L M$. By the Triangle Inequality, $32-25<A L<32+25$, so $A L$ is at least 8 and no greater than 56 . The number of possib... | [
"49"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,742 | Let $\mathcal{S}$ denote the set of all real polynomials $A(x)$ with leading coefficient 1 such that there exists a real polynomial $B(x)$ that satisfies
$$
\frac{1}{A(x)}+\frac{1}{B(x)}+\frac{1}{x+10}=\frac{1}{x}
$$
for all real numbers $x$ for which $A(x) \neq 0, B(x) \neq 0$, and $x \neq-10,0$. Compute $\sum_{A \in \mathcal{S}} A(10)$. | [
"For brevity, $P$ will be used to represent the polynomial $P(x)$, and let $\\operatorname{deg}(P)$ represent the degree of $P$. Rewrite the given condition as follows:\n\n$$\n\\begin{aligned}\n\\frac{1}{A(x)}+\\frac{1}{B(x)}+\\frac{1}{x+10}=\\frac{1}{x} & \\Longrightarrow \\frac{A+B}{A B}=\\frac{10}{x(x+10)} \\\\\... | [
"46760"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,744 | Let $T=688$. Let $a$ be the least nonzero digit in $T$, and let $b$ be the greatest digit in $T$. In square $N O R M, N O=b$, and points $P_{1}$ and $P_{2}$ lie on $\overline{N O}$ and $\overline{O R}$, respectively, so that $O P_{1}=O P_{2}=a$. A circle centered at $O$ has radius $a$, and quarter-circular arc $\widehat{P_{1} P_{2}}$ is drawn. There is a circle that is tangent to $\widehat{P_{1} P_{2}}$ and to sides $\overline{M N}$ and $\overline{M R}$. The radius of this circle can be written in the form $x-y \sqrt{2}$, where $x$ and $y$ are positive integers. Compute $x+y$. | [
"Let $r$ and $Q$ denote the respective radius and center of the circle whose radius is concerned. Let this circle be tangent to arc $\\widehat{P_{1} P_{2}}$ at point $P$, and let it be tangent to sides $\\overline{M N}$ and $\\overline{M R}$ at points $T_{1}$ and $T_{2}$, respectively.\n\n<img_3571>\n\nNote that $Q... | [
"36"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,745 | Let $T=36$. Square $A B C D$ has area $T$. Points $M, N, O$, and $P$ lie on $\overline{A B}$, $\overline{B C}, \overline{C D}$, and $\overline{D A}$, respectively, so that quadrilateral $M N O P$ is a rectangle with $M P=2$. Compute $M N$. | [
"Let $A M=a$ and $A P=b$, and let $s=\\sqrt{T}$ be the side length of square $A B C D$. Then $M B=s-a$ and $D P=s-b$. Using the right angles of $M N O P$ and complementary acute angles in triangles $A M P, B N M$, $C O N$, and $D P O$, note that\n\n$$\n\\angle A M P \\cong \\angle B N M \\cong \\angle C O N \\cong ... | [
"$6 \\sqrt{2}-2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,746 | In a game, a player chooses 2 of the 13 letters from the first half of the alphabet (i.e., A-M) and 2 of the 13 letters from the second half of the alphabet (i.e., N-Z). Aditya plays the game, and then Ayesha plays the game. Compute the probability that Aditya and Ayesha choose the same set of four letters. | [
"The number of ways to choose 2 distinct letters out of 13 is $\\frac{13 \\cdot 12}{2}=78$. The probability of matching on both halves is therefore $\\frac{1}{78^{2}}=\\frac{1}{6084}$."
] | [
"$\\frac{1}{6084}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,747 | Let $T=\frac{1}{6084}$. Compute the least positive integer $n$ such that when a fair coin is flipped $n$ times, the probability of it landing heads on all $n$ flips is less than $T$. | [
"The problem is equivalent to finding the least integer $n$ such that $\\frac{1}{2^{n}}<T$, or $2^{n}>\\frac{1}{T}=6084$. Because $2^{12}=4096$ and $2^{13}=8192$, the answer is $\\mathbf{1 3}$."
] | [
"13"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,748 | Let $T=13$. Compute the least integer $n>2023$ such that the equation $x^{2}-T x-n=0$ has integer solutions. | [
"The discriminant of the quadratic, $T^{2}+4 n$, must be a perfect square. Because $T$ and the discriminant have the same parity, and the leading coefficient of the quadratic is 1 , by the quadratic formula, the discriminant being a perfect square is sufficient to guarantee integer solutions. Before knowing $T$, no... | [
"2028"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,753 | In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other.
Find the maximum and minimum possible number of cromulent elements in a sequence of $n$ consecutive positive integers with $n=6$; | [
"First we prove that every sequence of five consecutive positive integers contains a cromulent element.\n\nProof: Consider a sequence of five consecutive integers. Exactly one number in such a sequence will be a multiple of 5 , but that number could also be a multiple of 2 and hence share a common factor with at le... | [
"1,2"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,754 | In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other.
Find the maximum and minimum possible number of cromulent elements in a sequence of $n$ consecutive positive integers with $n=7$. | [
"The minimum number is 1 and the maximum number is 3 . One example of a sequence of length 7 with one cromulent element is $4,5,6,7,8,9,10$, where 7 is the cromulent element. To show that it is not possible for such a sequence to have zero cromulent elements, consider two cases. If the sequence begins with an even ... | [
"1,3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,759 | For an integer $n \geq 4$, define $a_{n}$ to be the product of all real numbers that are roots to at least one quadratic polynomial whose coefficients are positive integers that sum to $n$. Compute
$$
\frac{a_{4}}{a_{5}}+\frac{a_{5}}{a_{6}}+\frac{a_{6}}{a_{7}}+\cdots+\frac{a_{2022}}{a_{2023}} .
$$ | [
"For an integer $n \\geq 4$, let $S_{n}$ denote the set of real numbers $x$ that are roots to at least one quadratic polynomial whose coefficients are positive integers that sum to $n$. (Note that $S_{n}$ is nonempty, as the polynomial $x^{2}+(n-2) x+1$ has a discriminant of $(n-2)^{2}-4$, which is nonnegative for ... | [
"-2019"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,760 | Suppose that $u$ and $v$ are distinct numbers chosen at random from the set $\{1,2,3, \ldots, 30\}$. Compute the probability that the roots of the polynomial $(x+u)(x+v)+4$ are integers. | [
"Assume without loss of generality that $u>v$. The condition that $(x+u)(x+v)+4$ has integer roots is equivalent to the discriminant $(u+v)^{2}-4(u v+4)=(u-v)^{2}-16$ being a perfect square. This is possible if and only if $u-v=4$ or $u-v=5$. There are $(30-4)+(30-5)=26+25=51$ such ordered pairs $(u, v)$, so the an... | [
"$\\frac{17}{145}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,761 | The degree-measures of the interior angles of convex hexagon TIEBRK are all integers in arithmetic progression. Compute the least possible degree-measure of the smallest interior angle in hexagon TIEBRK. | [
"The sum of the measures of the interior angles of a convex hexagon is $(6-2)\\left(180^{\\circ}\\right)=720^{\\circ}$. Let the measures of the angles be $a, a+d, \\ldots, a+5 d$. This implies that $6 a+15 d=720 \\rightarrow 2 a+5 d=240 \\rightarrow 5 d=240-2 a$. Note that $a+5 d<180 \\rightarrow 240-a<180 \\righta... | [
"65"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,762 | A six-digit natural number is "sort-of-decreasing" if its first three digits are in strictly decreasing order and its last three digits are in strictly decreasing order. For example, 821950 and 631631 are sort-of-decreasing but 853791 and 911411 are not. Compute the number of sort-of-decreasing six-digit natural numbers. | [
"If three distinct digits are chosen from the set of digits $\\{0,1,2, \\ldots, 9\\}$, then there is exactly one way to arrange them in decreasing order. There are $\\left(\\begin{array}{c}10 \\\\ 3\\end{array}\\right)=120$ ways to choose the first three digits and 120 ways to choose the last three digits. Thus the... | [
"14400"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,763 | For each positive integer $N$, let $P(N)$ denote the product of the digits of $N$. For example, $P(8)=8$, $P(451)=20$, and $P(2023)=0$. Compute the least positive integer $n$ such that $P(n+23)=P(n)+23$. | [
"One can verify that no single-digit positive integer $n$ satisfies the conditions of the problem.\n\nIf $n$ has two digits, then $n+23$ cannot be a three-digit number; this can be verified by checking the numbers $n \\geq 88$, because if $n<88$, then one of the digits of $n+23$ is 0 . Therefore both $n$ and $n+23$... | [
"34"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,764 | Compute the least integer value of the function
$$
f(x)=\frac{x^{4}-6 x^{3}+2 x^{2}-6 x+2}{x^{2}+1}
$$
whose domain is the set of all real numbers. | [
"$\\quad$ Use polynomial long division to rewrite $f(x)$ as\n\n$$\nf(x)=x^{2}-6 x+1+\\frac{1}{x^{2}+1}\n$$\n\nThe quadratic function $x^{2}-6 x+1=(x-3)^{2}-8$ has a minimum of -8 , achieved at $x=3$. The \"remainder term\" $\\frac{1}{x^{2}+1}$ is always positive. Thus $f(x)>-8$ for all $x$, so any integer value of ... | [
"-7"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,765 | Suppose that noncongruent triangles $A B C$ and $X Y Z$ are given such that $A B=X Y=10, B C=$ $Y Z=9$, and $\mathrm{m} \angle C A B=\mathrm{m} \angle Z X Y=30^{\circ}$. Compute $[A B C]+[X Y Z]$. | [
"Because triangles $A B C$ and $X Y Z$ are noncongruent yet have two adjacent sides and an angle in common, the two triangles are the two possibilities in the ambiguous case of the Law of Sines. Without loss of generality, let triangle $A B C$ have obtuse angle $C$ and triangle $X Y Z$ have acute angle $Z$ so that ... | [
"$25 \\sqrt{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,766 | The mean, median, and unique mode of a list of positive integers are three consecutive integers in some order. Compute the least possible sum of the integers in the original list. | [
"One possible list is $1,1,3,7$, which has mode 1 , median 2 , and mean 3 . The sum is $1+1+3+7=12$. A list with fewer than four numbers cannot produce a median and unique mode that are distinct from each other. To see this, first note that a list with one number has the same median and mode. In a list with two num... | [
"12"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,767 | David builds a circular table; he then carves one or more positive integers into the table at points equally spaced around its circumference. He considers two tables to be the same if one can be rotated so that it has the same numbers in the same positions as the other. For example, a table with the numbers $8,4,5$ (in clockwise order) is considered the same as a table with the numbers 4, 5,8 (in clockwise order), but both tables are different from a table with the numbers 8, 5, 4 (in clockwise order). Given that the numbers he carves sum to 17 , compute the number of different tables he can make. | [
"The problem calls for the number of ordered partitions of 17 , where two partitions are considered the same if they are cyclic permutations of each other. Because 17 is prime, each ordered partition of 17 into $n$ parts will be a cyclic permutation of exactly $n$ such partitions (including itself), unless $n=17$. ... | [
"7711"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,768 | In quadrilateral $A B C D, \mathrm{~m} \angle B+\mathrm{m} \angle D=270^{\circ}$. The circumcircle of $\triangle A B D$ intersects $\overline{C D}$ at point $E$, distinct from $D$. Given that $B C=4, C E=5$, and $D E=7$, compute the diameter of the circumcircle of $\triangle A B D$. | [
"Note that $\\mathrm{m} \\angle A+\\mathrm{m} \\angle C=90^{\\circ}$ in quadrilateral $A B C D$. Because quadrilateral $A B E D$ is cyclic, it follows that $\\mathrm{m} \\angle A D E+\\mathrm{m} \\angle A B E=180^{\\circ}$. Moreover, because $\\mathrm{m} \\angle A B E+\\mathrm{m} \\angle E B C+\\mathrm{m} \\angle A... | [
"$\\sqrt{130}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,770 | Let $i=\sqrt{-1}$. The complex number $z=-142+333 \sqrt{5} i$ can be expressed as a product of two complex numbers in multiple different ways, two of which are $(57-8 \sqrt{5} i)(-6+5 \sqrt{5} i)$ and $(24+\sqrt{5} i)(-3+14 \sqrt{5} i)$. Given that $z=-142+333 \sqrt{5} i$ can be written as $(a+b \sqrt{5} i)(c+d \sqrt{5} i)$, where $a, b, c$, and $d$ are positive integers, compute the lesser of $a+b$ and $c+d$. | [
"Multiply each of the given parenthesized expressions by its complex conjugate to obtain\n\n$$\n\\begin{aligned}\n142^{2}+5 \\cdot 333^{2} & =\\left(57^{2}+5 \\cdot 8^{2}\\right)\\left(6^{2}+5 \\cdot 5^{2}\\right) \\\\\n& =\\left(24^{2}+5 \\cdot 1^{2}\\right)\\left(3^{2}+5 \\cdot 14^{2}\\right) \\\\\n& =\\left(a^{2... | [
"17"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,771 | Parallelogram $A B C D$ is rotated about $A$ in the plane, resulting in $A B^{\prime} C^{\prime} D^{\prime}$, with $D$ on $\overline{A B^{\prime}}$. Suppose that $\left[B^{\prime} C D\right]=\left[A B D^{\prime}\right]=\left[B C C^{\prime}\right]$. Compute $\tan \angle A B D$. | [
"Editor's Note: It was pointed out that the conditions of the problem determine two possible values of $\\tan \\angle A B D$ : one based on $\\mathrm{m} \\angle A<90^{\\circ}$, and the other based on $\\mathrm{m} \\angle A>90^{\\circ}$. A complete solution is provided below. We thank Matthew Babbitt and Silas Johns... | [
"$\\sqrt{2}-1$,$\\frac{3-\\sqrt{2}}{7}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Geometry | Math | English |
2,772 | Compute the least integer greater than 2023 , the sum of whose digits is 17 . | [
"A candidate for desired number is $\\underline{2} \\underline{0} \\underline{X} \\underline{Y}$, where $X$ and $Y$ are digits and $X+Y=15$. To minimize this number, take $Y=9$. Then $X=6$, and the desired number is 2069 ."
] | [
"2069"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,773 | Let $T$ = 2069, and let $K$ be the sum of the digits of $T$. Let $r$ and $s$ be the two roots of the polynomial $x^{2}-18 x+K$. Compute $|r-s|$. | [
"Note that $|r-s|=\\sqrt{r^{2}-2 r s+s^{2}}=\\sqrt{(r+s)^{2}-4 r s}$. By Vieta's Formulas, $r+s=-(-18)$ and $r s=K$, so $|r-s|=\\sqrt{18^{2}-4 K}$. With $T=2069, K=17$, and the answer is $\\sqrt{324-68}=\\sqrt{256}=16$."
] | [
"16"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,775 | Let $T=$ 7, and let $K=9 T$. Let $A_{1}=2$, and for $n \geq 2$, let
$$
A_{n}= \begin{cases}A_{n-1}+1 & \text { if } n \text { is not a perfect square } \\ \sqrt{n} & \text { if } n \text { is a perfect square. }\end{cases}
$$
Compute $A_{K}$. | [
"Let $\\lfloor\\sqrt{n}\\rfloor=x$. Then $n$ can be written as $x^{2}+y$, where $y$ is an integer such that $0 \\leq y<2 x+1$. Let $m$ be the greatest perfect square less than or equal to $9 T$. Then the definition of the sequence and the previous observation imply that $A_{K}=A_{9 T}=\\sqrt{m}+(9 T-m)=\\lfloor\\sq... | [
"21"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,776 | Let $T=$ 21. The number $20^{T} \cdot 23^{T}$ has $K$ positive divisors. Compute the greatest prime factor of $K$. | [
"Write $20^{T} \\cdot 23^{T}$ as $2^{2 T} \\cdot 5^{T} \\cdot 23^{T}$. This number has $K=(2 T+1)(T+1)^{2}$ positive divisors. With $T=21, K=43 \\cdot 22^{2}$. The greatest prime factor of $K$ is $\\mathbf{4 3}$."
] | [
"43"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,777 | Let $T=43$. Compute the positive integer $n \neq 17$ for which $\left(\begin{array}{c}T-3 \\ 17\end{array}\right)=\left(\begin{array}{c}T-3 \\ n\end{array}\right)$. | [
"Using the symmetry property of binomial coefficients, the desired value of $n$ is $T-3-17=T-20$. With $T=43$, the answer is $\\mathbf{2 3}$."
] | [
"23"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,778 | Let $T=23$ . Compute the units digit of $T^{2023}+T^{20}-T^{23}$. | [
"Assuming that $T$ is a positive integer, because units digits of powers of $T$ cycle in groups of at most 4, the numbers $T^{2023}$ and $T^{23}$ have the same units digit, hence the number $T^{2023}-T^{23}$ has a units digit of 0 , and the answer is thus the units digit of $T^{20}$. With $T=23$, the units digit of... | [
"1"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,780 | Let $T=$ 3. Suppose that $T$ fair coins are flipped. Compute the probability that at least one tails is flipped. | [
"The probability of flipping all heads is $\\left(\\frac{1}{2}\\right)^{T}$, so the probability of flipping at least one tails is $1-\\frac{1}{2^{T}}$. With $T=3$, the desired probability is $1-\\frac{1}{8}=\\frac{7}{8}$."
] | [
"$\\frac{7}{8}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,781 | Let $T=$ $\frac{7}{8}$. The number $T$ can be expressed as a reduced fraction $\frac{m}{n}$, where $m$ and $n$ are positive integers whose greatest common divisor is 1 . The equation $x^{2}+(m+n) x+m n=0$ has two distinct real solutions. Compute the lesser of these two solutions. | [
"The left-hand side of the given equation can be factored as $(x+m)(x+n)$. The two solutions are therefore $-m$ and $-n$, so the answer is $\\min \\{-m,-n\\}$. With $T=\\frac{7}{8}, m=7, n=8$, and $\\min \\{-7,-8\\}$ is $\\mathbf{- 8}$."
] | [
"-8"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,782 | Let $T=$ -8, and let $i=\sqrt{-1}$. Compute the positive integer $k$ for which $(-1+i)^{k}=\frac{1}{2^{T}}$. | [
"Note that $(-1+i)^{2}=1+2 i-1=2 i$. Thus $(-1+i)^{4}=(2 i)^{2}=-4$, and $(-1+i)^{8}=(-4)^{2}=16$. The expression $\\frac{1}{2^{T}}$ is a power of 16 if $T$ is a negative multiple of 4 . With $T=-8, \\frac{1}{2^{-8}}=2^{8}=16^{2}=\\left((-1+i)^{8}\\right)^{2}=$ $(-1+i)^{16}$, so the desired value of $k$ is $\\mathb... | [
"16"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
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