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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
2,783 | Let $T=$ 16. Compute the value of $x$ that satisfies $\log _{4} T=\log _{2} x$. | [
"By the change of base rule and a property of $\\operatorname{logs}, \\log _{4} T=\\frac{\\log _{2} T}{\\log _{2} 4}=\\frac{\\log _{2} T}{2}=\\log _{2} \\sqrt{T}$. Thus $x=\\sqrt{T}$, and with $T=16, x=4$."
] | [
"4"
] | 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 |
2,784 | Let $T=$ 4. Pyramid $L E O J S$ is a right square pyramid with base $E O J S$, whose area is $T$. Given that $L E=5 \sqrt{2}$, compute $[L E O]$. | [
"Let the side length of square base $E O J S$ be $2 x$, and let $M$ be the midpoint of $\\overline{E O}$. Then $\\overline{L M} \\perp \\overline{E O}$, and $L M=\\sqrt{(5 \\sqrt{2})^{2}-x^{2}}$ by the Pythagorean Theorem. Thus $[L E O]=\\frac{1}{2} \\cdot 2 x \\sqrt{(5 \\sqrt{2})^{2}-x^{2}}=$\n\n\n\n$x \\sqrt{(5 \... | [
"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 |
2,785 | Let $T=$ 7. Compute the units digit of $T^{2023}+(T-2)^{20}-(T+10)^{23}$. | [
"Note that $T$ and $T+10$ have the same units digit. Because units digits of powers of $T$ cycle in groups of at most 4 , the numbers $T^{2023}$ and $(T+10)^{23}$ have the same units digit, hence the number $T^{2023}-(T+10)^{23}$ has a units digit of 0 , and the answer is thus the units digit of $(T-2)^{20}$. With ... | [
"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 |
2,786 | Let $r=1$ and $R=5$. A circle with radius $r$ is centered at $A$, and a circle with radius $R$ is centered at $B$. The two circles are internally tangent. Point $P$ lies on the smaller circle so that $\overline{B P}$ is tangent to the smaller circle. Compute $B P$. | [
"Draw radius $A P$ and note that $A P B$ is a right triangle with $\\mathrm{m} \\angle A P B=90^{\\circ}$. Note that $A B=R-r$ and $A P=r$, so by the Pythagorean Theorem, $B P=\\sqrt{(R-r)^{2}-r^{2}}=\\sqrt{R^{2}-2 R r}$. With $r=1$ and $R=5$, it follows that $B P=\\sqrt{\\mathbf{1 5}}$."
] | [
"$\\sqrt{15}$"
] | 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 | Geometry | Math | English |
2,787 | Compute the largest prime divisor of $15 !-13$ !. | [
"Factor 15 ! -13 ! to obtain $13 !(15 \\cdot 14-1)=13$ ! $\\cdot 209$. The largest prime divisor of 13 ! is 13 , so continue by factoring $209=11 \\cdot 19$. Thus the largest prime divisor of 15 ! - 13 ! is 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 | Number Theory | Math | English |
2,788 | Three non-overlapping squares of positive integer side lengths each have one vertex at the origin and sides parallel to the coordinate axes. Together, the three squares enclose a region whose area is 41 . Compute the largest possible perimeter of the region. | [
"Proceed in two steps: first, determine the possible sets of side lengths for the squares; then determine which arrangement of squares produces the largest perimeter. Let the side lengths of the squares be positive integers $m \\geq n \\geq p$. Then $m^{2}+n^{2}+p^{2}=41$, so $m \\leq 6$, and because $3^{2}+3^{2}+3... | [
"32"
] | 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 |
2,789 | A circle with center $O$ and radius 1 contains chord $\overline{A B}$ of length 1 , and point $M$ is the midpoint of $\overline{A B}$. If the perpendicular to $\overline{A O}$ through $M$ intersects $\overline{A O}$ at $P$, compute $[M A P]$. | [
"Draw auxiliary segment $\\overline{O B}$, as shown in the diagram below.\n\n<img_4031>\n\nTriangle $O A B$ is equilateral, so $\\mathrm{m} \\angle O A B=60^{\\circ}$. Then $\\triangle M A P$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle with hypotenuse $A M=1 / 2$. Thus $A P=1 / 4$ and $M P=\\sqrt{3} / 4$, s... | [
"$\\frac{\\sqrt{3}}{32}$"
] | 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 |
2,790 | $\quad$ Suppose that $p$ and $q$ are two-digit prime numbers such that $p^{2}-q^{2}=2 p+6 q+8$. Compute the largest possible value of $p+q$. | [
"Subtract from both sides and regroup to obtain $p^{2}-2 p-\\left(q^{2}+6 q\\right)=8$. Completing both squares yields $(p-1)^{2}-(q+3)^{2}=0$. The left side is a difference of two squares; factor to obtain $((p-1)+(q+3))((p-1)-(q+3))=0$, whence $(p+q+2)(p-q-4)=0$. For positive primes $p$ and $q$, the first factor ... | [
"162"
] | 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 |
2,791 | The four zeros of the polynomial $x^{4}+j x^{2}+k x+225$ are distinct real numbers in arithmetic progression. Compute the value of $j$. | [
"Let the four zeros be $p \\leq q \\leq r \\leq s$. The coefficient of $x^{3}$ is 0 , so $p+q+r+s=0$. The mean of four numbers in arithmetic progression is the mean of the middle two numbers, so $q=-r$. Then the common difference is $r-q=r-(-r)=2 r$, so $s=r+2 r=3 r$ and $p=q-2 r=-3 r$. Therefore the four zeros are... | [
"-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 |
2,792 | Compute the smallest positive integer $n$ such that
$$
n,\lfloor\sqrt{n}\rfloor,\lfloor\sqrt[3]{n}\rfloor,\lfloor\sqrt[4]{n}\rfloor,\lfloor\sqrt[5]{n}\rfloor,\lfloor\sqrt[6]{n}\rfloor,\lfloor\sqrt[7]{n}\rfloor, \text { and }\lfloor\sqrt[8]{n}\rfloor
$$
are distinct. | [
"Inverting the problem, the goal is to find seven positive integers $a<b<c<d<e<f<g$ and a positive integer $n$ such that $a^{8}, b^{7}, c^{6}, \\ldots, g^{2} \\leq n$ and $n<(a+1)^{8},(b+1)^{7}, \\ldots,(g+1)^{2}$. Proceed by cases starting with small values of $a$.\n\nIf $a=1$, then because $n<(a+1)^{8}, n<256$. B... | [
"4096"
] | 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 |
2,793 | If $n$ is a positive integer, then $n$ !! is defined to be $n(n-2)(n-4) \cdots 2$ if $n$ is even and $n(n-2)(n-4) \cdots 1$ if $n$ is odd. For example, $8 ! !=8 \cdot 6 \cdot 4 \cdot 2=384$ and $9 ! !=9 \cdot 7 \cdot 5 \cdot 3 \cdot 1=945$. Compute the number of positive integers $n$ such that $n !$ ! divides 2012!!. | [
"If $n$ is even and $n \\leq 2012$, then $n$ !! $\\mid 2012$ !! trivially, while if $n>2012,2012$ !! $<n$ !!, so $n$ !! cannot divide 2012!!. Thus there are a total of 1006 even values of $n$ such that $n$ !! | 2012!!. If $n$ is odd and $n<1006$, then $n$ !! | 2012!!. To show this, rearrange the terms of 2012!! and... | [
"1510"
] | 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 |
2,794 | On the complex plane, the parallelogram formed by the points $0, z, \frac{1}{z}$, and $z+\frac{1}{z}$ has area $\frac{35}{37}$, and the real part of $z$ is positive. If $d$ is the smallest possible value of $\left|z+\frac{1}{z}\right|$, compute $d^{2}$. | [
"As is usual, let $\\arg z$ refer to measure of the directed angle whose vertex is the origin, whose initial ray passes through 1 (i.e., the point $(1,0)$ ), and whose terminal ray passes through $z$. Then $\\arg 1 / z=-\\arg z$. Using the formula $a b \\sin \\gamma$ for the area of the parallelogram with sides $a$... | [
"$\\frac{50}{37}$"
] | 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 |
2,795 | One face of a $2 \times 2 \times 2$ cube is painted (not the entire cube), and the cube is cut into eight $1 \times 1 \times 1$ cubes. The small cubes are reassembled randomly into a $2 \times 2 \times 2$ cube. Compute the probability that no paint is showing. | [
"Call each $1 \\times 1 \\times 1$ cube a cubelet. Then four cubelets are each painted on one face, and the other four cubelets are completely unpainted and can be ignored. For each painted cubelet, the painted face can occur in six positions, of which three are hidden from the outside, so the probability that a pa... | [
"$\\frac{1}{16}$"
] | 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 | Combinatorics | Math | English |
2,796 | In triangle $A B C, A B=B C$. A trisector of $\angle B$ intersects $\overline{A C}$ at $D$. If $A B, A C$, and $B D$ are integers and $A B-B D=7$, compute $A C$. | [
"Let $E$ be the point where the other trisector of $\\angle B$ intersects side $\\overline{A C}$. Let $A B=B C=a$, and let $B D=B E=d$. Draw $X$ on $\\overline{B C}$ so that $B X=d$. Then $C X=7$.\n\n<img_3688>\n\nThe placement of point $X$ guarantees that $\\triangle B E X \\cong \\triangle B D E$ by Side-Angle-Si... | [
"146"
] | 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 |
2,797 | The rational number $r$ is the largest number less than 1 whose base-7 expansion consists of two distinct repeating digits, $r=0 . \underline{A} \underline{B} \underline{A} \underline{B} \underline{A} \underline{B} \ldots$ Written as a reduced fraction, $r=\frac{p}{q}$. Compute $p+q$ (in base 10). | [
"In base 7, the value of $r$ must be $0.656565 \\ldots=0 . \\overline{65}_{7}$. Then $100_{7} \\cdot r=65 . \\overline{65}_{7}$, and $\\left(100_{7}-1\\right) r=$ $65_{7}$. In base $10,65_{7}=6 \\cdot 7+5=47_{10}$ and $100_{7}-1=7^{2}-1=48_{10}$. Thus $r=47 / 48$, and $p+q=95$."
] | [
"95"
] | 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 |
2,798 | Let $T=95$. Triangle $A B C$ has $A B=A C$. Points $M$ and $N$ lie on $\overline{B C}$ such that $\overline{A M}$ and $\overline{A N}$ trisect $\angle B A C$, with $M$ closer to $C$. If $\mathrm{m} \angle A M C=T^{\circ}$, then $\mathrm{m} \angle A C B=U^{\circ}$. Compute $U$. | [
"Because $\\triangle A B C$ is isosceles with $A B=A C, \\mathrm{~m} \\angle A B C=U^{\\circ}$ and $\\mathrm{m} \\angle B A C=(180-2 U)^{\\circ}$. Therefore $\\mathrm{m} \\angle M A C=\\left(\\frac{180-2 U}{3}\\right)^{\\circ}=\\left(60-\\frac{2}{3} U\\right)^{\\circ}$. Then $\\left(60-\\frac{2}{3} U\\right)+U+T=18... | [
"75"
] | 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 |
2,799 | Let $T=75$. At Wash College of Higher Education (Wash Ed.), the entering class has $n$ students. Each day, two of these students are selected to oil the slide rules. If the entering class had two more students, there would be $T$ more ways of selecting the two slide rule oilers. Compute $n$. | [
"With $n$ students, Wash Ed. can choose slide-rule oilers in $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)=\\frac{n(n-1)}{2}$ ways. With $n+2$ students, there would be $\\left(\\begin{array}{c}n+2 \\\\ 2\\end{array}\\right)=\\frac{(n+2)(n+1)}{2}$ ways of choosing the oilers. The difference is $\\frac{(n+2)(... | [
"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 | Combinatorics | Math | English |
2,800 | Compute the least positive integer $n$ such that the set of angles
$$
\left\{123^{\circ}, 246^{\circ}, \ldots, n \cdot 123^{\circ}\right\}
$$
contains at least one angle in each of the four quadrants. | [
"The first angle is $123^{\\circ}$, which is in Quadrant II, the second $\\left(246^{\\circ}\\right)$ is in Quadrant III, and the third is in Quadrant I, because $3 \\cdot 123^{\\circ}=369^{\\circ} \\equiv 9^{\\circ} \\bmod 360^{\\circ}$. The missing quadrant is IV, which is $270^{\\circ}-246^{\\circ}=24^{\\circ}$ ... | [
"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 | Number Theory | Math | English |
2,801 | Let $T=11$. In ARMLvania, license plates use only the digits 1-9, and each license plate contains exactly $T-3$ digits. On each plate, all digits are distinct, and for all $k \leq T-3$, the $k^{\text {th }}$ digit is at least $k$. Compute the number of valid ARMLvanian license plates. | [
"There are 9 valid one-digit plates. For a two-digit plate to be valid, it has to be of the form $\\underline{A} \\underline{B}$, where $B \\in\\{2, \\ldots, 9\\}$, and either $A \\in\\{2, \\ldots, 9\\}$ with $A \\neq B$ or $A=1$. So there are 8 ways to choose $B$ and $8-1+1=8$ ways to choose $A$, for a total of $8... | [
"256"
] | 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 |
2,802 | Let $T=256$. Let $\mathcal{R}$ be the region in the plane defined by the inequalities $x^{2}+y^{2} \geq T$ and $|x|+|y| \leq \sqrt{2 T}$. Compute the area of region $\mathcal{R}$. | [
"The first inequality states that the point $(x, y)$ is outside the circle centered at the origin with radius $\\sqrt{T}$, while the second inequality states that $(x, y)$ is inside the tilted square centered at the origin with diagonal $2 \\sqrt{2 T}$. The area of the square is $4 \\cdot \\frac{1}{2}(\\sqrt{2 T})^... | [
"$1024-256 \\pi$"
] | 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 |
2,817 | Triangle $A B C$ has $\mathrm{m} \angle A>\mathrm{m} \angle B>\mathrm{m} \angle C$. The angle between the altitude and the angle bisector at vertex $A$ is $6^{\circ}$. The angle between the altitude and the angle bisector at vertex $B$ is $18^{\circ}$. Compute the degree measure of angle $C$. | [
"Let the feet of the altitudes from $A$ and $B$ be $E$ and $D$, respectively, and let $F$ and $G$ be the intersection points of the angle bisectors with $\\overline{A C}$ and $\\overline{B C}$, respectively, as shown below.\n\n<img_3386>\n\nThen $\\mathrm{m} \\angle G A E=6^{\\circ}$ and $\\mathrm{m} \\angle D B F=... | [
"$44$"
] | 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 | $^{\circ}$ | Numerical | null | Open-ended | Geometry | Math | English |
2,818 | Compute the number of ordered pairs of integers $(b, c)$, with $-20 \leq b \leq 20,-20 \leq c \leq 20$, such that the equations $x^{2}+b x+c=0$ and $x^{2}+c x+b=0$ share at least one root. | [
"Let $r$ be the common root. Then $r^{2}+b r+c=r^{2}+c r+b \\Rightarrow b r-c r=b-c$. So either $b=c$ or $r=1$. In the latter case, $1+b+c=0$, so $c=-1-b$.\n\nThere are 41 ordered pairs where $b=c$. If $c=-1-b$ and $-20 \\leq b \\leq 20$, then $-21 \\leq c \\leq 19$. Therefore there are 40 ordered pairs $(b,-1-b)$ ... | [
"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 |
2,819 | A seventeen-sided die has faces numbered 1 through 17, but it is not fair: 17 comes up with probability $1 / 2$, and each of the numbers 1 through 16 comes up with probability $1 / 32$. Compute the probability that the sum of two rolls is either 20 or 12. | [
"The rolls that add up to 20 are $17+3,16+4,15+5,14+6,13+7,12+8,11+9$, and $10+10$. Accounting for order, the probability of $17+3$ is $\\frac{1}{2} \\cdot \\frac{1}{32}+\\frac{1}{32} \\cdot \\frac{1}{2}=2 \\cdot \\frac{1}{2} \\cdot \\frac{1}{32}=\\frac{32}{1024}$. The combination $10+10$ has probability $\\frac{1}... | [
"$\\frac{7}{128}$"
] | 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 |
2,821 | Compute the number of ordered pairs of integers $(a, b)$ such that $1<a \leq 50,1<b \leq 50$, and $\log _{b} a$ is rational. | [
"Begin by partitioning $\\{2,3, \\ldots, 50\\}$ into the subsets\n\n$$\n\\begin{aligned}\nA & =\\{2,4,8,16,32\\} \\\\\nB & =\\{3,9,27\\} \\\\\nC & =\\{5,25\\} \\\\\nD & =\\{6,36\\} \\\\\nE & =\\{7,49\\} \\\\\nF & =\\text { all other integers between } 2 \\text { and } 50, \\text { inclusive. }\n\\end{aligned}\n$$\n... | [
"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 | Number Theory | Math | English |
2,822 | Suppose that 5-letter "words" are formed using only the letters A, R, M, and L. Each letter need not be used in a word, but each word must contain at least two distinct letters. Compute the number of such words that use the letter A more than any other letter. | [
"Condition on the number $n$ of A's that appear in the word; $n$ is at least two, because of the requirement that $\\mathbf{A}$ occur more often than any other letter, and $n$ is at most 4 , because of the requirement that there be at least two distinct letters. In the case $n=4$, there are 3 choices for the other ... | [
"165"
] | 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 |
2,823 | Positive integers $a_{1}, a_{2}, a_{3}, \ldots$ form an arithmetic sequence. If $a_{1}=10$ and $a_{a_{2}}=100$, compute $a_{a_{a_{3}}}$. | [
"Let $d$ be the common difference of the sequence. Then $a_{a_{2}}=a_{1}+\\left(a_{2}-1\\right) d=100 \\Rightarrow\\left(a_{2}-1\\right) d=$ 90. But $a_{2}=a_{1}+d=10+d$, so $(9+d) d=90$. Solving the quadratic yields $d=-15$ or $d=6$, but the requirement that $a_{i}$ be positive for all $i$ rules out the negative v... | [
"820"
] | 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 |
2,824 | The graphs of $y=x^{2}-|x|-12$ and $y=|x|-k$ intersect at distinct points $A, B, C$, and $D$, in order of increasing $x$-coordinates. If $A B=B C=C D$, compute $k$. | [
"First, note that both graphs are symmetric about the $y$-axis, so $C$ and $D$ must be reflections of $B$ and $A$, respectively, across the $y$-axis. Thus $x_{C}=-x_{B}$ and $y_{C}=y_{B}$, so $B C=2 x_{C}$. For $x<0$, the equations become $y=x^{2}+x-12$ and $y=-x-k$; setting the $x$-expressions equal to each other ... | [
"$10+2 \\sqrt{2}$"
] | 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 |
2,825 | The zeros of $f(x)=x^{6}+2 x^{5}+3 x^{4}+5 x^{3}+8 x^{2}+13 x+21$ are distinct complex numbers. Compute the average value of $A+B C+D E F$ over all possible permutations $(A, B, C, D, E, F)$ of these six numbers. | [
"There are $6 !=720$ permutations of the zeros, so the average value is the sum, $S$, divided by 720. Setting any particular zero as $A$ leaves $5 !=120$ ways to permute the other five zeros, so over the 720 permutations, each zero occupies the $A$ position 120 times. Similarly, fixing any ordered pair $(B, C)$ of ... | [
"$-\\frac{23}{60}$"
] | 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 |
2,827 | Let $N=\left\lfloor(3+\sqrt{5})^{34}\right\rfloor$. Compute the remainder when $N$ is divided by 100 . | [
"Let $\\alpha=3+\\sqrt{5}$ and $\\beta=3-\\sqrt{5}$, so that $N=\\left\\lfloor\\alpha^{34}\\right\\rfloor$, and let $M=\\alpha^{34}+\\beta^{34}$. When the binomials in $M$ are expanded, terms in which $\\sqrt{5}$ is raised to an odd power have opposite signs, and so cancel each other out. Therefore $M$ is an intege... | [
"47"
] | 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 |
2,828 | Let $A B C$ be a triangle with $\mathrm{m} \angle B=\mathrm{m} \angle C=80^{\circ}$. Compute the number of points $P$ in the plane such that triangles $P A B, P B C$, and $P C A$ are all isosceles and non-degenerate. Note: the approximation $\cos 80^{\circ} \approx 0.17$ may be useful. | [
"Focus on $\\triangle P B C$. Either $P B=P C$ or $P B=B C$ or $P C=B C$.\n\nIf $P B=P C$, then $P$ lies on the perpendicular bisector $l$ of side $\\overline{B C}$. Considering now $\\triangle P A B$, if $P A=P B$, then $P A=P C$, and $P$ must be the circumcenter of $\\triangle A B C$; call this location $P_{1}$. ... | [
"6"
] | 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 |
2,829 | If $\lceil u\rceil$ denotes the least integer greater than or equal to $u$, and $\lfloor u\rfloor$ denotes the greatest integer less than or equal to $u$, compute the largest solution $x$ to the equation
$$
\left\lfloor\frac{x}{3}\right\rfloor+\lceil 3 x\rceil=\sqrt{11} \cdot x
$$ | [
"Let $f(x)=\\left\\lfloor\\frac{x}{3}\\right\\rfloor+\\lceil 3 x\\rceil$. Observe that $f(x+3)=f(x)+1+9=f(x)+10$. Let $g(x)=f(x)-\\frac{10}{3} x$. Then $g$ is periodic, because $g(x+3)=f(x)+10-\\frac{10 x}{3}-\\frac{10 \\cdot 3}{3}=g(x)$. The graph of $g$ is shown below:\n\n<img_3987>\n\nBecause $g(x)$ is the (vert... | [
"$\\frac{189 \\sqrt{11}}{11}$"
] | 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 | Algebra | Math | English |
2,830 | If $x, y$, and $z$ are positive integers such that $x y=20$ and $y z=12$, compute the smallest possible value of $x+z$. | [
"Note that $x$ and $z$ can each be minimized by making $y$ as large as possible, so set $y=$ $\\operatorname{lcm}(12,20)=4$. Then $x=5, z=3$, and $x+z=\\mathbf{8}$."
] | [
"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 |
2,831 | Let $T=8$. Let $A=(1,5)$ and $B=(T-1,17)$. Compute the value of $x$ such that $(x, 3)$ lies on the perpendicular bisector of $\overline{A B}$. | [
"The midpoint of $\\overline{A B}$ is $\\left(\\frac{T}{2}, 11\\right)$, and the slope of $\\overleftrightarrow{A B}$ is $\\frac{12}{T-2}$. Thus the perpendicular bisector of $\\overline{A B}$ has slope $\\frac{2-T}{12}$ and passes through the point $\\left(\\frac{T}{2}, 11\\right)$. Thus the equation of the perpen... | [
"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 |
2,832 | Let T be a rational number. Let $N$ be the smallest positive $T$-digit number that is divisible by 33 . Compute the product of the last two digits of $N$. | [
"The sum of the digits of $N$ must be a multiple of 3 , and the alternating sum of the digits must be a multiple of 11 . Because the number of digits of $N$ is fixed, the minimum $N$ will have the alternating sum of its digits equal to 0 , and therefore the sum of the digits of $N$ will be even, so it must be 6 . T... | [
"6"
] | 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 |
2,834 | Let $T=15$. For complex $z$, define the function $f_{1}(z)=z$, and for $n>1, f_{n}(z)=$ $f_{n-1}(\bar{z})$. If $f_{1}(z)+2 f_{2}(z)+3 f_{3}(z)+4 f_{4}(z)+5 f_{5}(z)=T+T i$, compute $|z|$. | [
"Because $\\overline{\\bar{z}}=z$, it follows that $f_{n}(z)=z$ when $n$ is odd, and $f_{n}(z)=\\bar{z}$ when $n$ is even. Taking $z=a+b i$, where $a$ and $b$ are real, it follows that $\\sum_{k=1}^{5} k f_{k}(z)=15 a+3 b i$. Thus $a=\\frac{T}{15}, b=\\frac{T}{3}$, and $|z|=\\sqrt{a^{2}+b^{2}}=\\frac{|T| \\sqrt{26}... | [
"$\\sqrt{26}$"
] | 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 |
2,835 | Let $T=\sqrt{26}$. Compute the number of ordered pairs of positive integers $(a, b)$ with the property that $a b=T^{20} \cdot 210^{12}$, and the greatest common divisor of $a$ and $b$ is 1 . | [
"If the prime factorization of $a b$ is $p_{1}^{e_{1}} p_{2}^{e_{2}} \\ldots p_{k}^{e_{k}}$, where the $p_{i}$ 's are distinct primes and the $e_{i}$ 's are positive integers, then in order for $\\operatorname{gcd}(a, b)$ to equal 1 , each $p_{i}$ must be a divisor of exactly one of $a$ or $b$. Thus the desired num... | [
"32"
] | 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 |
2,836 | Let $T=32$. Given that $\sin \theta=\frac{\sqrt{T^{2}-64}}{T}$, compute the largest possible value of the infinite series $\cos \theta+\cos ^{2} \theta+\cos ^{3} \theta+\ldots$. | [
"Using $\\sin ^{2} \\theta+\\cos ^{2} \\theta=1$ gives $\\cos ^{2} \\theta=\\frac{64}{T^{2}}$, so to maximize the sum, take $\\cos \\theta=\\frac{8}{|T|}$. Using the formula for the sum of an infinite geometric series gives $\\frac{8 /|T|}{1-8 /|T|}=\\frac{8}{|T|-8}$. With $T=32$, the answer is $\\frac{8}{24}=\\fra... | [
"$\\frac{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 | Algebra | Math | English |
2,838 | Let $T=\frac{9}{17}$. When $T$ is expressed as a reduced fraction, let $m$ and $n$ be the numerator and denominator, respectively. A square pyramid has base $A B C D$, the distance from vertex $P$ to the base is $n-m$, and $P A=P B=P C=P D=n$. Compute the area of square $A B C D$. | [
"By the Pythagorean Theorem, half the diagonal of the square is $\\sqrt{n^{2}-(n-m)^{2}}=\\sqrt{2 m n-m^{2}}$. Thus the diagonal of the square is $2 \\sqrt{2 m n-m^{2}}$, and the square's area is $4 m n-2 m^{2}$. With $T=\\frac{9}{17}, m=9, n=17$, and the answer is 450 ."
] | [
"450"
] | 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 |
2,839 | Let $T=-14$, and let $d=|T|$. A person whose birthday falls between July 23 and August 22 inclusive is called a Leo. A person born in July is randomly selected, and it is given that her birthday is before the $d^{\text {th }}$ day of July. Another person who was also born in July is randomly selected, and it is given that his birthday is after the $d^{\text {th }}$ day of July. Compute the probability that exactly one of these people is a Leo. | [
"Note that there are 9 days in July in which a person could be a Leo (July 23-31). Let the woman (born before the $d^{\\text {th }}$ day of July) be called Carol, and let the man (born after the $d^{\\text {th }}$ day of July) be called John, and consider the possible values of $d$. If $d \\leq 21$, then Carol will... | [
"$\\frac{9}{17}$"
] | 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 | Combinatorics | Math | English |
2,840 | Let $T=-10$. Given that $\log _{2} 4^{8 !}+\log _{4} 2^{8 !}=6 ! \cdot T \cdot x$, compute $x$. | [
"Note that $4^{8 !}=2^{2 \\cdot 8 !}$, thus $\\log _{2} 4^{8 !}=2 \\cdot 8$ !. Similarly, $\\log _{4} 2^{8 !}=\\frac{8 !}{2}$. Thus $2 \\cdot 8 !+\\frac{8 !}{2}=$ $6 !\\left(2 \\cdot 7 \\cdot 8+7 \\cdot \\frac{8}{2}\\right)=6 ! \\cdot 140$. Thus $140=T x$, and with $T=-10, x=\\mathbf{- 1 4}$."
] | [
"-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 | Algebra | Math | English |
2,841 | Let $T=20$. For some real constants $a$ and $b$, the solution sets of the equations $x^{2}+(5 b-T-a) x=T+1$ and $2 x^{2}+(T+8 a-2) x=-10 b$ are the same. Compute $a$. | [
"Divide each side of the second equation by 2 and equate coefficients to obtain $5 b-T-a=$ $\\frac{T}{2}+4 a-1$ and $T+1=-5 b$. Thus $b=\\frac{T+1}{-5}$, and plugging this value into the first equation yields $a=-\\frac{T}{2}$. With $T=20$, the answer is $\\mathbf{- 1 0}$."
] | [
"-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 | Algebra | Math | English |
2,842 | Let T be a rational number, and let $K=T-2$. If $K$ workers can produce 9 widgets in 1 hour, compute the number of workers needed to produce $\frac{720}{K}$ widgets in 4 hours. | [
"Because $T$ workers produce 9 widgets in 1 hour, 1 worker will produce $\\frac{9}{T}$ widgets in 1 hour. Thus 1 worker will produce $\\frac{36}{T}$ widgets in 4 hours. In order to produce $\\frac{720}{T}$ widgets in 4 hours, it will require $\\frac{720 / T}{36 / T}=\\mathbf{2 0}$ workers (independent of $T$ )."
] | [
"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 | Algebra | Math | English |
2,843 | Let $T=2018$, and append the digits of $T$ to $\underline{A} \underline{A} \underline{B}$ (for example, if $T=17$, then the result would be $\underline{1} \underline{\underline{A}} \underline{A} \underline{B}$ ). If the resulting number is divisible by 11 , compute the largest possible value of $A+B$. | [
"Let $R$ be the remainder when $T$ is divided by 11 . Note that the alternating sum of the digits of the number must be divisible by 11 . This sum will be congruent $\\bmod 11$ to $B-A+A-R=$ $B-R$, thus $B=R$. Because $A$ 's value is irrelevant, to maximize $A+B$, set $A=9$ to yield $A+B=9+R$. For $T=2018, R=5$, an... | [
"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 | Algebra | Math | English |
2,844 | Given that April $1^{\text {st }}, 2012$ fell on a Sunday, what is the next year in which April $1^{\text {st }}$ will fall on a Sunday? | [
"Note that $365=7 \\cdot 52+1$. Thus over the next few years after 2012 , the day of the week for April $1^{\\text {st }}$ will advance by one day in a non-leap year, and it will advance by two days in a leap year. Thus in six years, the day of the week will have rotated a complete cycle, and the answer is 2018 ."
... | [
"2018"
] | 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 |
2,845 | Let $p$ be a prime number. If $p$ years ago, the ages of three children formed a geometric sequence with a sum of $p$ and a common ratio of 2 , compute the sum of the children's current ages. | [
"Let $x, 2 x$, and $4 x$ be the ages of the children $p$ years ago. Then $x+2 x+4 x=p$, so $7 x=p$. Since $p$ is prime, $x=1$. Thus the sum of the children's current ages is $(1+7)+(2+7)+(4+7)=\\mathbf{2 8}$."
] | [
"28"
] | 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 |
2,846 | Define a reverse prime to be a positive integer $N$ such that when the digits of $N$ are read in reverse order, the resulting number is a prime. For example, the numbers 5, 16, and 110 are all reverse primes. Compute the largest two-digit integer $N$ such that the numbers $N, 4 \cdot N$, and $5 \cdot N$ are all reverse primes. | [
"Because $N<100,5 \\cdot N<500$. Since no primes end in 4, it follows that $5 \\cdot N<400$, hence $N \\leq 79$. The reverses of $5 \\cdot 79=395,4 \\cdot 79=316$, and 79 are 593,613 , and 97 , respectively. All three of these numbers are prime, thus 79 is the largest two-digit integer $N$ for which $N$, $4 \\cdot ... | [
"79"
] | 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 |
2,847 | Some students in a gym class are wearing blue jerseys, and the rest are wearing red jerseys. There are exactly 25 ways to pick a team of three players that includes at least one player wearing each color. Compute the number of students in the class. | [
"Let $r$ and $b$ be the number of students wearing red and blue jerseys, respectively. Then either we choose two blues and one red or one blue and two reds. Thus\n\n$$\n\\begin{aligned}\n& \\left(\\begin{array}{l}\nb \\\\\n2\n\\end{array}\\right)\\left(\\begin{array}{l}\nr \\\\\n1\n\\end{array}\\right)+\\left(\\beg... | [
"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 |
2,848 | Point $P$ is on the hypotenuse $\overline{E N}$ of right triangle $B E N$ such that $\overline{B P}$ bisects $\angle E B N$. Perpendiculars $\overline{P R}$ and $\overline{P S}$ are drawn to sides $\overline{B E}$ and $\overline{B N}$, respectively. If $E N=221$ and $P R=60$, compute $\frac{1}{B E}+\frac{1}{B N}$. | [
"We observe that $\\frac{1}{B E}+\\frac{1}{B N}=\\frac{B E+B N}{B E \\cdot B N}$. The product in the denominator suggests that we compare areas. Let $[B E N]$ denote the area of $\\triangle B E N$. Then $[B E N]=\\frac{1}{2} B E \\cdot B N$, but because $P R=P S=60$, we can also write $[B E N]=[B E P]+[B N P]=\\fra... | [
"$\\frac{1}{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 |
2,849 | $\quad$ Compute all real values of $x$ such that $\log _{2}\left(\log _{2} x\right)=\log _{4}\left(\log _{4} x\right)$. | [
"If $y=\\log _{a}\\left(\\log _{a} x\\right)$, then $a^{a^{y}}=x$. Let $y=\\log _{2}\\left(\\log _{2} x\\right)=\\log _{4}\\left(\\log _{4} x\\right)$. Then $2^{2^{y}}=4^{4^{y}}=$ $\\left(2^{2}\\right)^{\\left(2^{2}\\right)^{y}}=2^{2^{2 y+1}}$, so $2 y+1=y, y=-1$, and $x=\\sqrt{\\mathbf{2}}$. (This problem is based... | [
"$\\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 | Algebra | Math | English |
2,850 | Let $k$ be the least common multiple of the numbers in the set $\mathcal{S}=\{1,2, \ldots, 30\}$. Determine the number of positive integer divisors of $k$ that are divisible by exactly 28 of the numbers in the set $\mathcal{S}$. | [
"We know that $k=2^{4} \\cdot 3^{3} \\cdot 5^{2} \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19 \\cdot 23 \\cdot 29$. It is not difficult to see that the set $\\mathcal{T}_{1}=\\left\\{\\frac{k}{2}, \\frac{k}{3}, \\frac{k}{5}, \\frac{k}{17}, \\frac{k}{19}, \\frac{k}{23}, \\frac{k}{29}\\right\\}$ comprises all div... | [
"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 |
2,851 | Let $A$ and $B$ be digits from the set $\{0,1,2, \ldots, 9\}$. Let $r$ be the two-digit integer $\underline{A} \underline{B}$ and let $s$ be the two-digit integer $\underline{B} \underline{A}$, so that $r$ and $s$ are members of the set $\{00,01, \ldots, 99\}$. Compute the number of ordered pairs $(A, B)$ such that $|r-s|=k^{2}$ for some integer $k$. | [
"Because $|(10 A+B)-(10 B+A)|=9|A-B|=k^{2}$, it follows that $|A-B|$ is a perfect square. $|A-B|=0$ yields 10 pairs of integers: $(A, B)=(0,0),(1,1), \\ldots,(9,9)$.\n\n$|A-B|=1$ yields 18 pairs: the nine $(A, B)=(0,1),(1,2), \\ldots,(8,9)$, and their reverses.\n\n$|A-B|=4$ yields 12 pairs: the six $(A, B)=(0,4),(1... | [
"42"
] | 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 |
2,852 | For $k \geq 3$, we define an ordered $k$-tuple of real numbers $\left(x_{1}, x_{2}, \ldots, x_{k}\right)$ to be special if, for every $i$ such that $1 \leq i \leq k$, the product $x_{1} \cdot x_{2} \cdot \ldots \cdot x_{k}=x_{i}^{2}$. Compute the smallest value of $k$ such that there are at least 2009 distinct special $k$-tuples. | [
"The given conditions imply $k$ equations. By taking the product of these $k$ equations, we have $\\left(x_{1} x_{2} \\ldots x_{k}\\right)^{k-1}=x_{1} x_{2} \\ldots x_{k}$. Thus it follows that either $x_{1} x_{2} \\ldots x_{k}=0$ or $x_{1} x_{2} \\ldots x_{k}= \\pm 1$. If $x_{1} x_{2} \\ldots x_{k}=0$, then some $... | [
"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 |
2,853 | A cylinder with radius $r$ and height $h$ has volume 1 and total surface area 12. Compute $\frac{1}{r}+\frac{1}{h}$. | [
"Since $\\pi r^{2} h=1$, we have $h=\\frac{1}{\\pi r^{2}}$ and $\\pi r^{2}=\\frac{1}{h}$. Consequently,\n\n$$\n2 \\pi r h+2 \\pi r^{2}=12 \\Rightarrow(2 \\pi r)\\left(\\frac{1}{\\pi r^{2}}\\right)+2\\left(\\frac{1}{h}\\right)=12 \\Rightarrow \\frac{2}{r}+\\frac{2}{h}=12 \\Rightarrow \\frac{1}{r}+\\frac{1}{h}=\\math... | [
"6"
] | 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 |
2,854 | If $6 \tan ^{-1} x+4 \tan ^{-1}(3 x)=\pi$, compute $x^{2}$. | [
"$\\quad$ Let $z=1+x i$ and $w=1+3 x i$, where $i=\\sqrt{-1}$. Then $\\tan ^{-1} x=\\arg z$ and $\\tan ^{-1}(3 x)=\\arg w$, where $\\arg z$ gives the measure of the angle in standard position whose terminal side passes through $z$. By DeMoivre's theorem, $6 \\tan ^{-1} x=\\arg \\left(z^{6}\\right)$ and $4 \\tan ^{-... | [
"$\\frac{15-8 \\sqrt{3}}{33}$"
] | 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 |
2,855 | A rectangular box has dimensions $8 \times 10 \times 12$. Compute the fraction of the box's volume that is not within 1 unit of any of the box's faces. | [
"Let the box be defined by the product of the intervals on the $x, y$, and $z$ axes as $[0,8] \\times$ $[0,10] \\times[0,12]$ with volume $8 \\times 10 \\times 12$. The set of points inside the box that are not within 1 unit of any face is defined by the product of the intervals $[1,7] \\times[1,9] \\times[1,11]$ w... | [
"$\\frac{1}{2}$"
] | 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 | Geometry | Math | English |
2,856 | Let $T=\frac{1}{2}$. Compute the largest real solution $x$ to $(\log x)^{2}-\log \sqrt{x}=T$. | [
"Let $u=\\log x$. Then the given equation can be rewritten as $u^{2}-\\frac{1}{2} u-T=0 \\rightarrow 2 u^{2}-u-2 T=0$. This quadratic has solutions $u=\\frac{1 \\pm \\sqrt{1+16 T}}{4}$. As we are looking for the largest real solution for $x$ (and therefore, for $u$ ), we want $u=\\frac{1+\\sqrt{1+16 T}}{4}=1$ when ... | [
"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 | Algebra | Math | English |
2,857 | Let $T=10$. Kay has $T+1$ different colors of fingernail polish. Compute the number of ways that Kay can paint the five fingernails on her left hand by using at least three colors and such that no two consecutive fingernails have the same color. | [
"There are $T+1$ possible colors for the first nail. Each remaining nail may be any color except that of the preceding nail, that is, there are $T$ possible colors. Thus, using at least two colors, there are $(T+1) T^{4}$ possible colorings. The problem requires that at least three colors be used, so we must subtra... | [
"109890"
] | 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 |
2,858 | Compute the number of ordered pairs $(x, y)$ of positive integers satisfying $x^{2}-8 x+y^{2}+4 y=5$. | [
"Completing the square twice in $x$ and $y$, we obtain the equivalent equation $(x-4)^{2}+(y+2)^{2}=$ 25 , which describes a circle centered at $(4,-2)$ with radius 5 . The lattice points on this circle are points 5 units up, down, left, or right of the center, or points 3 units away on one axis and 4 units away on... | [
"4"
] | 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 |
2,859 | Let $T=4$ and let $k=21+2 T$. Compute the largest integer $n$ such that $2 n^{2}-k n+77$ is a positive prime number. | [
"If $k$ is positive, there are only four possible factorizations of $2 n^{2}-k n+77$ over the integers, namely\n\n$$\n\\begin{aligned}\n& (2 n-77)(n-1)=2 n^{2}-79 n+77 \\\\\n& (2 n-1)(n-77)=2 n^{2}-145 n+77 \\\\\n& (2 n-11)(n-7)=2 n^{2}-25 n+77 \\\\\n& (2 n-7)(n-11)=2 n^{2}-29 n+77\n\\end{aligned}\n$$\n\n\n\nBecaus... | [
"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 |
2,860 | Let $T=12$. In triangle $A B C, B C=T$ and $\mathrm{m} \angle B=30^{\circ}$. Compute the number of integer values of $A C$ for which there are two possible values for side length $A B$. | [
"By the Law of Cosines, $(A C)^{2}=T^{2}+(A B)^{2}-2 T(A B) \\cos 30^{\\circ} \\rightarrow(A B)^{2}-2 T \\cos 30^{\\circ}(A B)+$ $\\left(T^{2}-(A C)^{2}\\right)=0$. This quadratic in $A B$ has two positive solutions when the discriminant and product of the roots are both positive. Thus $\\left(2 T \\cos 30^{\\circ}... | [
"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 |
2,861 | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
Compute the 3 -signature for 52341. | [
"$(312,123,231)$"
] | [
"$(312,123,231)$"
] | 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 | Tuple | null | Open-ended | Combinatorics | Math | English |
2,862 | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
Find another 5-label with the same 3-signature as in part (a). | [
"$41352,42351,51342$"
] | [
"$41352,42351,51342$"
] | 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 | Combinatorics | Math | English |
2,863 | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
Compute two other 6-labels with the same 4-signature as 462135. | [
"$352146,362145,452136,562134$"
] | [
"$352146,362145,452136,562134$"
] | 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 | Combinatorics | Math | English |
2,884 | In $\triangle A B C, D$ is on $\overline{A C}$ so that $\overline{B D}$ is the angle bisector of $\angle B$. Point $E$ is on $\overline{A B}$ and $\overline{C E}$ intersects $\overline{B D}$ at $P$. Quadrilateral $B C D E$ is cyclic, $B P=12$ and $P E=4$. Compute the ratio $\frac{A C}{A E}$. | [
"Let $\\omega$ denote the circle that circumscribes quadrilateral $B C D E$. Draw in line segment $\\overline{D E}$. Note that $\\angle D P E$ and $\\angle C P B$ are congruent, and $\\angle D E C$ and $\\angle D B C$ are congruent, since they cut off the same arc of $\\omega$. Therefore, $\\triangle B C P$ and $\\... | [
"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 |
2,887 | Let $N$ be a six-digit number formed by an arrangement of the digits $1,2,3,3,4,5$. Compute the smallest value of $N$ that is divisible by 264 . | [
"Note that $264=3 \\cdot 8 \\cdot 11$, so we will need to address all these factors. Because the sum of the digits is 18 , it follows that 3 divides $N$, regardless of how we order the digits of $N$. In order for 8 to divide $N$, we need $N$ to end in $\\underline{O} 12, \\underline{O} 52, \\underline{E} 32$, or $\... | [
"135432"
] | 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 |
2,888 | In triangle $A B C, A B=4, B C=6$, and $A C=8$. Squares $A B Q R$ and $B C S T$ are drawn external to and lie in the same plane as $\triangle A B C$. Compute $Q T$. | [
"Set $\\mathrm{m} \\angle A B C=x$ and $\\mathrm{m} \\angle T B Q=y$. Then $x+y=180^{\\circ}$ and so $\\cos x+\\cos y=0$. Applying the Law of Cosines to triangles $A B C$ and $T B Q$ gives $A C^{2}=A B^{2}+B C^{2}-2 A B \\cdot B C \\cos x$ and $Q T^{2}=B T^{2}+B Q^{2}-2 B T \\cdot B Q \\cos y$, which, after substit... | [
"$2 \\sqrt{10}$"
] | 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 |
2,890 | An ellipse in the first quadrant is tangent to both the $x$-axis and $y$-axis. One focus is at $(3,7)$, and the other focus is at $(d, 7)$. Compute $d$. | [
"See the diagram below. The center of the ellipse is $C=\\left(\\frac{d+3}{2}, 7\\right)$. The major axis of the ellipse is the line $y=7$, and the minor axis is the line $x=\\frac{d+3}{2}$. The ellipse is tangent to the coordinate axes at $T_{x}=\\left(\\frac{d+3}{2}, 0\\right)$ and $T_{y}=(0,7)$. Let $F_{1}=(3,7)... | [
"$\\frac{49}{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 |
2,891 | Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7} A_{8}$ be a regular octagon. Let $\mathbf{u}$ be the vector from $A_{1}$ to $A_{2}$ and let $\mathbf{v}$ be the vector from $A_{1}$ to $A_{8}$. The vector from $A_{1}$ to $A_{4}$ can be written as $a \mathbf{u}+b \mathbf{v}$ for a unique ordered pair of real numbers $(a, b)$. Compute $(a, b)$. | [
"We can scale the octagon so that $A_{1} A_{2}=\\sqrt{2}$. Because the exterior angle of the octagon is $45^{\\circ}$, we can place the octagon in the coordinate plane with $A_{1}$ being the origin, $A_{2}=(\\sqrt{2}, 0)$, and $A_{8}=(1,1)$.\n\n<img_3693>\n\nThen $A_{3}=(1+\\sqrt{2}, 1)$ and $A_{4}=(1+\\sqrt{2}, 1+... | [
"$\\quad(2+\\sqrt{2}, 1+\\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 |
2,892 | Compute the integer $n$ such that $2009<n<3009$ and the sum of the odd positive divisors of $n$ is 1024 . | [
"Suppose that $n=2^{k} p_{1}^{a_{1}} \\cdots p_{r}^{a_{r}}$, where the $p_{i}$ are distinct odd primes, $k$ is a nonnegative integer, and $a_{1}, \\ldots, a_{r}$ are positive integers. Then the sum of the odd positive divisors of $n$ is equal to\n\n$$\n\\prod_{i=1}^{r}\\left(1+p_{i}+\\cdots+p_{i}^{a_{i}}\\right)=\\... | [
"2604"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,893 | Points $A, R, M$, and $L$ are consecutively the midpoints of the sides of a square whose area is 650. The coordinates of point $A$ are $(11,5)$. If points $R, M$, and $L$ are all lattice points, and $R$ is in Quadrant I, compute the number of possible ordered pairs $(x, y)$ of coordinates for point $R$. | [
"Write $x=11+c$ and $y=5+d$. Then $A R^{2}=c^{2}+d^{2}=\\frac{1}{2} \\cdot 650=325$. Note that $325=18^{2}+1^{2}=17^{2}+6^{2}=15^{2}+10^{2}$. Temporarily restricting ourselves to the case where $c$ and $d$ are both positive, there are three classes of solutions: $\\{c, d\\}=\\{18,1\\},\\{c, d\\}=\\{17,6\\}$, or $\\... | [
"10"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,894 | The taxicab distance between points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by
$$
d\left(\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)\right)=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|+\left|z_{1}-z_{2}\right| .
$$
The region $\mathcal{R}$ is obtained by taking the cube $\{(x, y, z): 0 \leq x, y, z \leq 1\}$ and removing every point whose taxicab distance to any vertex of the cube is less than $\frac{3}{5}$. Compute the volume of $\mathcal{R}$. | [
"For a fixed vertex $V$ on the cube, the locus of points on or inside the cube that are at most $\\frac{3}{5}$ away from $V$ form a corner at $V$ (that is, the right pyramid $V W_{1} W_{2} W_{3}$ in the figure shown at left below, with equilateral triangular base $W_{1} W_{2} W_{3}$ and three isosceles right triang... | [
"$\\frac{179}{250}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,895 | $\quad$ Let $a$ and $b$ be real numbers such that
$$
a^{3}-15 a^{2}+20 a-50=0 \quad \text { and } \quad 8 b^{3}-60 b^{2}-290 b+2575=0
$$
Compute $a+b$. | [
"Each cubic expression can be depressed - that is, the quadratic term can be eliminated-by substituting as follows. Because $(a-p)^{3}=a^{3}-3 a^{2} p+3 a p^{2}-p^{3}$, setting $p=-\\frac{(-15)}{3}=5$ and substituting $c+p=a$ transforms the expression $a^{3}-15 a^{2}+20 a-50$ into the equivalent expression $(c+5)^{... | [
"$\\frac{15}{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 |
2,896 | For a positive integer $n$, define $s(n)$ to be the sum of $n$ and its digits. For example, $s(2009)=2009+2+0+0+9=2020$. Compute the number of elements in the set $\{s(0), s(1), s(2), \ldots, s(9999)\}$. | [
"If $s(10 x)=a$, then the values of $s$ over $\\{10 x+0,10 x+1, \\ldots, 10 x+9\\}$ are $a, a+2, a+4, \\ldots, a+18$. Furthermore, if $x$ is not a multiple of 10 , then $s(10(x+1))=a+11$. This indicates that the values of $s$ \"interweave\" somewhat from one group of 10 to the next: the sets alternate between even ... | [
"9046"
] | 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 |
2,897 | Quadrilateral $A R M L$ is a kite with $A R=R M=5, A M=8$, and $R L=11$. Compute $A L$. | [
"Let $K$ be the midpoint of $\\overline{A M}$. Then $A K=K M=8 / 2=4, R K=\\sqrt{5^{2}-4^{2}}=3$, and $K L=11-3=8$. Thus $A L=\\sqrt{A K^{2}+K L^{2}}=\\sqrt{4^{2}+8^{2}}=4 \\sqrt{5}$."
] | [
"$4 \\sqrt{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 |
2,898 | Let $T=4 \sqrt{5}$. If $x y=\sqrt{5}, y z=5$, and $x z=T$, compute the positive value of $x$. | [
"Multiply the three given equations to obtain $x^{2} y^{2} z^{2}=5 T \\sqrt{5}$. Thus $x y z= \\pm \\sqrt[4]{125 T^{2}}$, and the positive value of $x$ is $x=x y z / y z=\\sqrt[4]{125 T^{2}} / 5=\\sqrt[4]{T^{2} / 5}$. With $T=4 \\sqrt{5}$, we have $x=\\mathbf{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 | Algebra | Math | English |
2,899 | $\quad$ Let $T=2$. In how many ways can $T$ boys and $T+1$ girls be arranged in a row if all the girls must be standing next to each other? | [
"First choose the position of the first girl, starting from the left. There are $T+1$ possible positions, and then the positions for the girls are all determined. There are $(T+1)$ ! ways to arrange the girls, and there are $T$ ! ways to arrange the boys, for a total of $(T+1) \\cdot(T+1) ! \\cdot T !=$ $((T+1) !)^... | [
"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 | Combinatorics | Math | English |
2,900 | $\triangle A B C$ is on a coordinate plane such that $A=(3,6)$, $B=(T, 0)$, and $C=(2 T-1,1-T)$. Let $\ell$ be the line containing the altitude to $\overline{B C}$. Compute the $y$-intercept of $\ell$. | [
"The slope of $\\overleftrightarrow{B C}$ is $\\frac{(1-T)-0}{(2 T-1)-T}=-1$, and since $\\ell$ is perpendicular to $\\overleftrightarrow{B C}$, the slope of $\\ell$ is 1. Because $\\ell$ passes through $A=(3,6)$, the equation of $\\ell$ is $y=x+3$, and its $y$-intercept is 3 (independent of $T$ )."
] | [
"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 |
2,901 | Let $T=3$. In triangle $A B C, A B=A C-2=T$, and $\mathrm{m} \angle A=60^{\circ}$. Compute $B C^{2}$. | [
"By the Law of Cosines, $B C^{2}=A B^{2}+A C^{2}-2 \\cdot A B \\cdot A C \\cdot \\cos A=T^{2}+(T+2)^{2}-2 \\cdot T \\cdot(T+2) \\cdot \\frac{1}{2}=$ $T^{2}+2 T+4$. With $T=3$, the answer is 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 |
2,902 | Let $T=19$. Let $\mathcal{S}_{1}$ denote the arithmetic sequence $0, \frac{1}{4}, \frac{1}{2}, \ldots$, and let $\mathcal{S}_{2}$ denote the arithmetic sequence $0, \frac{1}{6}, \frac{1}{3}, \ldots$ Compute the $T^{\text {th }}$ smallest number that occurs in both sequences $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$. | [
"$\\mathcal{S}_{1}$ consists of all numbers of the form $\\frac{n}{4}$, and $\\mathcal{S}_{2}$ consists of all numbers of the form $\\frac{n}{6}$, where $n$ is a nonnegative integer. Since $\\operatorname{gcd}(4,6)=2$, the numbers that are in both sequences are of the form $\\frac{n}{2}$, and the $T^{\\text {th }}$... | [
"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 | Algebra | Math | English |
2,903 | $\quad$ Let $T=9$. An integer $n$ is randomly selected from the set $\{1,2,3, \ldots, 2 T\}$. Compute the probability that the integer $\left|n^{3}-7 n^{2}+13 n-6\right|$ is a prime number. | [
"Let $P(n)=n^{3}-7 n^{2}+13 n-6$, and note that $P(n)=(n-2)\\left(n^{2}-5 n+3\\right)$. Thus $|P(n)|$ is prime if either $|n-2|=1$ and $\\left|n^{2}-5 n+3\\right|$ is prime or if $\\left|n^{2}-5 n+3\\right|=1$ and $|n-2|$ is prime. Solving $|n-2|=1$ gives $n=1$ or 3 , and solving $\\left|n^{2}-5 n+3\\right|=1$ give... | [
"$\\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 | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,904 | Let $A=\frac{1}{9}$, and let $B=\frac{1}{25}$. In $\frac{1}{A}$ minutes, 20 frogs can eat 1800 flies. At this rate, in $\frac{1}{B}$ minutes, how many flies will 15 frogs be able to eat? | [
"In $\\frac{1}{A}$ minutes, 1 frog can eat $1800 / 20=90$ flies; thus in $\\frac{1}{B}$ minutes, 1 frog can eat $\\frac{A}{B} \\cdot 90$ flies. Thus in $\\frac{1}{B}$ minutes, 15 frogs can eat $15 \\cdot 90 \\cdot \\frac{A}{B}$ flies. With $A=\\frac{1}{9}$ and $B=\\frac{1}{25}$, this simplifies to $15 \\cdot 250=\\... | [
"3750"
] | 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 |
2,905 | Let $T=5$. If $|T|-1+3 i=\frac{1}{z}$, compute the sum of the real and imaginary parts of $z$. | [
"Let $t=|T|$. Note that $z=\\frac{1}{t-1+3 i}=\\frac{1}{t-1+3 i} \\cdot \\frac{t-1-3 i}{t-1-3 i}=\\frac{t-1-3 i}{t^{2}-2 t+10}$. Thus the sum of the real and imaginary parts of $z$ is $\\frac{t-1}{t^{2}-2 t+10}+\\frac{-3}{t^{2}-2 t+10}=\\frac{|T|-4}{|T|^{2}-2|T|+10}$. With $T=5$, the answer is $\\frac{1}{25}$."
] | [
"$\\frac{1}{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 | Algebra | Math | English |
2,907 | Let $T=10$. Ann spends 80 seconds climbing up a $T$ meter rope at a constant speed, and she spends 70 seconds climbing down the same rope at a constant speed (different from her upward speed). Ann begins climbing up and down the rope repeatedly, and she does not pause after climbing the length of the rope. After $T$ minutes, how many meters will Ann have climbed in either direction? | [
"In 150 seconds (or 2.5 minutes), Ann climbs up and down the entire rope. Thus in $T$ minutes, she makes $\\left\\lfloor\\frac{T}{2.5}\\right\\rfloor$ round trips, and therefore climbs $2 T\\left\\lfloor\\frac{T}{2.5}\\right\\rfloor$ meters. After making all her round trips, there are $t=60\\left(T-2.5\\left\\lfloo... | [
"80"
] | 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 |
2,908 | Let $T=800$. Simplify $2^{\log _{4} T} / 2^{\log _{16} 64}$. | [
"Note that $2^{\\log _{4} T}=4^{\\left(\\frac{1}{2} \\log _{4} T\\right)}=4^{\\log _{4} T^{\\frac{1}{2}}}=\\sqrt{T}$. Letting $\\log _{16} 64=x$, we see that $2^{4 x}=2^{6}$, thus $x=\\frac{3}{2}$, and $2^{x}=\\sqrt{8}$. Thus the given expression equals $\\sqrt{\\frac{T}{8}}$, and with $T=800$, this is equal to 10 ... | [
"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 | Algebra | Math | English |
2,909 | Let $P(x)=x^{2}+T x+800$, and let $r_{1}$ and $r_{2}$ be the roots of $P(x)$. The polynomial $Q(x)$ is quadratic, it has leading coefficient 1, and it has roots $r_{1}+1$ and $r_{2}+1$. Find the sum of the coefficients of $Q(x)$. | [
"Let $Q(x)=x^{2}+A x+B$. Then $A=-\\left(r_{1}+1+r_{2}+1\\right)$ and $B=\\left(r_{1}+1\\right)\\left(r_{2}+1\\right)$. Thus the sum of the coefficients of $Q(x)$ is $1+\\left(-r_{1}-r_{2}-2\\right)+\\left(r_{1} r_{2}+r_{1}+r_{2}+1\\right)=r_{1} r_{2}$. Note that $T=-\\left(r_{1}+r_{2}\\right)$ and $800=r_{1} r_{2}... | [
"800"
] | 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 |
2,910 | Let $T=12$. Equilateral triangle $A B C$ is given with side length $T$. Points $D$ and $E$ are the midpoints of $\overline{A B}$ and $\overline{A C}$, respectively. Point $F$ lies in space such that $\triangle D E F$ is equilateral and $\triangle D E F$ lies in a plane perpendicular to the plane containing $\triangle A B C$. Compute the volume of tetrahedron $A B C F$. | [
"The volume of tetrahedron $A B C F$ is one-third the area of $\\triangle A B C$ times the distance from $F$ to $\\triangle A B C$. Since $D$ and $E$ are midpoints, $D E=\\frac{B C}{2}=\\frac{T}{2}$, and the distance from $F$ to $\\triangle A B C$ is $\\frac{T \\sqrt{3}}{4}$. Thus the volume of $A B C F$ is $\\frac... | [
"108"
] | 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 |
2,911 | In triangle $A B C, A B=5, A C=6$, and $\tan \angle B A C=-\frac{4}{3}$. Compute the area of $\triangle A B C$. | [
"Let $s=\\sin \\angle B A C$. Then $s>0$ and $\\frac{s}{-\\sqrt{1-s^{2}}}=-\\frac{4}{3}$, which gives $s=\\frac{4}{5}$. The area of triangle $A B C$ is therefore $\\frac{1}{2} \\cdot A B \\cdot A C \\cdot \\sin \\angle B A C=\\frac{1}{2} \\cdot 5 \\cdot 6 \\cdot \\frac{4}{5}=\\mathbf{1 2}$."
] | [
"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 | Geometry | Math | English |
2,912 | Compute the number of positive integers less than 25 that cannot be written as the difference of two squares of integers. | [
"Suppose $n=a^{2}-b^{2}=(a+b)(a-b)$, where $a$ and $b$ are integers. Because $a+b$ and $a-b$ differ by an even number, they have the same parity. Thus $n$ must be expressible as the product of two even integers or two odd integers. This condition is sufficient for $n$ to be a difference of squares, because if $n$ i... | [
"6"
] | 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 |
2,913 | For digits $A, B$, and $C,(\underline{A} \underline{B})^{2}+(\underline{A} \underline{C})^{2}=1313$. Compute $A+B+C$. | [
"Because $10 A \\leq \\underline{A} \\underline{B}<10(A+1), 200 A^{2}<(\\underline{A} \\underline{B})^{2}+(\\underline{A} \\underline{C})^{2}<200(A+1)^{2}$. So $200 A^{2}<$ $1313<200(A+1)^{2}$, and $A=2$. Note that $B$ and $C$ must have opposite parity, so without loss of generality, assume that $B$ is even. Consid... | [
"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 |
2,914 | Points $P, Q, R$, and $S$ lie in the interior of square $A B C D$ such that triangles $A B P, B C Q$, $C D R$, and $D A S$ are equilateral. If $A B=1$, compute the area of quadrilateral $P Q R S$. | [
"$P Q R S$ is a square with diagonal $\\overline{R P}$. Extend $\\overline{R P}$ to intersect $\\overline{A B}$ and $\\overline{C D}$ at $M$ and $N$ respectively, as shown in the diagram below.\n\n<img_3457>\n\nThen $\\overline{M P}$ is an altitude of $\\triangle A B P$ and $\\overline{R N}$ is an altitude of $\\tr... | [
"$2-\\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 |
2,915 | For real numbers $\alpha, B$, and $C$, the zeros of $T(x)=x^{3}+x^{2}+B x+C \operatorname{are~}^{2} \alpha$, $\cos ^{2} \alpha$, and $-\csc ^{2} \alpha$. Compute $T(5)$. | [
"Use the sum of the roots formula to obtain $\\sin ^{2} \\alpha+\\cos ^{2} \\alpha+-\\csc ^{2} \\alpha=-1$, so $\\csc ^{2} \\alpha=2$, and $\\sin ^{2} \\alpha=\\frac{1}{2}$. Therefore $\\cos ^{2} \\alpha=\\frac{1}{2}$. T(x) has leading coefficient 1 , so by the factor theorem, $T(x)=\\left(x-\\frac{1}{2}\\right)\\l... | [
"$\\frac{567}{4}$"
] | 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 |
2,916 | Let $\mathcal{R}$ denote the circular region bounded by $x^{2}+y^{2}=36$. The lines $x=4$ and $y=3$ partition $\mathcal{R}$ into four regions $\mathcal{R}_{1}, \mathcal{R}_{2}, \mathcal{R}_{3}$, and $\mathcal{R}_{4}$. $\left[\mathcal{R}_{i}\right]$ denotes the area of region $\mathcal{R}_{i}$. If $\left[\mathcal{R}_{1}\right]>\left[\mathcal{R}_{2}\right]>\left[\mathcal{R}_{3}\right]>\left[\mathcal{R}_{4}\right]$, compute $\left[\mathcal{R}_{1}\right]-\left[\mathcal{R}_{2}\right]-\left[\mathcal{R}_{3}\right]+\left[\mathcal{R}_{4}\right]$. | [
"Draw the lines $x=-4$ and $y=-3$, creating regions $\\mathcal{R}_{21}, \\mathcal{R}_{22}, \\mathcal{R}_{11}, \\mathcal{R}_{12}, \\mathcal{R}_{13}, \\mathcal{R}_{14}$ as shown below.\n\n<img_3593>\n\n\n\nThen $\\left[\\mathcal{R}_{21}\\right]=\\left[\\mathcal{R}_{4}\\right]=\\left[\\mathcal{R}_{13}\\right],\\left[\... | [
"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 |
2,917 | Let $x$ be a real number in the interval $[0,360]$ such that the four expressions $\sin x^{\circ}, \cos x^{\circ}$, $\tan x^{\circ}, \cot x^{\circ}$ take on exactly three distinct (finite) real values. Compute the sum of all possible values of $x$. | [
"If the four expressions take on three different values, exactly two of the expressions must have equal values. There are $\\left(\\begin{array}{l}4 \\\\ 2\\end{array}\\right)=6$ cases to consider:\n\nCase 1: $\\sin x^{\\circ}=\\cos x^{\\circ}$ : Then $\\tan x^{\\circ}=\\cot x^{\\circ}=1$, violating the condition t... | [
"990"
] | 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 |
2,918 | Let $a_{1}, a_{2}, a_{3}, \ldots$ be an arithmetic sequence, and let $b_{1}, b_{2}, b_{3}, \ldots$ be a geometric sequence. The sequence $c_{1}, c_{2}, c_{3}, \ldots$ has $c_{n}=a_{n}+b_{n}$ for each positive integer $n$. If $c_{1}=1, c_{2}=4, c_{3}=15$, and $c_{4}=2$, compute $c_{5}$. | [
"Let $a_{2}-a_{1}=d$ and $\\frac{b_{2}}{b_{1}}=r$. Using $a=a_{1}$ and $b=b_{1}$, write the system of equations:\n\n$$\n\\begin{aligned}\na+b & =1 \\\\\n(a+d)+b r & =4 \\\\\n(a+2 d)+b r^{2} & =15 \\\\\n(a+3 d)+b r^{3} & =2 .\n\\end{aligned}\n$$\n\nSubtract the first equation from the second, the second from the thi... | [
"61"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,919 | In square $A B C D$ with diagonal $1, E$ is on $\overline{A B}$ and $F$ is on $\overline{B C}$ with $\mathrm{m} \angle B C E=\mathrm{m} \angle B A F=$ $30^{\circ}$. If $\overline{C E}$ and $\overline{A F}$ intersect at $G$, compute the distance between the incenters of triangles $A G E$ and $C G F$. | [
"Let $M$ be the midpoint of $\\overline{A G}$, and $I$ the incenter of $\\triangle A G E$ as shown below.\n\n<img_3715>\n\nBecause $\\frac{A B}{A C}=\\sin 45^{\\circ}$ and $\\frac{E B}{A B}=\\frac{E B}{B C}=\\tan 30^{\\circ}$,\n\n$$\n\\begin{aligned}\nA E & =A B-E B=A B\\left(1-\\tan 30^{\\circ}\\right) \\\\\n& =\\... | [
"$4-2 \\sqrt{3}$"
] | 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 |
2,920 | Let $a, b, m, n$ be positive integers with $a m=b n=120$ and $a \neq b$. In the coordinate plane, let $A=(a, m), B=(b, n)$, and $O=(0,0)$. If $X$ is a point in the plane such that $A O B X$ is a parallelogram, compute the minimum area of $A O B X$. | [
"The area of parallelogram $A O B X$ is given by the absolute value of the cross product $|\\langle a, m\\rangle \\times\\langle b, n\\rangle|=|a n-m b|$. Because $m=\\frac{120}{a}$ and $n=\\frac{120}{b}$, the desired area of $A O B X$ equals $120\\left|\\frac{a}{b}-\\frac{b}{a}\\right|$. Note that the function $f(... | [
"44"
] | 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 |
2,921 | Let $\mathcal{S}$ be the set of integers from 0 to 9999 inclusive whose base- 2 and base- 5 representations end in the same four digits. (Leading zeros are allowed, so $1=0001_{2}=0001_{5}$ is one such number.) Compute the remainder when the sum of the elements of $\mathcal{S}$ is divided by 10,000. | [
"The remainders of an integer $N$ modulo $2^{4}=16$ and $5^{4}=625$ uniquely determine its remainder modulo 10000. There are only 16 strings of four 0's and 1's. In addition, because 16 and 625 are relatively prime, it will be shown below that for each such string $s$, there exists exactly one integer $x_{s}$ in th... | [
"6248"
] | 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 |
2,922 | If $A, R, M$, and $L$ are positive integers such that $A^{2}+R^{2}=20$ and $M^{2}+L^{2}=10$, compute the product $A \cdot R \cdot M \cdot L$. | [
"The only positive integers whose squares sum to 20 are 2 and 4 . The only positive integers whose squares sum to 10 are 1 and 3 . Thus $A \\cdot R=8$ and $M \\cdot L=3$, so $A \\cdot R \\cdot M \\cdot L=\\mathbf{2 4}$."
] | [
"24"
] | 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 |
2,924 | Let $T=49$. Compute the last digit, in base 10, of the sum
$$
T^{2}+(2 T)^{2}+(3 T)^{2}+\ldots+\left(T^{2}\right)^{2}
$$ | [
"Let $S$ be the required sum. Factoring $T^{2}$ from the sum yields\n\n$$\n\\begin{aligned}\nS & =T^{2}\\left(1+4+9+\\ldots+T^{2}\\right) \\\\\n& =T^{2}\\left(\\frac{T(T+1)(2 T+1)}{6}\\right) \\\\\n& =\\frac{T^{3}(T+1)(2 T+1)}{6} .\n\\end{aligned}\n$$\n\nFurther analysis makes the final computation simpler. If $T \... | [
"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 |
2,925 | A fair coin is flipped $n$ times. Compute the smallest positive integer $n$ for which the probability that the coin has the same result every time is less than $10 \%$. | [
"After the first throw, the probability that the succeeding $n-1$ throws have the same result is $\\frac{1}{2^{n-1}}$. Thus $\\frac{1}{2^{n-1}}<\\frac{1}{10} \\Rightarrow 2^{n-1}>10 \\Rightarrow n-1 \\geq 4$, so $n=5$ is the smallest possible value."
] | [
"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 | Combinatorics | Math | English |
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