id int64 | subfield string | context null | question string | solution list | final_answer list | is_multiple_answer bool | unit string | answer_type string | error string | answer string |
|---|---|---|---|---|---|---|---|---|---|---|
2,764 | Number Theory | null | 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"
] | false | null | Numerical | null | \boxed{-7} |
2,727 | Geometry | null | Let $T=T N Y W R$. 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"
] | false | null | Numerical | null | \boxed{184} |
2,344 | Algebra | null | The geometric sequence with $n$ terms $t_{1}, t_{2}, \ldots, t_{n-1}, t_{n}$ has $t_{1} t_{n}=3$. Also, the product of all $n$ terms equals 59049 (that is, $t_{1} t_{2} \cdots t_{n-1} t_{n}=59049$ ). Determine the value of $n$.
(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.) | [
"Suppose that the first term in the geometric sequence is $t_{1}=a$ and the common ratio in the sequence is $r$.\n\nThen the sequence, which has $n$ terms, is $a, a r, a r^{2}, a r^{3}, \\ldots, a r^{n-1}$.\n\nIn general, the $k$ th term is $t_{k}=a r^{k-1}$; in particular, the $n$th term is $t_{n}=a r^{n-1}$.\n\nS... | [
"20"
] | false | null | Numerical | null | \boxed{20} |
2,582 | Combinatorics | null | Eight people, including triplets Barry, Carrie and Mary, are going for a trip in four canoes. Each canoe seats two people. The eight people are to be randomly assigned to the four canoes in pairs. What is the probability that no two of Barry, Carrie and Mary will be in the same canoe? | [
"Among a group of $n$ people, there are $\\frac{n(n-1)}{2}$ ways of choosing a pair of these people:\n\nThere are $n$ people that can be chosen first.\n\nFor each of these $n$ people, there are $n-1$ people that can be chosen second.\n\nThis gives $n(n-1)$ orderings of two people.\n\nEach pair is counted twice (giv... | [
"$\\frac{4}{7}$"
] | false | null | Numerical | null | \boxed{\frac{4}{7}} |
2,831 | Geometry | null | 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"
] | false | null | Numerical | null | \boxed{20} |
2,212 | Algebra | null | Given a positive integer $n \geq 2$, determine the largest positive integer $N$ for which there exist $N+1$ real numbers $a_{0}, a_{1}, \ldots, a_{N}$ such that
(1) $a_{0}+a_{1}=-\frac{1}{n}$, and
(2) $\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ for $1 \leq k \leq N-1$. | [
"$\\left(a_{k}+a_{k-1}\\right)\\left(a_{k}+a_{k+1}\\right)=a_{k-1}-a_{k+1}$ is equivalent to $\\left(a_{k}+a_{k-1}+1\\right)\\left(a_{k}+a_{k+1}-1\\right)=-1$. Let $b_{k}=a_{k}+a_{k+1}$. Thus we need $b_{0}, b_{1}, \\ldots$ the following way: $b_{0}=-\\frac{1}{n}$ and $\\left(b_{k-1}+1\\right)\\left(b_{k}-1\\right)... | [
"$N=n$"
] | false | null | Expression | null | \boxed{N=n} |
3,013 | Algebra | null | Consider the system of equations
$$
\begin{aligned}
& \log _{4} x+\log _{8}(y z)=2 \\
& \log _{4} y+\log _{8}(x z)=4 \\
& \log _{4} z+\log _{8}(x y)=5 .
\end{aligned}
$$
Given that $x y z$ can be expressed in the form $2^{k}$, compute $k$. | [
"Note that for $n>0, \\log _{4} n=\\log _{64} n^{3}$ and $\\log _{8} n=\\log _{64} n^{2}$. Adding together the three given equations and using both the preceding facts and properties of logarithms yields\n\n$$\n\\begin{aligned}\n& \\log _{4}(x y z)+\\log _{8}\\left(x^{2} y^{2} z^{2}\\right)=11 \\\\\n\\Longrightarro... | [
"$\\frac{66}{7}$"
] | false | null | Numerical | null | \boxed{\frac{66}{7}} |
2,778 | Number Theory | null | 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"
] | false | null | Numerical | null | \boxed{1} |
2,772 | Number Theory | null | 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"
] | false | null | Numerical | null | \boxed{2069} |
2,238 | Algebra | null | Determine all integer values of $x$ such that $\left(x^{2}-3\right)\left(x^{2}+5\right)<0$. | [
"Since $x^{2} \\geq 0$ for all $x, x^{2}+5>0$. Since $\\left(x^{2}-3\\right)\\left(x^{2}+5\\right)<0, x^{2}-3<0$, so $x^{2}<3$ or $-\\sqrt{3}<x<\\sqrt{3}$. Thus $x=-1,0,1$."
] | [
"-1,0,1"
] | true | null | Numerical | null | \boxed{-1,0,1} |
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