Can LLMs Reliably Grade Olympiad Proofs? A Study of Mathematical Verification with LLMs
Paper • 2610.36767 • Published
Datasets:
global_idx int64 | problem_idx string | problem string | awarded_points int64 | solution string | sample_solution string | justification string | grading_scheme list |
|---|---|---|---|---|---|---|---|
1 | 1 | Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n\gt N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$. | 2 |
We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the follow... | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the followi... | Incorrect floor and modulo swap operation, resulting in loss of 3 points under item "Interchanging the floor and mod". Removed the critical observation that the numerator ($n^{k-i} \pmod{2^{i+1}n}$) is a multiple of $n$. Since the solution does not mention this observation, the bound on the expression is incomplete, an... | [
{
"desc": "Uses the base-2n representation of $n^k$ to find a closed form expression of each digit $a_i=\\left\\lfloor \\frac{n^k}{(2n)^i}\\right\\rfloor mod\\ (2n)$",
"points": 1,
"title": "Initial closed form expression for each digit"
},
{
"desc": "Transforms the digit expression into a easil... |
2 | 1 | Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n\gt N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$. | 1 | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the followi... | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the followi... |
The identity $\lfloor \frac{x}{y} \rfloor \pmod m = \lfloor \frac{x \pmod {my}}{y} \rfloor$ was applied incorrectly. The bounding logic now fails. With the modulus reduced to $2n$, the numerator $n^{k-i} \pmod{2n}$ evaluates to $n$ (for odd $n$). While this technically yields $a_i = \lfloor n/2^i \rfloor$, the subsequ... | [
{
"desc": "Uses the base-2n representation of $n^k$ to find a closed form expression of each digit $a_i=\\left\\lfloor \\frac{n^k}{(2n)^i}\\right\\rfloor mod\\ (2n)$",
"points": 1,
"title": "Initial closed form expression for each digit"
},
{
"desc": "Transforms the digit expression into a easil... |
3 | 1 | Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n\gt N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$. | 2 | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the following... | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the followi... | Paraphrased proof that follows sample solution and markscheme. | [
{
"desc": "Uses the base-2n representation of $n^k$ to find a closed form expression of each digit $a_i=\\left\\lfloor \\frac{n^k}{(2n)^i}\\right\\rfloor mod\\ (2n)$",
"points": 1,
"title": "Initial closed form expression for each digit"
},
{
"desc": "Transforms the digit expression into a easil... |
4 | 1 | Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n\gt N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$. | 6 |
We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the following:$$ a_i=\left\lfloor\frac{n^k}{(2n)^i}\right\rfloor\text{... | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the followi... | No $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. Results in loss of point in the last markscheme item "Finishing". | [
{
"desc": "Uses the base-2n representation of $n^k$ to find a closed form expression of each digit $a_i=\\left\\lfloor \\frac{n^k}{(2n)^i}\\right\\rfloor mod\\ (2n)$",
"points": 1,
"title": "Initial closed form expression for each digit"
},
{
"desc": "Transforms the digit expression into a easil... |
5 | 1 | Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n\gt N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$. | 1 | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the following:$$ a_i=\left\lfloor\frac{n^k}{(2n)^i}\right\rfloor\text{ ... | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the followi... | Incorrect floor and modulo swap operation, resulting in loss of 3 points under item "Interchanging the floor and mod". Another point lost for not mentioning $n^k\lt (2n)^k$ condition under item
"Finishing" of the markscheme. | [
{
"desc": "Uses the base-2n representation of $n^k$ to find a closed form expression of each digit $a_i=\\left\\lfloor \\frac{n^k}{(2n)^i}\\right\\rfloor mod\\ (2n)$",
"points": 1,
"title": "Initial closed form expression for each digit"
},
{
"desc": "Transforms the digit expression into a easil... |
6 | 1 | Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n\gt N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$. | 5 |
We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the follow... | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the followi... | No critical observation that the numerator ($n^{k-i} \pmod{2^{i+1}n}$) is a multiple of $n$. Since the solution does not mention this observation, the bound on the expression is incomplete, and proof does not have sufficient evidence, resulting in 2 point deduction under item "Bounding the expression".
| [
{
"desc": "Uses the base-2n representation of $n^k$ to find a closed form expression of each digit $a_i=\\left\\lfloor \\frac{n^k}{(2n)^i}\\right\\rfloor mod\\ (2n)$",
"points": 1,
"title": "Initial closed form expression for each digit"
},
{
"desc": "Transforms the digit expression into a easil... |
7 | 1 | Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n\gt N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$. | 4 | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the followi... | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the followi... | Incorrect floor and modulo swap operation, resulting in loss of 3 points under item "Interchanging the floor and mod". | [
{
"desc": "Uses the base-2n representation of $n^k$ to find a closed form expression of each digit $a_i=\\left\\lfloor \\frac{n^k}{(2n)^i}\\right\\rfloor mod\\ (2n)$",
"points": 1,
"title": "Initial closed form expression for each digit"
},
{
"desc": "Transforms the digit expression into a easil... |
8 | 1 | Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n\gt N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$. | 0 | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the following... | We claim one such $N$ is $\boxed{N=2^{k-1}(d+1)}$. Begin by letting $a_i$ be the digit at the $i$th position from the right (0-indexed) of $n^k$ in base $2n$; i.e., $a_0$ is the units digit. Obviously, $a_i=0$ for all $i\ge k$, since $n^k\lt (2n)^k$. We WTS $a_i\gt d$ for each $i=0,1,\dots,k-1$, and observe the followi... | Proof uses base-n, instead of 2n, so the construction is wrong. Subsequent steps are based on flawed logic. | [
{
"desc": "Uses the base-2n representation of $n^k$ to find a closed form expression of each digit $a_i=\\left\\lfloor \\frac{n^k}{(2n)^i}\\right\\rfloor mod\\ (2n)$",
"points": 1,
"title": "Initial closed form expression for each digit"
},
{
"desc": "Transforms the digit expression into a easil... |
9 | 2 | Let $n$ and $k$ be positive integers with $k\lt n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$... | 7 | Assume for the sake of contradiction that $P$ has $n$ real roots. Without loss of generality, let $P$ be monic, such that $P(x)=\prod_{i=1}^n (x-r_i)$. We may assume $k=n-1$; otherwise, we simply consider the sub-polynomial formed by the first $k+1$ roots instead of $P$.
By the problem statement, the polynomial $\frac... | Assume FTSOC $P$ has $n$ real roots. WLOG $P$ has leading coefficient $1$ so let $P(x)=:(x-r_1)\cdots(x-r_n)$. WLOG $k=n-1$, as otherwise we can take $Q:=(x-r_1)\cdots(x-r_{k+1})$ and deal with $Q$ instead of $P$.
By the assertion, each $\frac{P(x)}{x-r_i}$ for $1\leq i\leq n$ has a coefficient equal to $0$. Since $r_... | Paraphrased proof that follows sample solution and markscheme. | [
{
"desc": "Identifies 2 polynomials which share a coefficient of 0 at the same power of $x$.",
"points": 1,
"title": "Pigeonhole principle"
},
{
"desc": "Finds a divisor polynomial that has 2 adjacent zero coefficients.",
"points": 2,
"title": "Adjacent zero coefficients"
},
{
"d... |
10 | 2 | Let $n$ and $k$ be positive integers with $k\lt n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$... | 0 | Assume for the sake of contradiction that $P$ has $n$ real roots. Without loss of generality, let $P$ be monic, such that $P(x)=\prod_{i=1}^n (x-r_i)$. We may assume $k=n-1$; otherwise, we simply consider the sub-polynomial formed by the first $k+1$ roots instead of $P$.By the problem statement, the polynomial $\frac{P... | Assume FTSOC $P$ has $n$ real roots. WLOG $P$ has leading coefficient $1$ so let $P(x)=:(x-r_1)\cdots(x-r_n)$. WLOG $k=n-1$, as otherwise we can take $Q:=(x-r_1)\cdots(x-r_{k+1})$ and deal with $Q$ instead of $P$.
By the assertion, each $\frac{P(x)}{x-r_i}$ for $1\leq i\leq n$ has a coefficient equal to $0$. Since $r_... | Wrong application of Pigeonhole principle, which results in wrong proof logic. | [
{
"desc": "Identifies 2 polynomials which share a coefficient of 0 at the same power of $x$.",
"points": 1,
"title": "Pigeonhole principle"
},
{
"desc": "Finds a divisor polynomial that has 2 adjacent zero coefficients.",
"points": 2,
"title": "Adjacent zero coefficients"
},
{
"d... |
11 | 2 | Let $n$ and $k$ be positive integers with $k\lt n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$... | 4 |
Assume for the sake of contradiction that $P$ has $n$ real roots. Without loss of generality, let $P$ be monic, such that $P(x)=\prod_{i=1}^n (x-r_i)$. We may assume $k=n-1$; otherwise, we simply consider the sub-polynomial formed by the first $k+1$ roots instead of $P$.By the problem statement, the polynomial $\frac{... | Assume FTSOC $P$ has $n$ real roots. WLOG $P$ has leading coefficient $1$ so let $P(x)=:(x-r_1)\cdots(x-r_n)$. WLOG $k=n-1$, as otherwise we can take $Q:=(x-r_1)\cdots(x-r_{k+1})$ and deal with $Q$ instead of $P$.
By the assertion, each $\frac{P(x)}{x-r_i}$ for $1\leq i\leq n$ has a coefficient equal to $0$. Since $r_... | Solution identifies missing terms and double root at 0; however, it claims that double root does not contradict Rolle's theorem, which is false. | [
{
"desc": "Identifies 2 polynomials which share a coefficient of 0 at the same power of $x$.",
"points": 1,
"title": "Pigeonhole principle"
},
{
"desc": "Finds a divisor polynomial that has 2 adjacent zero coefficients.",
"points": 2,
"title": "Adjacent zero coefficients"
},
{
"d... |
12 | 2 | Let $n$ and $k$ be positive integers with $k\lt n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$... | 1 |
Assume for the sake of contradiction that $P$ has $n$ real roots. Without loss of generality, let $P$ be monic, such that $P(x)=\prod_{i=1}^n (x-r_i)$. We may assume $k=n-1$; otherwise, we simply consider the sub-polynomial formed by the first $k+1$ roots instead of $P$.By the problem statement, the polynomial $\frac{... | Assume FTSOC $P$ has $n$ real roots. WLOG $P$ has leading coefficient $1$ so let $P(x)=:(x-r_1)\cdots(x-r_n)$. WLOG $k=n-1$, as otherwise we can take $Q:=(x-r_1)\cdots(x-r_{k+1})$ and deal with $Q$ instead of $P$.
By the assertion, each $\frac{P(x)}{x-r_i}$ for $1\leq i\leq n$ has a coefficient equal to $0$. Since $r_... | Point awarded only for Pigeon's hole principle, since linear combination $r_nR(x) - r_{n-1}Q(x)$ is incorrect (missing an $x$ term). | [
{
"desc": "Identifies 2 polynomials which share a coefficient of 0 at the same power of $x$.",
"points": 1,
"title": "Pigeonhole principle"
},
{
"desc": "Finds a divisor polynomial that has 2 adjacent zero coefficients.",
"points": 2,
"title": "Adjacent zero coefficients"
},
{
"d... |
13 | 2 | Let $n$ and $k$ be positive integers with $k\lt n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$... | 3 | Assume for the sake of contradiction that $P$ has $n$ real roots. Without loss of generality, let $P$ be monic, such that $P(x)=\prod_{i=1}^n (x-r_i)$. We may assume $k=n-1$; otherwise, we simply consider the sub-polynomial formed by the first $k+1$ roots instead of $P$.By the problem statement, the polynomial $\frac{P... | Assume FTSOC $P$ has $n$ real roots. WLOG $P$ has leading coefficient $1$ so let $P(x)=:(x-r_1)\cdots(x-r_n)$. WLOG $k=n-1$, as otherwise we can take $Q:=(x-r_1)\cdots(x-r_{k+1})$ and deal with $Q$ instead of $P$.
By the assertion, each $\frac{P(x)}{x-r_i}$ for $1\leq i\leq n$ has a coefficient equal to $0$. Since $r_... | Solution fails to establish multiplicity of 2 necessary to derive a contradiction at an inductive step. The rest of the proof therefore is invalid. | [
{
"desc": "Identifies 2 polynomials which share a coefficient of 0 at the same power of $x$.",
"points": 1,
"title": "Pigeonhole principle"
},
{
"desc": "Finds a divisor polynomial that has 2 adjacent zero coefficients.",
"points": 2,
"title": "Adjacent zero coefficients"
},
{
"d... |
14 | 2 | Let $n$ and $k$ be positive integers with $k\lt n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$... | 5 |
Assume for the sake of contradiction that $P$ has $n$ real roots. Without loss of generality, let $P$ be monic, such that $P(x)=\prod_{i=1}^n (x-r_i)$. We may assume $k=n-1$; otherwise, we simply consider the sub-polynomial formed by the first $k+1$ roots instead of $P$.By the problem statement, the polynomial $\frac{... | Assume FTSOC $P$ has $n$ real roots. WLOG $P$ has leading coefficient $1$ so let $P(x)=:(x-r_1)\cdots(x-r_n)$. WLOG $k=n-1$, as otherwise we can take $Q:=(x-r_1)\cdots(x-r_{k+1})$ and deal with $Q$ instead of $P$.
By the assertion, each $\frac{P(x)}{x-r_i}$ for $1\leq i\leq n$ has a coefficient equal to $0$. Since $r_... | Solution attempts to apply Rolle's theorem, but takes $\alpha$-th derivative, instead of $\alpha-1$-th derivative, resulting in a 2 point deduction. | [
{
"desc": "Identifies 2 polynomials which share a coefficient of 0 at the same power of $x$.",
"points": 1,
"title": "Pigeonhole principle"
},
{
"desc": "Finds a divisor polynomial that has 2 adjacent zero coefficients.",
"points": 2,
"title": "Adjacent zero coefficients"
},
{
"d... |
15 | 2 | Let $n$ and $k$ be positive integers with $k\lt n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$... | 2 | Assume for the sake of contradiction that $P$ has $n$ real roots. Without loss of generality, let $P$ be monic, such that $P(x)=\prod_{i=1}^n (x-r_i)$. We may assume $k=n-1$; otherwise, we simply consider the sub-polynomial formed by the first $k+1$ roots instead of $P$.By the problem statement, the polynomial $\frac{P... | Assume FTSOC $P$ has $n$ real roots. WLOG $P$ has leading coefficient $1$ so let $P(x)=:(x-r_1)\cdots(x-r_n)$. WLOG $k=n-1$, as otherwise we can take $Q:=(x-r_1)\cdots(x-r_{k+1})$ and deal with $Q$ instead of $P$.
By the assertion, each $\frac{P(x)}{x-r_i}$ for $1\leq i\leq n$ has a coefficient equal to $0$. Since $r_... | Missing $x^{\alpha-1}$ term necessary to establish root distinctness. | [
{
"desc": "Identifies 2 polynomials which share a coefficient of 0 at the same power of $x$.",
"points": 1,
"title": "Pigeonhole principle"
},
{
"desc": "Finds a divisor polynomial that has 2 adjacent zero coefficients.",
"points": 2,
"title": "Adjacent zero coefficients"
},
{
"d... |
16 | 2 | Let $n$ and $k$ be positive integers with $k\lt n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$... | 7 |
Assume for the sake of contradiction that $P$ has $n$ real roots. Without loss of generality, let $P$ be monic, such that $P(x)=\prod_{i=1}^n (x-r_i)$. We may assume $k=n-1$; otherwise, we simply consider the sub-polynomial formed by the first $k+1$ roots instead of $P$.By the problem statement, the polynomial $\frac{... | Assume FTSOC $P$ has $n$ real roots. WLOG $P$ has leading coefficient $1$ so let $P(x)=:(x-r_1)\cdots(x-r_n)$. WLOG $k=n-1$, as otherwise we can take $Q:=(x-r_1)\cdots(x-r_{k+1})$ and deal with $Q$ instead of $P$.
By the assertion, each $\frac{P(x)}{x-r_i}$ for $1\leq i\leq n$ has a coefficient equal to $0$. Since $r_... | Number of possible powers is wrong: $n-1$ instead of $n-2$, since the proof does not take into account the constant term. | [
{
"desc": "Identifies 2 polynomials which share a coefficient of 0 at the same power of $x$.",
"points": 1,
"title": "Pigeonhole principle"
},
{
"desc": "Finds a divisor polynomial that has 2 adjacent zero coefficients.",
"points": 2,
"title": "Adjacent zero coefficients"
},
{
"d... |
17 | 3 | Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of... | 0 |
The winning strategy belongs to Alice. We define a ``Bob-set'' $V \subseteq \mathbb{R}^2$ as a collection of points where no three are collinear and the pairwise distance between any two points is at least $1$. The core of the argument rests on the following property.
\textbf{Claim:} Construct a ``Bob-graph'' with ve... | The answer is that Alice wins. Let's define a Bob-set $V$ to be a set of points in the plane with no three collinear and with all distances at least $1$. The point of the problem is to prove the following fact.
Claim: Given a Bob-set $V \subseteq {\mathbb R}^2$, consider the Bob-graph with vertex set $V$ defined as fol... | Contradiction in the proof of connectivity is wrong, since the specified sequence is bounded by an increasing value (instead of decreasing), which does not force $\delta_i^2<0$. | [
{
"desc": "The given construction is correct and induces a graph that is connected.",
"points": 3,
"title": "Connectivity of construction"
},
{
"desc": "The given construction induces a graph where the shortest path between any two nodes is finite.",
"points": 1,
"title": "Finiteness of ... |
18 | 3 | Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of... | 7 | Alice has a winning strategy: Choose $\mathcal{S}$ to be the set of points strictly outside the disk with diameter $PQ$. Then two cities $A$ and $B$ have a road between them if and only if every other city is outside the closed disk with diameter $AB$. Note that this condition is equivalent to the angle $\angle ACB$ be... | The answer is that Alice wins. Let's define a Bob-set $V$ to be a set of points in the plane with no three collinear and with all distances at least $1$. The point of the problem is to prove the following fact.
Claim: Given a Bob-set $V \subseteq {\mathbb R}^2$, consider the Bob-graph with vertex set $V$ defined as fol... | Paraphrased proof that follows sample solution and markscheme. | [
{
"desc": "The given construction is correct and induces a graph that is connected.",
"points": 3,
"title": "Connectivity of construction"
},
{
"desc": "The given construction induces a graph where the shortest path between any two nodes is finite.",
"points": 1,
"title": "Finiteness of ... |
19 | 3 | Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of... | 1 | The winning strategy belongs to Alice. We define a ``Bob-set'' $V \subseteq \mathbb{R}^2$ as a collection of points where no three are collinear and the pairwise distance between any two points is at least $1$. The core of the argument rests on the following property.
\textbf{Claim:} Construct a ``Bob-graph'' with ver... | The answer is that Alice wins. Let's define a Bob-set $V$ to be a set of points in the plane with no three collinear and with all distances at least $1$. The point of the problem is to prove the following fact.
Claim: Given a Bob-set $V \subseteq {\mathbb R}^2$, consider the Bob-graph with vertex set $V$ defined as fol... | The construction of this proof is wrong. This proof operates on Visibility graph, not Gabriel Graph (proof mentions line segment $\overline{ab}$, not a closed disk with diameter $\overline{ab}$). The only correct part relies on finiteness of this graph; whereas connectivity and planarity proof are wrong. | [
{
"desc": "The given construction is correct and induces a graph that is connected.",
"points": 3,
"title": "Connectivity of construction"
},
{
"desc": "The given construction induces a graph where the shortest path between any two nodes is finite.",
"points": 1,
"title": "Finiteness of ... |
20 | 3 | Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of... | 4 |
The winning strategy belongs to Alice. We define a ``Bob-set'' $V \subseteq \mathbb{R}^2$ as a collection of points where no three are collinear and the pairwise distance between any two points is at least $1$. The core of the argument rests on the following property.
\textbf{Claim:} Construct a ``Bob-graph'' with ve... | The answer is that Alice wins. Let's define a Bob-set $V$ to be a set of points in the plane with no three collinear and with all distances at least $1$. The point of the problem is to prove the following fact.
Claim: Given a Bob-set $V \subseteq {\mathbb R}^2$, consider the Bob-graph with vertex set $V$ defined as fol... | The final step of the proof incorrectly argues that an acute angle in a convex quadrilateral places the vertex inside the disk, which is false, since acute angle places the vertex outside the circle. This error causes loss of 2 points under "Planarity" section of the proof. | [
{
"desc": "The given construction is correct and induces a graph that is connected.",
"points": 3,
"title": "Connectivity of construction"
},
{
"desc": "The given construction induces a graph where the shortest path between any two nodes is finite.",
"points": 1,
"title": "Finiteness of ... |
21 | 3 | Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of... | 3 | The winning strategy belongs to Alice. We define a ``Bob-set'' $V \subseteq \mathbb{R}^2$ as a collection of points where no three are collinear and the pairwise distance between any two points is at least $1$. The core of the argument rests on the following property.
\textbf{Claim:} Construct a ``Bob-graph'' with ver... | The answer is that Alice wins. Let's define a Bob-set $V$ to be a set of points in the plane with no three collinear and with all distances at least $1$. The point of the problem is to prove the following fact.
Claim: Given a Bob-set $V \subseteq {\mathbb R}^2$, consider the Bob-graph with vertex set $V$ defined as fol... | Inequality sign in "Planarity" section of the proof is flipped, which invalidates the argument and finishing step, resulting in 4 point loss. | [
{
"desc": "The given construction is correct and induces a graph that is connected.",
"points": 3,
"title": "Connectivity of construction"
},
{
"desc": "The given construction induces a graph where the shortest path between any two nodes is finite.",
"points": 1,
"title": "Finiteness of ... |
22 | 3 | Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of... | 2 | The winning strategy belongs to Alice. We define a ``Bob-set'' $V \subseteq \mathbb{R}^2$ as a collection of points where no three are collinear and the pairwise distance between any two points is at least $1$. The core of the argument rests on the following property.
\textbf{Claim:} Construct a ``Bob-graph'' with ver... | The answer is that Alice wins. Let's define a Bob-set $V$ to be a set of points in the plane with no three collinear and with all distances at least $1$. The point of the problem is to prove the following fact.
Claim: Given a Bob-set $V \subseteq {\mathbb R}^2$, consider the Bob-graph with vertex set $V$ defined as fol... | Arguing that the angle at a new point is acute, invalidates the proof; since the proof revolves around an ascending inequality. The rest of the proof is therefore invalid. "Planarity" section of the proof is the only correct conclusion that can be derived from the solution. | [
{
"desc": "The given construction is correct and induces a graph that is connected.",
"points": 3,
"title": "Connectivity of construction"
},
{
"desc": "The given construction induces a graph where the shortest path between any two nodes is finite.",
"points": 1,
"title": "Finiteness of ... |
23 | 3 | Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of... | 6 | The winning strategy belongs to Alice. We define a ``Bob-set'' $V \subseteq \mathbb{R}^2$ as a collection of points where no three are collinear and the pairwise distance between any two points is at least $1$. The core of the argument rests on the following property.
\textbf{Claim:} Construct a ``Bob-graph'' with ver... | The answer is that Alice wins. Let's define a Bob-set $V$ to be a set of points in the plane with no three collinear and with all distances at least $1$. The point of the problem is to prove the following fact.
Claim: Given a Bob-set $V \subseteq {\mathbb R}^2$, consider the Bob-graph with vertex set $V$ defined as fol... | The proof claims that connectivity alone creates the barrier. This is a logical gap: a graph can be connected but not separate the plane (if edges could cross). A point is deducted according to "Finishing" mark scheme item. | [
{
"desc": "The given construction is correct and induces a graph that is connected.",
"points": 3,
"title": "Connectivity of construction"
},
{
"desc": "The given construction induces a graph where the shortest path between any two nodes is finite.",
"points": 1,
"title": "Finiteness of ... |
24 | 3 | Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of... | 0 |
The winning strategy belongs to Bob. We define a ``Bob-set'' $V \subseteq \mathbb{R}^2$ as a collection of points where no three are collinear and the pairwise distance between any two points is at least $1$. The core of the argument rests on the following property.
\textbf{Claim:} Construct a ``Bob-graph'' with vert... | The answer is that Alice wins. Let's define a Bob-set $V$ to be a set of points in the plane with no three collinear and with all distances at least $1$. The point of the problem is to prove the following fact.
Claim: Given a Bob-set $V \subseteq {\mathbb R}^2$, consider the Bob-graph with vertex set $V$ defined as fol... | The solution is completely wrong, since it claims that Bob has a winning strategy. | [
{
"desc": "The given construction is correct and induces a graph that is connected.",
"points": 3,
"title": "Connectivity of construction"
},
{
"desc": "The given construction induces a graph where the shortest path between any two nodes is finite.",
"points": 1,
"title": "Finiteness of ... |
25 | 4 | Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$. | 3 |
Consider the line passing through $C$ perpendicular to $BC$. Let this line intersect the circumcircle $(ABC)$ at point $C'$. Define $M$ as the midpoint of the segment $CC'$. We assert that $M$ is the circumcenter of $\triangle AFP$. Note that $P$ is a point on $(ABC)$.
Observe that the quadrilateral $AC'CP$ forms an ... | Consider the line perpendicular to $BC$ at $C$. Let this line meet $(ABC)$ at $C'$. Let $M$ be the midpoint of $CC'$. We claim $M$ is the circumcenter of $AFP$. Note that $P$ lies on $(ABC)$.
Clearly $AC'CP$ is an isosceles trapezoid, so the perpendicular bisector of $AP$ meets $CC'$ at $M$. Now, since $\angle BCC' = ... | The proof shows that $AC'\perp CF$, not parallel. This results in a loss of points for construction. | [
{
"desc": "Constructs useful points. Is only granted if the given construction gives you useful properties that can be used to reach the solution.",
"points": 1,
"title": "Additional construction"
},
{
"desc": "Proves useful properties of the construction that involves some symmetry along the $(... |
26 | 4 | Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$. | 6 | Consider the line passing through $C$ perpendicular to $BC$. Let this line intersect the circumcircle $(ABC)$ at point $C'$. Define $M$ as the midpoint of the segment $CC'$. We assert that $M$ is the circumcenter of $\triangle AFP$. Note that $P$ is a point on $(ABC)$.Observe that the quadrilateral $AC'CP$ forms an iso... | Consider the line perpendicular to $BC$ at $C$. Let this line meet $(ABC)$ at $C'$. Let $M$ be the midpoint of $CC'$. We claim $M$ is the circumcenter of $AFP$. Note that $P$ lies on $(ABC)$.
Clearly $AC'CP$ is an isosceles trapezoid, so the perpendicular bisector of $AP$ meets $CC'$ at $M$. Now, since $\angle BCC' = ... | Minor mistake in the final argument, since equality of segments hold due to RHS congruence, not SAS. This results in a 1 point deduction under finalizing. | [
{
"desc": "Constructs useful points. Is only granted if the given construction gives you useful properties that can be used to reach the solution.",
"points": 1,
"title": "Additional construction"
},
{
"desc": "Proves useful properties of the construction that involves some symmetry along the $(... |
27 | 4 | Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$. | 2 |
Consider the line passing through $C$ perpendicular to $BC$. Let this line intersect the circumcircle $(ABC)$ at point $C'$. Define $M$ as the midpoint of the segment $CC'$. We assert that $M$ is the circumcenter of $\triangle AFP$. Note that $P$ is a point on $(ABC)$.
Observe that the quadrilateral $AC'CP$ forms a t... | Consider the line perpendicular to $BC$ at $C$. Let this line meet $(ABC)$ at $C'$. Let $M$ be the midpoint of $CC'$. We claim $M$ is the circumcenter of $AFP$. Note that $P$ lies on $(ABC)$.
Clearly $AC'CP$ is an isosceles trapezoid, so the perpendicular bisector of $AP$ meets $CC'$ at $M$. Now, since $\angle BCC' = ... | The proof relies on isosceles trapezoid, but never mentions it. Solution incorrectly implies that $AX' \perp CF$, and claims to use SAS congruence for final result, instead of RHS. | [
{
"desc": "Constructs useful points. Is only granted if the given construction gives you useful properties that can be used to reach the solution.",
"points": 1,
"title": "Additional construction"
},
{
"desc": "Proves useful properties of the construction that involves some symmetry along the $(... |
28 | 4 | Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$. | 7 | Consider the line passing through $C$ perpendicular to $BC$. Let this line intersect the circumcircle $(ABC)$ at point $C'$. Define $M$ as the midpoint of the segment $CC'$. We assert that $M$ is the circumcenter of $\triangle AFP$. Note that $P$ is a point on $(ABC)$.
Observe that the quadrilateral $AC'CP$ forms an i... | Consider the line perpendicular to $BC$ at $C$. Let this line meet $(ABC)$ at $C'$. Let $M$ be the midpoint of $CC'$. We claim $M$ is the circumcenter of $AFP$. Note that $P$ lies on $(ABC)$.
Clearly $AC'CP$ is an isosceles trapezoid, so the perpendicular bisector of $AP$ meets $CC'$ at $M$. Now, since $\angle BCC' = ... | Paraphrased proof that follows sample solution and markscheme. | [
{
"desc": "Constructs useful points. Is only granted if the given construction gives you useful properties that can be used to reach the solution.",
"points": 1,
"title": "Additional construction"
},
{
"desc": "Proves useful properties of the construction that involves some symmetry along the $(... |
29 | 4 | Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$. | 4 |
Consider the line passing through $C$ perpendicular to $BC$. Let this line intersect the circumcircle $(ABC)$ at point $C'$. Define $M$ as the midpoint of the segment $CC'$. We assert that $M$ is the circumcenter of $\triangle AFP$. Note that $P$ is a point on $(ABC)$.
Observe that the quadrilateral $AC'CP$ forms an ... | Consider the line perpendicular to $BC$ at $C$. Let this line meet $(ABC)$ at $C'$. Let $M$ be the midpoint of $CC'$. We claim $M$ is the circumcenter of $AFP$. Note that $P$ lies on $(ABC)$.
Clearly $AC'CP$ is an isosceles trapezoid, so the perpendicular bisector of $AP$ meets $CC'$ at $M$. Now, since $\angle BCC' = ... | The proof does not provide enough evidence that $AC'CF$ is a parallelogram, which disrupts the rest of the proof. | [
{
"desc": "Constructs useful points. Is only granted if the given construction gives you useful properties that can be used to reach the solution.",
"points": 1,
"title": "Additional construction"
},
{
"desc": "Proves useful properties of the construction that involves some symmetry along the $(... |
30 | 4 | Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$. | 1 | Consider the line passing through $C$ perpendicular to $BC$. Let this line intersect the circumcircle $(ABC)$ at point $C'$. Define $M$ as the midpoint of the segment $AC'$. We assert that $M$ is the circumcenter of $\triangle AFP$. Note that $P$ is a point on $(ABC)$.
Observe that the quadrilateral $AC'CP$ forms an i... | Consider the line perpendicular to $BC$ at $C$. Let this line meet $(ABC)$ at $C'$. Let $M$ be the midpoint of $CC'$. We claim $M$ is the circumcenter of $AFP$. Note that $P$ lies on $(ABC)$.
Clearly $AC'CP$ is an isosceles trapezoid, so the perpendicular bisector of $AP$ meets $CC'$ at $M$. Now, since $\angle BCC' = ... | Invalid construction of point $M$ on segment $AC'$, not $CC'$. Therefore, the rest of the proof has no base. | [
{
"desc": "Constructs useful points. Is only granted if the given construction gives you useful properties that can be used to reach the solution.",
"points": 1,
"title": "Additional construction"
},
{
"desc": "Proves useful properties of the construction that involves some symmetry along the $(... |
31 | 4 | Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$. | 2 |
Consider the line passing through $C$ perpendicular to $BC$. Let this line intersect the circumcircle $(ABC)$ at point $C'$. Define $M$ as the midpoint of the segment $CC'$. We assert that $M$ is the circumcenter of $\triangle AFP$. Note that $P$ is a point on $(ABC)$.Observe that the quadrilateral $AC'CP$ forms an is... | Consider the line perpendicular to $BC$ at $C$. Let this line meet $(ABC)$ at $C'$. Let $M$ be the midpoint of $CC'$. We claim $M$ is the circumcenter of $AFP$. Note that $P$ lies on $(ABC)$.
Clearly $AC'CP$ is an isosceles trapezoid, so the perpendicular bisector of $AP$ meets $CC'$ at $M$. Now, since $\angle BCC' = ... | Construction contains $C'$ and $M$, but the rest of the proof is completely False. | [
{
"desc": "Constructs useful points. Is only granted if the given construction gives you useful properties that can be used to reach the solution.",
"points": 1,
"title": "Additional construction"
},
{
"desc": "Proves useful properties of the construction that involves some symmetry along the $(... |
32 | 4 | Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$. | 4 | Consider the line passing through $C$ perpendicular to $BC$. Let this line intersect the circumcircle $(ABC)$ at point $C'$. Define $M$ as the midpoint of the segment $CC'$. We assert that $M$ is the circumcenter of $\triangle AFP$. Note that $P$ is a point on $(ABC)$.
Observe that the quadrilateral $AC'CP$ forms an i... | Consider the line perpendicular to $BC$ at $C$. Let this line meet $(ABC)$ at $C'$. Let $M$ be the midpoint of $CC'$. We claim $M$ is the circumcenter of $AFP$. Note that $P$ lies on $(ABC)$.
Clearly $AC'CP$ is an isosceles trapezoid, so the perpendicular bisector of $AP$ meets $CC'$ at $M$. Now, since $\angle BCC' = ... | Solution lacks a rigorous justification of $M$ being a circumcenter, resulting in a 2 point loss under "Properties of construction" item. | [
{
"desc": "Constructs useful points. Is only granted if the given construction gives you useful properties that can be used to reach the solution.",
"points": 1,
"title": "Additional construction"
},
{
"desc": "Proves useful properties of the construction that involves some symmetry along the $(... |
33 | 5 | Determine, with proof, all positive integers $k$ such that$$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for all positive integers $n.$ | 7 | We assert that the solution is $\boxed{\text{all even integers}}$.
\textbf{Proof that odd $k$ fail:}
Consider the case $n=2$. The condition requires $2 \mid \sum_{i=0}^{1} \binom{1}{i}^k = 1^k + 1^k = 2$.
Wait, looking closer at the original problem context implicitly provided:
If we use the original substitution $n-1... | We claim that the answer is $\boxed{\text{all even integers}}.$
Proof that odd $k$ fail: Just take $n = 2$ and get $2^k + 2$ is divisible by $3,$ which implies $2^k \equiv 1 \pmod{3}$ This can only happen if $k$ is even.
Proof that even $k$ work: Substitute $n-1$ in place for $n$ in the problem; we then must prove th... | Paraphrased proof that follows sample solution and markscheme. | [
{
"desc": "Uses Chinese Remainder Theorem to reduce the problem to a system of equations modulo $p_l^{e_l}$.",
"points": 1,
"title": "Reduction to powers of prime divisors of n+1"
},
{
"desc": "Represents the term $n \\choose i$ as a product of fractions $\\frac{n-j}{j}$ and finds the remainder ... |
34 | 5 | Determine, with proof, all positive integers $k$ such that$$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for all positive integers $n.$ | 5 | We assert that the solution is \boxed{\text{all even integers}}.
Proof that odd $k$ fail: Consider the case $n=2$. The condition requires $2 \mid \sum_{i=0}^{1} \binom{1}{i}^k = 1^k + 1^k = 2$.
If we use the original substitution $n-1 \to n$, for $n=2$ in the original problem, we specifically look at the divisibility ... | We claim that the answer is $\boxed{\text{all even integers}}.$
Proof that odd $k$ fail: Just take $n = 2$ and get $2^k + 2$ is divisible by $3,$ which implies $2^k \equiv 1 \pmod{3}$ This can only happen if $k$ is even.
Proof that even $k$ work: Substitute $n-1$ in place for $n$ in the problem; we then must prove th... | Reduction of the product into a binomial coefficient for the case where $p_l$ divides $j$ is off by one, which introduces a small error into inductive hypothesis. | [
{
"desc": "Uses Chinese Remainder Theorem to reduce the problem to a system of equations modulo $p_l^{e_l}$.",
"points": 1,
"title": "Reduction to powers of prime divisors of n+1"
},
{
"desc": "Represents the term $n \\choose i$ as a product of fractions $\\frac{n-j}{j}$ and finds the remainder ... |
35 | 5 | Determine, with proof, all positive integers $k$ such that$$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for all positive integers $n.$ | 2 | We assert that the solution is all even integers.
Proof that odd $k$ fail: Consider the case $n=2$. The condition requires $2 \mid \sum_{i=0}^{1} \binom{1}{i}^k = 1^k + 1^k = 2$. If we use the original substitution $n-1 \to n$, for $n=2$ in the original problem, we specifically look at the divisibility condition. Take... | We claim that the answer is $\boxed{\text{all even integers}}.$
Proof that odd $k$ fail: Just take $n = 2$ and get $2^k + 2$ is divisible by $3,$ which implies $2^k \equiv 1 \pmod{3}$ This can only happen if $k$ is even.
Proof that even $k$ work: Substitute $n-1$ in place for $n$ in the problem; we then must prove th... | In the modular arithmetic evaluation for $p_l \nmid j$, the step is $(-1)^k \equiv -1$, not $(-1)^k \equiv 1$, which produces an incorrect sequence with alternating signs that depend on $i$. | [
{
"desc": "Uses Chinese Remainder Theorem to reduce the problem to a system of equations modulo $p_l^{e_l}$.",
"points": 1,
"title": "Reduction to powers of prime divisors of n+1"
},
{
"desc": "Represents the term $n \\choose i$ as a product of fractions $\\frac{n-j}{j}$ and finds the remainder ... |
36 | 5 | Determine, with proof, all positive integers $k$ such that$$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for all positive integers $n.$ | 3 | We assert that the solution is \boxed{\text{all even integers}}.
Proof that odd $k$ fail: Consider the case $n=2$. The condition requires $2 \mid \sum_{i=0}^{1} \binom{1}{i}^k = 1^k + 1^k = 2$.If we use the original substitution $n-1 \to n$, for $n=2$ in the original problem, we specifically look at the divisibility c... | We claim that the answer is $\boxed{\text{all even integers}}.$
Proof that odd $k$ fail: Just take $n = 2$ and get $2^k + 2$ is divisible by $3,$ which implies $2^k \equiv 1 \pmod{3}$ This can only happen if $k$ is even.
Proof that even $k$ work: Substitute $n-1$ in place for $n$ in the problem; we then must prove th... | The proof attempts to use the CRT on modulo $p_l$ instead of the prime power $p_l^{e_l}$. This results in invalid application of the theorem and subsequent arguments. | [
{
"desc": "Uses Chinese Remainder Theorem to reduce the problem to a system of equations modulo $p_l^{e_l}$.",
"points": 1,
"title": "Reduction to powers of prime divisors of n+1"
},
{
"desc": "Represents the term $n \\choose i$ as a product of fractions $\\frac{n-j}{j}$ and finds the remainder ... |
37 | 5 | Determine, with proof, all positive integers $k$ such that$$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for all positive integers $n.$ | 4 | We assert that the solution is \boxed{\text{all even integers}}.
Proof that odd $k$ fail: Consider the case $n=2$. The condition requires $2 \mid \sum_{i=0}^{1} \binom{1}{i}^k = 1^k + 1^k = 2$.Wait, looking closer at the original problem context implicitly provided:If we use the original substitution $n-1 \to n$, for ... | We claim that the answer is $\boxed{\text{all even integers}}.$
Proof that odd $k$ fail: Just take $n = 2$ and get $2^k + 2$ is divisible by $3,$ which implies $2^k \equiv 1 \pmod{3}$ This can only happen if $k$ is even.
Proof that even $k$ work: Substitute $n-1$ in place for $n$ in the problem; we then must prove th... | Incorrect application of Lucas's Theorem (it only works for modulo of primes, not prime powers) that results in loss of 3 points for representing $\binom{n}{i}$ as a product of fractions. | [
{
"desc": "Uses Chinese Remainder Theorem to reduce the problem to a system of equations modulo $p_l^{e_l}$.",
"points": 1,
"title": "Reduction to powers of prime divisors of n+1"
},
{
"desc": "Represents the term $n \\choose i$ as a product of fractions $\\frac{n-j}{j}$ and finds the remainder ... |
38 | 5 | Determine, with proof, all positive integers $k$ such that$$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for all positive integers $n.$ | 6 | We assert that the solution is \boxed{\text{all even integers}}.
Proof that odd $k$ fail: Consider the case $n=2$. The condition requires $2 \mid \sum_{i=0}^{1} \binom{1}{i}^k = 1^k + 1^k = 2$. If we use the original substitution $n-1 \to n$, for $n=2$ in the original problem, we specifically look at the divisibility ... | We claim that the answer is $\boxed{\text{all even integers}}.$
Proof that odd $k$ fail: Just take $n = 2$ and get $2^k + 2$ is divisible by $3,$ which implies $2^k \equiv 1 \pmod{3}$ This can only happen if $k$ is even.
Proof that even $k$ work: Substitute $n-1$ in place for $n$ in the problem; we then must prove th... | Proof is correct, however indexing of $j'$ begins from zero (resulting in division by zero). This results in a 1 point deduction due to technical mistake. | [
{
"desc": "Uses Chinese Remainder Theorem to reduce the problem to a system of equations modulo $p_l^{e_l}$.",
"points": 1,
"title": "Reduction to powers of prime divisors of n+1"
},
{
"desc": "Represents the term $n \\choose i$ as a product of fractions $\\frac{n-j}{j}$ and finds the remainder ... |
39 | 5 | Determine, with proof, all positive integers $k$ such that$$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for all positive integers $n.$ | 0 | We assert that the solution is \boxed{\text{all even integers}}.
Proof that odd $k$ fail: Consider the case $n=2$. The condition requires $2 \mid \sum_{i=0}^{1} \binom{1}{i}^k = 1^k + 1^k = 2$. Take $n=2$. Then $2^k+2$ must be divisible by $3$. This implies $2^k \equiv 1 \pmod{3}$, which holds if and only if $k$ is ev... | We claim that the answer is $\boxed{\text{all even integers}}.$
Proof that odd $k$ fail: Just take $n = 2$ and get $2^k + 2$ is divisible by $3,$ which implies $2^k \equiv 1 \pmod{3}$ This can only happen if $k$ is even.
Proof that even $k$ work: Substitute $n-1$ in place for $n$ in the problem; we then must prove th... | In the beginning the proof incorrectly asserts that $n$ is $k$-good if its sum divisible by $k$ ($n$ in the correct proof). Target modulus is $p_l$, not $p_l^{e_l}$. In the modular arithmetic step, the proof incorrectly substitutes $n \equiv 0 \pmod{p_l}$ to yield $\left(\frac{n}{j}\right)^k \equiv 0$. And derivation o... | [
{
"desc": "Uses Chinese Remainder Theorem to reduce the problem to a system of equations modulo $p_l^{e_l}$.",
"points": 1,
"title": "Reduction to powers of prime divisors of n+1"
},
{
"desc": "Represents the term $n \\choose i$ as a product of fractions $\\frac{n-j}{j}$ and finds the remainder ... |
40 | 5 | Determine, with proof, all positive integers $k$ such that$$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for all positive integers $n.$ | 1 | We assert that the solution is \boxed{\text{all even integers}}.
Proof that odd $k$ fail: Consider the case $n=2$. The condition requires $2 \mid \sum_{i=0}^{1} \binom{1}{i}^k = 1^k + 1^k = 2$. If we use the original substitution $n-1 \to n$, for $n=2$ in the original problem, we get $2^k + 2$ is divisible by $3$. Thi... | We claim that the answer is $\boxed{\text{all even integers}}.$
Proof that odd $k$ fail: Just take $n = 2$ and get $2^k + 2$ is divisible by $3,$ which implies $2^k \equiv 1 \pmod{3}$ This can only happen if $k$ is even.
Proof that even $k$ work: Substitute $n-1$ in place for $n$ in the problem; we then must prove th... | Proof is missing a critical proof segment required for substituting $n \equiv 0$ into the numerator of a fraction. Since the proof stands on this error, subsequent steps are invalid. | [
{
"desc": "Uses Chinese Remainder Theorem to reduce the problem to a system of equations modulo $p_l^{e_l}$.",
"points": 1,
"title": "Reduction to powers of prime divisors of n+1"
},
{
"desc": "Represents the term $n \\choose i$ as a product of fractions $\\frac{n-j}{j}$ and finds the remainder ... |
End of preview. Expand in Data Studio
TROJAN-Bench-H
A benchmark for LLM proof graders. It holds 48 proofs with human-injected errors: each one is a reference solution that a person edited to introduce mistakes, labeled with the score it should receive. It is the human-made counterpart of TROJAN-Bench-S.
This dataset accompanies the paper Can LLMs Reliably Grade Olympiad Proofs? A Study of Mathematical Verification with LLMs. Code is available at https://github.com/aziksh-ospanov/TROJAN.
It covers the six USAMO 2025 problems, with 8 proofs each. Problems and rubrics come from MathArena. Every problem is worth 7 points, and scores range from 0 to 7.
| Field | Description |
|---|---|
global_idx |
Row id (1–48) |
problem_idx |
USAMO 2025 problem number ("1"–"6") |
problem |
Problem statement |
awarded_points |
Ground-truth total score |
solution |
Error-injected proof to be graded |
sample_solution |
Reference solution |
grading_scheme |
Rubric: title, desc, points (no per-criterion ground truth) |
justification |
Which errors were introduced and why they cost points (hide this from the judge) |
from datasets import load_dataset
ds = load_dataset("aziksh/TROJAN-Bench-H", split="test")
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