Find all pairs of non-zero natural numbers $(k, n)$ for which

$$
1!+2!+\cdots+k!=1+2+\cdots+n
$$
In the entire solution, we will denote $A_{k}$ as the expression $1!+2!+\cdots+k$ and $B_{n}$ as the expression $1+2+\cdots+n$. As with any problem of this type, we start by looking for potential solutions for small values of $k$ and $n$. For this purpose, it is useful to recognize in $B_{n}$ the sum of the terms of an arithmetic sequence, so that $B_{n}=n(n+1) / 2$. In particular, if $(k, n)$ is a solution, we know that

$$
n^{2} \leqslant 2 B_{n}=2 A_{k}<(n+1)^{2},
$$

so that $n=\left\lfloor\sqrt{2 A_{k}}\right\rfloor$.
We then calculate the first values of $A_{k}$: for example, when $k$ varies from 1 to 5, $A_{k}$ successively takes the values $1,3,9,33$, and 153, and the integer $n=\left\lfloor\sqrt{2 A_{k}}\right\rfloor$ then successively takes the values $1,2,4,8$, and 17, so that $B_{n}$ takes the values $1,3,10,36$, and 153. This initial study, although tedious towards the end, allowed us to identify three solution pairs: $(k, n)=(1,1),(2,2)$, and $(5,17)$.
Moreover, it would be impractical to compute $\left\lfloor 2 \sqrt{A_{\ell}}\right\rfloor$ by hand when $\ell \geqslant 6$, and we need to find a more general approach when $\ell$ is large. For example, when $\ell \leqslant k$, the integer $\ell$! divides the integer $k!=A_{k}-A_{k-1}$, so that $A_{k} \equiv A_{\ell-1}(\bmod \ell!)$. We therefore undertake to study the sequence $\left(A_{k}\right)$ modulo $\ell$! or $\ell$ for small values of $\ell$, for example when $\ell$ is a prime number.
After an unfruitful study of the cases $\ell=2,3$, and 5, we observe that $k!\equiv 1,2,6,3,1,6$ $(\bmod 7)$ when $k$ varies from 1 to 6, and thus that $A_{k} \equiv A_{6} \equiv 1+2+6+3+1+6 \equiv 5(\bmod 7)$ when $k \geqslant 6$. On the other hand, $B_{n} \equiv 1,3,6,3,1,0,0(\bmod 7)$ when $n$ varies from 1 to 7. Since 2 is invertible modulo 7, we know that $B_{n}(\bmod 7)$ depends only on $n(\bmod 7)$. This means in particular that $B_{n} \not \equiv 5 \equiv A_{k}(\bmod 7)$ as soon as $k \geqslant 6$ and $n \geqslant 1$, proving at the same time that $(k, n)$ is not a solution.
In conclusion, the solutions are the pairs $(k, n)=(1,1),(2,2)$, and $(5,17)$.
Comment from the graders: The problem was rather difficult and was well approached. However, some candidates did not look for solutions, or missed the solution $(5,17)$: it is important not to hesitate to spend time on these initial proofs, which also help to build intuition. Here, once the solution $(5,17)$ was found, one could observe that any modulus dividing 5! would not allow progress towards the solution. It was therefore necessary to look for a modulus not dividing 5!; here, 7 was the easiest modulus, but 13 and 25 also worked.