Selection Examination

We first compute the entries of the following matrix

$k$	1	2	3	4	5	6	7	8	9	10
$k!$	1	2	6	24	120	720	5040	40320	362880	3628800
$1!+2!+\cdots +k!$	1	3	9	33	153	873	5913	46233	409113	4037913

Obviously, the pairs $(k,n)=(1,1)$ and $(k,n)=(2,2)$ are solutions. We will show that the unique solution with $k>2$ is $(k,n)=(5,17)$.

We observe that since $k!$ is divided by $100$ for $k\ge 10$, the sum $1!+2!+\cdots +k!$ leaves a remainder $13$ when divided by $100$ for $k\ge 9$. If equality holds $$1!+2!+\cdots +k!=1+2+\cdots +n=\frac{n(n+1)}{2}$$ for some $k\ge 10$, then there is some natural number $m$ such that $$n(n+1)=100m+26$$ or $$n^2+n-100m-26=0.$$ The discriminant is $\Delta =400m+105$ and it should be a perfect square. On the other hand we cannot have $$x^2\equiv 5(\mathrm{mod}100),$$ as otherwise $25$ would divide $5$. Therefore the given relation cannot hold for $k\ge 9$, as well as, for $k=7$. Since $$9=3^2,153=3^2\cdot 17,873=3^2\cdot 97\text{ and }46233=3^2\cdot 11\cdot 467,$$ we observe that $2\cdot 153=18\cdot 17$, a product of two consecutive integers $k>2$. Therefore the only solution is $(k,n)=(5,17)$.