Track / Stage 5 / 59 of 400 #659 of 1964

Problem 659

AIME late

Number theory Difficulty 5.2 Find the answer

12. (2004 Czech and Slovak Mathematical Olympiad) Find the positive integer $n$ , such that $\frac{n}{1!}+\frac{n}{2!}+\cdots+\frac{n}{n!}$ is an integer.

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Official solution

12. When $n=1,2,3$ , the sums are $1,3,5$ , respectively, which are integers.

When $n>3$ ,
$\begin{aligned} & \frac{n}{1!}+\frac{n}{2!}+\cdots+\frac{n}{(n-2)!}+\frac{n}{(n-1)!}+\frac{n}{n!} \\ = & \frac{n(n-1) \cdots 2+n(n-1) \cdots 3+\cdots+n(n-1)+n+1}{(n-1)!} . \end{aligned}$

For it to be an integer, the numerator must be divisible by $n-1$ , so $n+1$ must be divisible by $n-1$ , which means $2[=n+1-(n-1)]$ must be divisible by $n-1$ . Therefore, $n-1 \in\{1,2\}$ , which is impossible.
Thus, the sum is an integer if and only if $n \in\{1,2,3\}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.