Track / Stage 5 / 17 of 400 #617 of 1964

Problem 617

AIME late

Number theory Difficulty 5.0 Prove it Romanian Mathematical Olympiad · Romania

Show that every non-empty subset of the set
$Y = \{113! + 2, 113! + 3, \dots, 113! + 15\}$
has the sum of its elements a composite number.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

A non-empty subset of $Y$ has the sum of its elements $S = 113! + s$ , where $2 \leq s \leq 2 + 3 + \dots + 15$ , whence $2 \leq s \leq 119$ .

If $2 \leq s \leq 113$ , then $S = 113! + s = Ms > s$ .

If $s \in \{114, 116, 118\}$ , then $S = M2$ .

If $s = 115$ , then $S = M5$ .

If $s = 117$ , then $S = M3$ .

If $s = 119$ , then $S = M7$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.