Track / Stage 4 / 31 of 340 #291 of 1964

Problem 291

AMC 12 late, AIME early

Number theory Difficulty 4.6 Find the answer

6. If $2n+1, 20n+1 \left(n \in \mathbf{N}_{+}\right)$ are powers of the same positive integer, then all possible values of $n$ are

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

6. 4 .

According to the problem, we know that $(2 n+1) \mid(20 n+1)$ . Then $(2 n+1) \mid[10(2 n+1)-(20 n+1)]=9$ . Therefore, $n \in\{1,4\}$ . Upon verification, $n=4$ .

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