Track / Stage 4 / 262 of 340 #522 of 1964

Problem 522

AMC 12 late, AIME early

Number theory Difficulty 4.9 Find the answer

$16.2 .26 *$ Find the smallest positive integer $n$ , such that the last three digits of its cube are 888.

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Given that the unit digit of $n^{3}$ is 8, we know $n=10 k+2$ . Thus, $n^{3}=(10 k+2)^{3}=$ $1000 k^{3}+600 k^{2}+120 k+8 \equiv 88(\bmod 100)$ , therefore $k=4(\bmod 5)$ . Let $k=5 m+4(m$ be a non-negative integer), then $n^{3} \equiv 600 \times 4^{2}+120 \times(5 m+4)+8 \equiv 88+600 m \equiv 888(\bmod 1000)$ , hence $m=3, n=192$ .

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