PUMaC 2012 · 代数（B 组） · 第 1 题

PUMaC 2012 — Algebra (Division B) — Problem 1

解析

[ 3 ] Find the largest n such that the last nonzero digit of n ! is 1. i j Solution: For n > 2 if we express n ! = 2 5 m where, i, j, m ∈ N such that m is not divisible by i − j j 2 or 5, we can easily observe that i > j . Thus, we can write n ! = 2 10 m which means that the last non-zero digit of n ! is always going to be an even number. Thus, the only possible values for n are 0 and 1, which both give the last non-zero digit equal to 1. Because 1 is greater, the final answer is 1. Answer: 1 1000 ∑