PUMaC 2013 · 数论（B 组） · 第 2 题

PUMaC 2013 — Number Theory (Division B) — Problem 2

[ 3 ] What is the smallest positive integer n such that 2013 ends in 001 (i.e. the rightmost n three digits of 2013 are 001)?

解析

[ 3 ] What is the smallest positive integer n such that 2013 ends in 001 (i.e. the rightmost n three digits of 2013 are 001)? Solution Firstly, 100 100 2013 ≡ 13 100 = (10 + 3) 100 ≡ 3 10 ≡ 49 5 ≡ 401 5 = (400 + 1) ≡ 1 (mod 1000) yields n | 100. Next 50 50 2013 ≡ (10 + 3) ( ) 50 50 49 2 48 ≡ 3 + 50 · 10 · 3 + · 10 · 3 2 50 ≡ 3 5 ≡ 49 2 ≡ 807 ≡ 249 (mod 1000) and 20 20 2013 ≡ (10 + 3) 20 19 ≡ 3 + 20 · 10 · 3 2 ≡ 49 + 400 ≡ 801 (mod 1000) shows that n = 100.