HMMT 二月 2013 · 冲刺赛 · 第 12 题

HMMT February 2013 — Guts Round — Problem 12

[ 6 ] For how many integers 1 ≤ k ≤ 2013 does the decimal representation of k end with a 1? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND Organization Team Team ID# n +1 n +1 5 +2

解析

[ 6 ] For how many integers 1 ≤ k ≤ 2013 does the decimal representation of k end with a 1? Answer: 202 We claim that this is only possible if k has a units digit of 1. Clearly, it is true in k these cases. Additionally, k cannot have a units digit of 1 when k has a units digit of 2 , 4 , 5 , 6, or 8. If k k has a units digit of 3 or 7, then k has a units digit of 1 if and only if 4 | k , a contradiction. Similarly, k if k has a units digit of 9, then k has a units digit of 1 if and only if 2 | k , also a contradiction. Since there are 202 integers between 1 and 2013, inclusive, with a units digit of 1, there are 202 such k which fulfill our criterion. n +1 n +1 5 +2