HMMT 二月 2025 · 冲刺赛 · 第 20 题
HMMT February 2025 — Guts Round — Problem 20
题目详情
- [11] Compute the 100th smallest positive multiple of 7 whose digits in base 10 are all strictly less than 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2025, February 15, 2025 — GUTS ROUND Organization Team Team ID#
解析
- [11] Compute the 100th smallest positive multiple of 7 whose digits in base 10 are all strictly less than
Proposed by: Srinivas Arun Answer: 221221 Solution: We construct an order-preserving bijection between positive multiples of 7 in base 10 whose digits are all less than 3 and positive multiples of 7. For any multiple of 7 in base 10 with digits all less than 3, interpret it as a base 3 number and convert it to a base 10 decimal, which will be a multiple of 7. For example, 1022 would map to 1022 = 35 . 3 10 We first show that this is a valid mapping. Let a . . . a a be an arbitrary multiple of 7 (in base n 1 0