HMMT 十一月 2025 · 冲刺赛 · 第 29 题

HMMT November 2025 — Guts Round — Problem 29

[15] Compute the smallest positive integer multiple of 10001 with all of its digits distinct (when written in base 10). ◦

解析

[15] Compute the smallest positive integer multiple of 10001 with all of its digits distinct (when written in base 10). Proposed by: Derek Liu Answer: 2650134987 Solution: Any multiple of 10001 with 8 or fewer digits can be written as the same 4-digit number repeated twice (possibly with leading 0s), so it is guaranteed to have repeated digits. Therefore, we consider multiples with > 8 digits. Any multiple of 10001 with 9 or 10 digits can be written in the form abc , where b and c are 4-digit numbers (with leading zeros allowed) and a is a 1 or 2-digit number such that 10001 | a − b + c . Hence, a + c = b or a + c = b + 10001. © 2025 HMMT We cannot have a + c = b + 10001, otherwise c > 9900 and has repeated 9’s. Thus, a + c = b . Suppose b starts with the digit d ; then, c cannot start with d , but c ≥ b − 99, so c must start with the digit e = d − 1. Furthermore, b ≥ d 012 and c ≤ e 987 because b and c have distinct digits, so a ≥ d 012 − e 987 = 25. If a starts with a 2, then b ≥ d 013 to not repeat 2, but then a would have to be at least d 013 − e 987 = 26. Thus, a ≥ 26. There is exactly one solution with a = 26, namely 26 + 4987 = 5013, so the desired multiple is 2650134987 . ◦