HMMT 二月 2026 · 冲刺赛 · 第 14 题
HMMT February 2026 — Guts Round — Problem 14
题目详情
- [9] There exists exactly one ordered pair of positive integers ( m, n ) , both greater than 1 , with the property ( ) m that, when written out in base 10 , m · n = ABCD and = CDAB for distinct nonzero digits A , B , n C , and D . Compute m + n .
解析
- [9] There exists exactly one ordered pair of positive integers ( m, n ) , both greater than 1 , with the ( ) m property that, when written out in base 10 , m · n = ABCD and = CDAB for distinct nonzero n digits A , B , C , and D . Compute m + n . Proposed by: Andrew Brahms Answer: 136 Solution: We begin with the following crucial claim. Claim 1. We have n = m − 2 . ( ) m Proof. Note that m > n since = CDAB ≥ 1111 . Furthermore, mn = ABCD ≥ 1111 , so n √ m > 1111 > 33 . We continue by showing that any pair ( m, n ) not satisfying n = m − 2 can’t lead to a solution. • We are given that n ̸ = 1 . ( ) m • If n = m − 1 , then = m . Hence, m = CDAB ≥ 1111 , and therefore n 10000 ≥ ABCD = mn = m ( m − 1) ≥ 1111 · 1110 , which is clearly a contradiction. ( ) ( ) m 370 • If n ∈ { 2 , 3 } , then m ≥ ABCD/ 3 ≥ 1111 / 3 > 370 . However, CDAB = ≥ > 10000 n 2 clearly leads to a contradiction. ( ) ( ) ( ) m m 33 • If 4 ≤ n ≤ m − 4 , then CDAB = ≥ ≥ > 10000 . n 4 4 • Finally, we consider the case n = m − 3 . We must have that ( ) m ( m − 4)( m − 5) m = − m ( m − 3) = CDAB − ABCD = 99 · ( CD − AB ) . 6 m − 3 ( ) 41 Therefore, 27 | m ( m − 4)( m − 5) . Since m > 33 , and m ≤ 40 (as = 10660 > CDAB ), the 3 numbers m , m − 4 , and m − 5 lie in the interval [29 , 40] . However, 3 can divide at most one of them, which leads to a contradiction as no integer in the interval [29 , 40] is divisible by 27 . ©2026 HMMT ( ) m Following Claim 1, we must have that n = m − 2 . Let AB = k , and CD = ℓ . Thus, = 100 ℓ + k , n m ( m − 1) and mn = 100 k + ℓ . Hence, we have that = 100 l + k and m ( m − 2) = 100 k + l . Adding these 2 equations yields m (3 m − 5) = 101( k + l ) . 2 Therefore, 101 | m (3 m − 5) . If 101 | m , then either m = 101 which doesn’t lead to a solution, or m ≥ 202 , in which case m ( m − 2) ≥ 202 · 200 > 10000 > ABCD . Hence, 101 | 3 m − 5 . If m > 101 , then m ( m − 2) ≥ 102 · 100 > 10000 , which leads to a contradiction. If m ≤ 101 , then m ≡ 5 / 3 ≡ 69 (mod 101) = ⇒ m = 69 . Therefore, m = 69 and n = 67 , which leads to a solution with A = 4 , B = 6 , C = 2 , and D = 3 as ( ) m mn = 4623 and = 2346 . Finally, we compute m + n = 69 + 67 = 136 . n Remark. Note that the above solution never uses the fact that the digits A , B , C , and D are distinct.