HMMT 二月 2026 · 冲刺赛 · 第 1 题

HMMT February 2026 — Guts Round — Problem 1

[5] Let H , M , and T be (not necessarily distinct) digits such that H is nonzero and HM M T = HT M × HT . Compute the only possible value of the four-digit positive integer HM M T .

解析

[5] Let H , M , and T be (not necessarily distinct) digits such that H is nonzero and HM M T = HT M × HT . Compute the only possible value of the four-digit positive integer HM M T . Proposed by: Jackson Dryg, Jason Mao Answer: 1000 Solution: Note that 2 ( H + 1) · 1000 > HM M T = HT M × HT ≥ ( H · 100) × ( H · 10) = H · 1000 . 2 Since H > 2 H − 1 ≥ H + 1 for H ≥ 2 , we must have H = 1 . Then, the equation becomes 2 1000 + 110 M + T = (100 + 10 T + M ) · (10 + T ) = 1000 + 200 T + 10 T + 10 M + M T. The above is equivalent to (100 − T ) M = T (10 T + 199) . (1) Clearly, if M = 0 , then T must be zero, and if T = 0 , then M must be zero. This leads to the solution ( H, M, T ) = (1 , 0 , 0) , when the equation reads 1000 = 100 · 10 . We will show that no other solution exists. For the sake if contradiction, assume that M = 0 ̸ and T ̸ = 0 leads to a valid solution. Then, both sides of (1) are nonzero. We set d = gcd(100 − T, 10 T + 199) . Since 10 T + 199 | (100 − T ) M , we must have (10 T + 199) /d | M and (199 + 10 T ) /d ≤ M . Therefore, 199 + 10 T 199 d ≥ ≥ > 22 . M 9 On the other hand, we have that d | 10 · (100 − T ) + (199 + 10 T ) = 1199 . Since 1199 = 11 × 109 , the only possible choices for d are 109 and 1199 . However, d | 100 − T < 109 , which is a contradiction. Thus, the only solution is ( H, M, T ) = (1 , 0 , 0) , and so HM M T = 1000 .