HMMT 二月 2005 · 冲刺赛 · 第 24 题

HMMT February 2005 — Guts Round — Problem 24

[9] In the base 10 arithmetic problem HMMT + GUT S = ROUND , each distinct letter represents a different digit, and leading zeroes are not allowed. What is the maximum possible value of ROUND ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

解析

In the base 10 arithmetic problem HM M T + GU T S = ROU N D , each distinct letter represents a different digit, and leading zeroes are not allowed. What is the maximum possible value of ROU N D ? Solution: 16352 Clearly R = 1, and from the hundreds column, M = 0 or 9. Since H + G = 9 + O or 10 + O , it is easy to see that O can be at most 7, in which case H and G must be 8 and 9, so M = 0. But because of the tens column, we must have S + T ≥ 10, and in fact since D cannot be 0 or 1, S + T ≥ 12, which is impossible given the remaining choices. Therefore, O is at most 6. Suppose O = 6 and M = 9. Then we must have H and G be 7 and 8. With the remaining digits 0, 2, 3, 4, and 5, we must have in the ones column that T and S are 2 and 3, which leaves no possibility for N . If instead M = 0, then H and G are 7 and