HMMT 二月 2005 · 冲刺赛 · 第 24 题

英文解析

24. In the base 10 arithmetic problem HM M T + GU T S = ROU N D , each distinct letterrepresents a different digit, and leading zeroes are not allowed. What is the maximumpossible 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 infact since D cannot be 0 or 1, S + T ≥ 12, which is impossible given the remainingchoices. Therefore, O is at most 6.
    Suppose O = 6 and M = 9. Then we must have H and G be 7 and 8. With theremaining 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