AMC 8 1997 Problem 23
AMC8 1997 年第 23 题
整除与因数★★★★☆
There are positive integers that have these properties: The sum of the squares of their digits is equal to 50 Each digit is larger than the one on it's left The product of the digits of the largest integer with both properties is
存在满足以下性质的正整数：各位数字的平方和等于 50，且每位数字都比它左边的数字大。同时满足这两个性质的最大整数，其各位数字之积是多少？
- A.7
- B.25
- C.36
- D.48
- E.60
正确答案 · Correct Answer
C
解析 · Solution
解法1
常规解法 · 按位数枚举
各位严格递增，故互不相同且非零。先试四位 ：
（）。z=7 太大（其余至少 1+4+9=14，和超 50）；z=6 给 ，唯一解 1、2、3，得 1236； 凑不出更大四位数。五位最小平方和 1+4+9+16+25=55>50，不可能。故最大为 1236，数字积 。
巧解 · 先锁位数再锁最高位
要数最大就位数尽量多，但五位递增的最小平方和已是 55，超过 50，排除；故为四位且让最大位 尽量大。z=7 超界，z=6 时余下 唯一解 1、2、3，得 1236，积 36。
Solution 1
Elementary · enumerate by digit count
The digits strictly increase, so they are distinct and nonzero. For a four-digit :
(since ): z=7 is too big (the rest are at least 1+4+9=14), z=6 gives with the only solution 1, 2, 3 giving 1236, and smaller cannot give a larger number. Five digits force a minimum square-sum 1+4+9+16+25=55>50. So the largest is 1236 and the digit product is .
Slick · fix the digit count, then the leading digit
To maximize the number use as many digits as possible, but five increasing digits give minimum square-sum 55, over 50; so it is four digits with as large as possible. Since z=7 overshoots, take z=6, leaving whose only solution is 1, 2, 3. The number is 1236, product 36.
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