1997 AMC 8 第 23 题

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23.

有一些正整数满足:

I. 它们各位数字的平方和为 5050,并且

II. 每个数字都大于它左边的数字。

满足这两个条件的最大整数,其各位数字的乘积是

There are positive integers that have these properties:

I. the sum of the squares of their digits is 50,50, and

II. each digit is larger than the one to its left.

The product of the digits of the largest integer with both properties is

77

2525

3636

4848

6060

答案:C
知识点:数字分类讨论
难度评级:1670
解答:

如果这个数有五位,最小的递增正数字会使平方和为 12+22+32+42+52=551^2+2^2+3^2+4^2+5^2=55,已经太大。因此最大的有效数最多有四位。

对于四位数 abcdabcd,其中 0<a<b<c<d0<a<b<c<d,末位不能是 77 或更大,因为即使是 12+22+32+72=63>501^2+2^2+3^2+7^2=63>50

尝试 d=6d=6,其余数字的平方和必须是 5036=1450-36=14,而 12+22+32=141^2+2^2+3^2=14。这给出有效数 12361236

任何以 55 或更小数字结尾的数都小于 12361236,所以最大的有效整数是 12361236

它的各位数字乘积为 1236=361\cdot2\cdot3\cdot6=36

所以正确答案是 C

If the number had five digits, the smallest possible increasing positive digits would give square-sum 12+22+32+42+52=55,1^2+2^2+3^2+4^2+5^2=55, already too large. So the largest valid number has at most four digits.

For a four-digit number abcdabcd with 0<a<b<c<d0<a<b<c<d, the last digit cannot be 77 or larger, since even 12+22+32+72=63>501^2+2^2+3^2+7^2=63>50.

Trying d=6,d=6, the remaining squares must sum to 5036=14,50-36=14, and 12+22+32=14.1^2+2^2+3^2=14. This gives the valid number 1236.1236.

Any number ending with 55 or less is smaller than 1236,1236, so the largest valid integer is 1236.1236.

The product of its digits is 1236=36.1\cdot2\cdot3\cdot6=36.

Thus, C is the correct answer.

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