1997 AMC 8 Problem 23

Attempt Problem 23 of the 1997 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1997 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

23.

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
Answer: C
Concepts:digitscasework
Small hint:
Try the largest possible final digit and work downward
Big hint:
More digits make the minimum square-sum too large
Solution:
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 50−36=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 1⋅2⋅3⋅6=36.1\cdot2\cdot3\cdot6=36.
Thus, C is the correct answer.
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