Question

There are positive integers that have these properties:

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

$7$

$25$

$36$

$48$

$60$

Solution:

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

For a four-digit number $abcd$ with $0<a<b<c<d$ , the last digit cannot be $7$ or larger, since even $1^2+2^2+3^2+7^2=63>50$ .

Trying $d=6,$ the remaining squares must sum to $50-36=14,$ and $1^2+2^2+3^2=14.$ This gives the valid number $1236.$

Any number ending with $5$ or less is smaller than $1236,$ so the largest valid integer is $1236.$

The product of its digits is $1\cdot2\cdot3\cdot6=36.$

Thus, C is the correct answer.

1997 AMC 8 Problem 23

Solution:

Problem 23 in Other Years