1997 AJHSME Problems/Problem 23
Problem
There are positive integers that have these properties:
- the sum of the squares of their digits is 50, and
- each digit is larger than the one to its left.
The product of the digits of the largest integer with both properties is
Solution
Five digit numbers will have a minimum of as the sum of their squares if the five digits are distinct and non-zero. If there is a zero, it will be forced to the left by rule #2.
No digit will be greater than , as .
Trying four digit numbers , we have with
will not work, since the other digits must be at least , and the sum of the squares would be over .
will give . will work, giving the number . No other number with will work, as and would each have to be greater.
will give . forces and , which has a leading zero. Smaller will force all the numbers to the smallest values, and will give a sum of squares that is too small.
can only give the number , which does not satisfy the condition of the problem.
Thus, the number in question is , and the product of the digits is , giving as the answer.
See also
1997 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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