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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
