Difference between revisions of "1997 AJHSME Problems/Problem 23"
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* [[Mathematics competition resources]] | * [[Mathematics competition resources]] |
Revision as of 23:11, 31 July 2011
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 |