Problem:
Let x and y be two-digit integers such that y is obtained by reversing the digits of x. The integers x and y satisfy x2−y2=m2 for some positive integer m. What is x+y+m ?
Answer Choices:
A. 88
B. 112
C. 116
D. 144
E. 154
Solution:
By the given conditions, it follows that x>y. Let x=10a+b and y=10b+a, where a>b. Then
m2=x2−y2=(10a+b)2−(10b+a)2=99a2−99b2=99(a2−b2).
Since 99(a2−b2) must be a perfect square,
a2−b2=(a+b)(a−b)=11k2
for some positive integer k. Because a and b are distinct digits, we have a−b≤ 9−1=8 and a+b≤9+8=17. It follows that a+b=11,a−b=k2, and k is either 1 or 2 .
If k=2, then (a,b)=(15/2,7/2), which is impossible. Thus k=1 and (a,b)=(6,5). This gives x=65,y=56,m=33, and x+y+m=154.
The problems on this page are the property of the MAA's American Mathematics Competitions