Problem:
The arithmetic mean of two distinct positive integers x and y is a two-digit integer. The geometric mean of x and y is obtained by reversing the digits of the arithmetic mean. What is ∣x−y∣ ?
Answer Choices:
A. 24
B. 48
C. 54
D. 66
E. 70 Solution:
Let the arithmetic and geometric means of x and y be 10a+b and 10b+a, respectively. Then
2x+y=10a+b⇒(x+y)2=400a2+80ab+4b2
and
xy=10b+a⇒xy=100b2+20ab+a2
So
(x−y)2=(x+y)2−4xy=396(a2−b2)=11⋅62⋅(a+b)(a−b)
Because x and y are distinct, a and b are distinct digits, and the last expression is a perfect square if and only if a+b=11 and a−b is a perfect square. The cases a−b=1,4, and 9 give solutions (a,b)=(6,5),(7.5,3.5), and (10,1), respectively. Because a and b are digits only the first solution is valid. Thus (x−y)2=11⋅62⋅11=662 and ∣x−y∣=66. Note that the given conditions are satisfied if {x,y}={32,98}.