Problem: If xxx and yyy are positive integers for which 2x3y=12962^{x} 3^{y}=12962x3y=1296, what is the value of x+yx+yx+y ?
Answer Choices:
A. 888 B. 999 C. 101010 D. 111111 E. 121212 Solution:
Note that 1296=64=24341296=6^{4}=2^{4} 3^{4}1296=64=2434, so x=y=4x=y=4x=y=4 and x+y=8x+y=\boxed{8}x+y=8​.
The problems on this page are the property of the MAA's American Mathematics Competitions