Problem:
Let x and y be positive integers such that 7x5=11y13. The minimum possible value of x has a prime factorization acbd. What is a+b+c+d ?
Answer Choices:
A. 30
B. 31
C. 32
D. 33
E. 34
Solution:
We have
11y13=7x5=7(acbd)5=7a5cb5d
The minimum value of x is obtained when neither x nor y contains prime factors other than 7 and 11. Therefore we may assume that a=7 and b=11, so x= 7c11d and 7x5=75c+1115d. Letting y=7m11n we obtain 11y13=713m1113n+1. Hence 75c+1115d=713m1113n+1. The smallest positive integer solutions are c=5,d=8,m=2, and n=3. Thus a+b+c+d=7+11+5+8=31​.
The problems on this page are the property of the MAA's American Mathematics Competitions