Problem:
In the expansion of (ax+b)2000, where a and b are relatively prime positive integers, the coefficients of x2 and x3 are equal. Find a+b.
Solution:
The problem statement implies that
2⋅12000⋅1999a2b1998=3⋅2⋅12000⋅1999⋅1998a3b1997
This is equivalent to b=666a. Because a and b have no common divisors greater than 1, it follows that a=1,b=666, and a+b=667.
The problems on this page are the property of the MAA's American Mathematics Competitions