Problem:
How many pairs of positive integers (a,b) with a+b≤100 satisfy the equation
a−1+ba+b−1=13?
Answer Choices:
A. 1
B. 5
C. 7
D. 9
E. 13
Solution:
Multiply the numerator and denominator of the given fraction by ab to obtain
b+ab2a2b+a=b(1+ab)a(ab+1)=ba=13.
Thus a=13b; and a+b≤100 implies 14b≤100, so 0<b≤7. For each of the seven possible values of b=1,2,3,4,5,6,7, the pair (13b,b) is a solution.