1995 AIME Problems/Problem 8

Problem

For how many ordered pairs of positive integers $(x,y),$ with $y<x\le 100,$ are both $\frac xy$ and $\frac{x+1}{y+1}$ integers?

Solution

Since $y|x$ , $y+1|x+1$ , then $\text{gcd}\,(y,x)=y$ and $\text{gcd}\,(y+1,x+1)=y+1$ . By the Euclidean Algorithm, these can be rewritten as $\text{gcd}\,(y,x-y)=y$ and $\text{gcd}\, (y+1,x-y)=y+1$ , which implies that $y,y+1 | x-y$ . Also, $\text{gcd}\,(y,y+1) = 1$ , so $\frac{x-y}{y(y+1)}$ must be an integer. Substituting the upper bound of $x \le 100$ , it follows that there are $\left\lfloor\frac{100-y}{y(y+1)} \right\rfloor$ satisfactory positive integers. The answer is $\sum_{y=1}^{99} \left\lfloor\frac{100-y}{y(y+1)} \right\rfloor = \boxed{085}$

1995 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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1995 AIME Problems/Problem 8

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