# Difference between revisions of "1992 AHSME Problems/Problem 13"

## Problem

How many pairs of positive integers (a,b) with $a+b\le 100$ satisfy the equation

$$\frac{a+b^{-1}}{a^{-1}+b}=13?$$

$\text{(A) } 1\quad \text{(B) } 5\quad \text{(C) } 7\quad \text{(D) } 9\quad \text{(E) } 13$

## Solution 1

$\fbox{C}$ We can rewrite the left-hand side as $\frac{a^2b+a}{b+ab^2}$ by multiplying the numerator and denominator by $ab$. Now we can multiply up by the denominator to give $a^2b+a = 13b + 13ab^2 \implies ab(a-13b)=13b-a = -(a-13b) \implies (ab+1)(a-13b) = 0$. Now, as $a$ and $b$ are positive, we cannot have $ab+1=0$, so we must have $a-13b=0 \implies a=13b$, so the solutions are $(13,1), (26,2), (39,3), (52,4), (65,5), (78,6), (91,7)$, which is 7 ordered pairs.

## Solution 2

The equation is equivalent to $\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=13$ or $\frac{\frac{ab+1}{b}}{\frac{ab+1}{a}}=13$. The $ab+1$s cancel out leaving us with $\frac{a}{b}=13$. This is asking us for the positive multiples of $13$ up to $100$, and we know that $13 \cdot 7 = 91$, so our answer is $\fbox{C}$.