# 1997 AJHSME Problems/Problem 19

## Problem

If the product $\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9$, what is the sum of $a$ and $b$?

$\text{(A)}\ 11 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 37$

## Solution 1

Notice that the numerator of the first fraction cancels out the denominator of the second fraction, and the numerator of the second fraction cancels out the denominator of the third fraction, and so on.

The only numbers left will be $a$ in the numerator from the last fraction and $2$ in the denominator from the first fraction. (The $b$ will cancel with the numerator of the preceeding number.) Thus, $\frac{a}{2} = 9$, and $a=18$.

Since the numerator is always one more than the denominator, $b = a-1 = 17$, and $a+ b = 18 + 17 = 35$, giving an answer of $\boxed{D}$

## Solution 2

Find a pattern. If $a=4$, then the expression is just the first two terms, which is $2$.

If $a=5$, then the expression is the first three terms, giving $2.5$.

If $a=6$, the expression is the first four terms, giving $3$.

If $a=7$, the expression will be the first five terms, giving $3.5$.

Conjecture that the expression is always going to equal $\frac{a}{2}$, and thus when $a=18$, the expression will be $9$, as desired.

As above, when $a=18$, $b=17$, and the sum is $35$, or $\boxed{D}$