# 1994 AJHSME Problems/Problem 20

## Problem

Let $W,X,Y$ and $Z$ be four different digits selected from the set $\{ 1,2,3,4,5,6,7,8,9\}.$

If the sum $\dfrac{W}{X} + \dfrac{Y}{Z}$ is to be as small as possible, then $\dfrac{W}{X} + \dfrac{Y}{Z}$ must equal $\text{(A)}\ \dfrac{2}{17} \qquad \text{(B)}\ \dfrac{3}{17} \qquad \text{(C)}\ \dfrac{17}{72} \qquad \text{(D)}\ \dfrac{25}{72} \qquad \text{(E)}\ \dfrac{13}{36}$

## Solution $$\frac{W}{X} + \frac{Y}{Z} = \frac{WZ+XY}{XZ}$$

Small fractions have small numerators and large denominators. To maximize the denominator, let $X=8$ and $Z=9$. $$\frac{9W+8Y}{72}$$

To minimize the numerator, let $W=1$ and $Y=2$. $$\frac{9+16}{72} = \boxed{\text{(D)}\rightarrow \frac{25}{72}}$$

## Solution 2

To make the smallest fraction, you need the lowest numerator and the highest denominator. So, take the first $2$ and last $2$ digits of the set, which are $1,2$ and $8,9$. Balance the equations to be "even". Since $1$ is smaller than $2$, put it over $8$. You get $\frac{1}{8}+\frac{2}{9}=\frac{25}{72}$, or $\boxed{D}$.

-goldenn

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 