# 1985 AHSME Problems/Problem 5

## Problem

Which terms must be removed from the sum

$$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}$$

if the sum of the remaining terms is to equal $1$?

$\mathrm{(A)\ } \frac{1}{4}\text{ and }\frac{1}{8} \qquad \mathrm{(B) \ }\frac{1}{4}\text{ and }\frac{1}{12} \qquad \mathrm{(C) \ } \frac{1}{8}\text{ and }\frac{1}{12} \qquad \mathrm{(D) \ } \frac{1}{6}\text{ and }\frac{1}{10} \qquad \mathrm{(E) \ }\frac{1}{8}\text{ and }\frac{1}{10}$

## Solution

We compute the entire sum as \begin{align*}\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12} &= \frac{60+30+20+15+12+10}{120} \\ &= \frac{147}{120} \\ &= \frac{49}{40},\end{align*} so the two terms to be removed must sum to $\frac{49}{40}-1 = \frac{9}{40}$. That is, $\frac{9}{40} = \frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$ for some $x,y \in \{2,4,6,8,10,12\}$.

Since $9$ and $40$ are coprime, we deduce that $xy$ is a multiple of $40$, so $x$ or $y$ must be a multiple of $5$. It follows that one of them must be $10$, and we obtain \begin{align*}\frac{9}{40}-\frac{1}{10} &= \frac{9-4}{40} \\ &= \frac{5}{40} \\ &= \frac{1}{8},\end{align*} so the two terms we must remove are $\boxed{\text{(E)} \ \frac{1}{8}\text{ and }\frac{1}{10}}$.