# 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

First, we sum all of the terms to see how much more than $1$ the entire sum is. $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}$ $\frac{60}{120}+\frac{30}{120}+\frac{20}{120}+\frac{15}{120}+\frac{12}{120}+\frac{10}{120}$ $\frac{147}{120}$ $\frac{49}{40}$

So we want two of the terms that sum to $\frac{49}{40}-1=\frac{9}{40}$.

Consider $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{9}{40}$. Therefore, we must have $40$ as a factor of $xy$. Notice that $10$ is the only possible value of $x$ and $y$ that's a multiple of $5$, so one of $x$ or $y$ must be $10$. Now subtract $\frac{9}{40}-\frac{1}{10}=\frac{9}{40}-\frac{4}{40}=\frac{5}{40}=\frac{1}{8}$, so the two fractions we must remove are $\frac{1}{8}\text{ and }\frac{1}{10}, \boxed{\text{E}}$.

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