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 $\boxed{\text{(E)} \ \frac{1}{8}\text{ and }\frac{1}{10}}$ .

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1985 AHSME Problems/Problem 5

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