1965 AHSME Problems/Problem 5

Problem

When the repeating decimal $0.363636\ldots$ is written in simplest fractional form, the sum of the numerator and denominator is:

$\textbf{(A)}\ 15 \qquad \textbf{(B) }\ 45 \qquad \textbf{(C) }\ 114 \qquad \textbf{(D) }\ 135 \qquad \textbf{(E) }\ 150$

Solution

We let $x=0.\overline{36}$ . Thus, $100x=36.\overline{36}$ . We find that $100x-x=99x=36.\overline{36}-0.\overline{36}=36$ , or $x=\frac{36}{99}=\frac{4}{11}$ . Since $4+11=15$ , the answer is $\boxed{\textbf{(A)}\ 15}$ .

1965 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1965 AHSME Problems/Problem 5

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