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 $\boxed{\textbf{(A)}\ 15}$.

See Also

1965 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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