1960 AHSME Problems/Problem 8

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Problem

The number $2.5252525\ldots$ can be written as a fraction. When reduced to lowest terms the sum of the numerator and denominator of this fraction is:

$\textbf{(A) }7\qquad \textbf{(B) }29\qquad \textbf{(C) }141\qquad \textbf{(D) }349\qquad \textbf{(E) }\text{none of these}$

Solution

Solution 1

Let $x = 2.5252525\ldots$ , so \[100x = 252.52525\ldots\] \[x = 2.5252525\ldots\] \[99x = 250\] \[x = \frac{250}{99}\]

The sum of the numerator and the denominator is $349$, so the answer is $\boxed{\textbf{(D)}}$ .

Solution 2

The number $2.5252525\ldots$ can also be written as \[2+\frac{5}{10} + \frac{2}{100} + \frac{5}{1000} + \frac{2}{10,000} \ldots\] This is really two infinite series -- one with first term $2$ and common difference $\frac{1}{100}$ and one with first term $\frac{5}{10}$ and common difference $\frac{1}{100}$. Apply the infinite series formula $S = \frac{a_1}{1-r}$ on both series. \[\frac{2}{1-\frac{1}{100}} + \frac{\frac{5}{10}}{1-\frac{1}{100}}\] \[\frac{2}{\frac{99}{100}} + \frac{\frac{5}{10}}{\frac{99}{100}}\] \[2 \cdot \frac{100}{99} + \frac{5}{10} \cdot \frac{100}{99}\] \[\frac{200}{99} + \frac{50}{99}\] \[\frac{250}{99}\]

The sum of the numerator and the denominator is $349$, so the answer is $\boxed{\textbf{(D)}}$ .

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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