1960 AHSME Problems/Problem 8

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 $\boxed{\textbf{(D)}}$ .

1960 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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1960 AHSME Problems/Problem 8

Contents

Problem

Solution

Solution 1

Solution 2

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