# Difference between revisions of "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 $\boxed{\textbf{(D)}}$ .