Difference between revisions of "1960 AHSME Problems/Problem 8"

Latest revision as of 18:02, 17 May 2018

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 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

@@ Line 34: / Line 34: @@
 ==See Also==
-{{AHSME box|year=1960|num-b=7|num-a=9}}
+{{AHSME 40p box|year=1960|num-b=7|num-a=9}}
+[[Category:Introductory Number Theory Problems]]

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