Difference between revisions of "1970 AHSME Problems/Problem 10"

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== Problem ==
 
== Problem ==
  
Let <math>F=.48181\cdots</math> be an infinite repeating decimal with the digits <math>8</math> and *1* repeating. When <math>F</math> is written as a fraction in lowest terms, the denominator exceeds the numerator by
+
Let <math>F=.48181\cdots</math> be an infinite repeating decimal with the digits <math>8</math> and <math>1</math> repeating. When <math>F</math> is written as a fraction in lowest terms, the denominator exceeds the numerator by
  
 
<math>\text{(A) } 13\quad
 
<math>\text{(A) } 13\quad

Revision as of 17:35, 1 October 2014

Problem

Let $F=.48181\cdots$ be an infinite repeating decimal with the digits $8$ and $1$ repeating. When $F$ is written as a fraction in lowest terms, the denominator exceeds the numerator by

$\text{(A) } 13\quad \text{(B) } 14\quad \text{(C) } 29\quad \text{(D) } 57\quad \text{(E) } 126$

Solution

$\fbox{A}$

See also

1970 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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
All AHSME Problems and Solutions

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