Difference between revisions of "1965 AHSME Problems/Problem 5"
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We let <math>x=0.\overline{36}</math>. Thus, <math>100x=36.\overline{36}</math>. We find that <math>100x-x=99x=36.\overline{36}-0.\overline{36}=36</math>, or <math>x=\frac{36}{99}=\frac{4}{11}</math>. Since <math>4+11=15</math>, the answer is <math>\boxed{\textbf{(A)}\ 15}</math>. | We let <math>x=0.\overline{36}</math>. Thus, <math>100x=36.\overline{36}</math>. We find that <math>100x-x=99x=36.\overline{36}-0.\overline{36}=36</math>, or <math>x=\frac{36}{99}=\frac{4}{11}</math>. Since <math>4+11=15</math>, the answer is <math>\boxed{\textbf{(A)}\ 15}</math>. | ||
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| + | ==See Also== | ||
| + | {{AHSME box|year=1965|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:AHSME]][[Category:AHSME Problems]] | ||
Latest revision as of 20:17, 10 January 2023
Problem
When the repeating decimal 
is written in simplest fractional form, the sum of the numerator and denominator is:
Solution
We let 
. Thus, 
. We find that 
, or 
. Since 
, the answer is 
.
See Also
| 1965 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
