Difference between revisions of "1970 AHSME Problems/Problem 10"
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== Solution == | == Solution == | ||
| − | <math>\fbox{ | + | Multiplying by <math>100</math> gives <math>100F = 48.181818...</math>. Subtracting the first equation from the second gives <math>99F = 47.7</math>, and all the other repeating parts cancel out. This gives <math>F = \frac{47.7}{99} = \frac{477}{990} = \frac{159}{330} = \frac{53}{110}</math>. Subtracting the numerator from the denominator gives <math>\fbox{D}</math>. |
== See also == | == See also == | ||
| − | {{AHSME box|year=1970|num-b=9|num-a=11}} | + | {{AHSME 35p box|year=1970|num-b=9|num-a=11}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:21, 13 July 2019
Problem
Let 
be an infinite repeating decimal with the digits 
and 
repeating. When 
is written as a fraction in lowest terms, the denominator exceeds the numerator by
Solution
Multiplying by 
gives 
. Subtracting the first equation from the second gives 
, and all the other repeating parts cancel out. This gives 
. Subtracting the numerator from the denominator gives 
.
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| 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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
