Difference between revisions of "2002 AMC 12A Problems/Problem 20"
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Another way to convert the decimal into a fraction (no idea what it's called). We have <cmath>100(0.\overline{ab}) = ab.\overline{ab}</cmath> <cmath>99(0.\overline{ab}) = 100(0.\overline{ab}) - 0.\overline{ab} = ab.\overline{ab} - 0.\overline{ab} = ab</cmath> <cmath>0.\overline{ab} = \frac{ab}{99}</cmath> | Another way to convert the decimal into a fraction (no idea what it's called). We have <cmath>100(0.\overline{ab}) = ab.\overline{ab}</cmath> <cmath>99(0.\overline{ab}) = 100(0.\overline{ab}) - 0.\overline{ab} = ab.\overline{ab} - 0.\overline{ab} = ab</cmath> <cmath>0.\overline{ab} = \frac{ab}{99}</cmath> | ||
| − | Continuing in the same way, we have 5 different possibilities for the denomenator. <math>\mathrm {(B)}</math> | + | where <math>a, b</math> are digits. Continuing in the same way, we have 5 different possibilities for the denomenator. <math>\mathrm {(B)}</math> |
| + | |||
| + | ~ Nafer | ||
== See Also == | == See Also == | ||
Revision as of 21:16, 1 July 2019
Problem
Suppose that 
and 
are digits, not both nine and not both zero, and the repeating decimal 
is expressed as a fraction in lowest terms. How many different denominators are possible?
Solution
Solution 1
The repeating decimal is equal to
![\[\frac{10a+b}{100} + \frac{10a+b}{10000} + \cdots = (10a+b)\cdot\left(\frac 1{10^2} + \frac 1{10^4} + \cdots \right) = (10a+b) \cdot \frac 1{99} = \frac{10a+b}{99}\]](http://latex.artofproblemsolving.com/7/5/e/75e79d4e1c4a5a5a280b24504155a203de95908f.png)
When expressed in lowest terms, the denominator of this fraction will always be a divisor of the number 
. This gives us the possibilities 
. As and are not both nine and not both zero, the denominator 
can not be achieved, leaving us with 
possible denominators.
(The other ones are achieved e.g. for 
equal to 
, 
, 
, 
, and , respectively.)
Solution 2
Another way to convert the decimal into a fraction (no idea what it's called). We have ![\[100(0.\overline{ab}) = ab.\overline{ab}\]](http://latex.artofproblemsolving.com/8/7/0/8707e53b3d4768c851c5a79f6de7e4e1c9b50ff5.png)
![\[99(0.\overline{ab}) = 100(0.\overline{ab}) - 0.\overline{ab} = ab.\overline{ab} - 0.\overline{ab} = ab\]](http://latex.artofproblemsolving.com/c/4/c/c4c34df03e28538a13094fec999535943fe7c461.png)
![\[0.\overline{ab} = \frac{ab}{99}\]](http://latex.artofproblemsolving.com/0/d/b/0dba88361c68a63ea9cfe47d53c29c700459dd47.png)
where 
are digits. Continuing in the same way, we have 5 different possibilities for the denomenator. 
~ Nafer
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 19 |
Followed by Problem 21 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
