Difference between revisions of "2002 AMC 12A Problems/Problem 20"
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=== Solution 2 === | === Solution 2 === | ||
| − | Another way to convert the decimal into a fraction (no idea what it's called). We have < | + | Another way to convert the decimal into a fraction (no idea what it's called). We have <cmath>100\cdot 0.\overline{ab} = ab.\overline{ab}</cmath> <cmath>99\cdot 0.\overline{ab} = 100\cdot 0.\overline{ab} - 0.\overline{ab} = ab</cmath> <cmath>0.\overline{ab} = ab/99</cmath> |
== See Also == | == See Also == | ||
Revision as of 20:26, 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 ![\[100\cdot 0.\overline{ab} = ab.\overline{ab}\]](http://latex.artofproblemsolving.com/3/e/6/3e642048f50e20bb608f54f45222c32b8f844f7b.png)
![\[99\cdot 0.\overline{ab} = 100\cdot 0.\overline{ab} - 0.\overline{ab} = ab\]](http://latex.artofproblemsolving.com/2/4/a/24a78d3b9c0bf6b221df1ba6548bdc716c52f8df.png)
![\[0.\overline{ab} = ab/99\]](http://latex.artofproblemsolving.com/9/4/4/944c61952f730a0f28e6321647900ff7be5942ef.png)
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. 
