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 ( | + | Another way to convert the decimal into a fraction (simplifying, I guess?). 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> |
− | + | where <math>a, b</math> are digits. Continuing in the same way by looking at the factors of 99, we have 5 different possibilities for the denomenator. <math>\boxed{(C)}</math> | |
+ | |||
+ | ~ Nafer | ||
+ | ~ edit by SpeedCuber7 | ||
+ | |||
+ | === Solution 3 === | ||
+ | Since <math>\frac{1}{99}=0.\overline{01}</math>, we know that <math>0.\overline{ab} = \frac{ab}{99}</math>. From here, we wish to find the number of factors of <math>99</math>, which is <math>6</math>. However, notice that <math>1</math> is not a possible denominator, so our answer is <math>6-1=\boxed{5}</math>. | ||
+ | <cmath></cmath> | ||
+ | ~AopsUser101 | ||
== See Also == | == See Also == |
Revision as of 16:10, 6 September 2021
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
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 (simplifying, I guess?). We have
where
are digits. Continuing in the same way by looking at the factors of 99, we have 5 different possibilities for the denomenator.
~ Nafer ~ edit by SpeedCuber7
Solution 3
Since , we know that
. From here, we wish to find the number of factors of
, which is
. However, notice that
is not a possible denominator, so our answer is
.
~AopsUser101
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.