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> | 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 | ~ Nafer | ||
+ | ~ edit by SpeedCuber7 | ||
=== Solution 3 === | === Solution 3 === |
Revision as of 15:10, 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.