Difference between revisions of "2002 AMC 12A Problems/Problem 20"
Megaboy6679 (talk | contribs) m (→Solution 1) |
|||
(4 intermediate revisions by 4 users not shown) | |||
Line 37: | Line 37: | ||
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> | 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 | + | 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 denominator. <math>\boxed{(C)}</math> |
~ Nafer | ~ Nafer | ||
~ edit by SpeedCuber7 | ~ edit by SpeedCuber7 | ||
+ | ~ edit by PojoDotCom | ||
=== Solution 3 === | === Solution 3 === | ||
Line 46: | Line 47: | ||
<cmath></cmath> | <cmath></cmath> | ||
~AopsUser101 | ~AopsUser101 | ||
+ | === Solution 4 (Alcumus) === | ||
+ | Since <math>0.\overline{ab} = \frac{ab}{99}</math>, the denominator must be a factor of <math>99 = 3^2 \cdot 11</math>. The factors of <math>99</math> are <math>1,</math> <math>3,</math> <math>9,</math> <math>11,</math> <math>33,</math> and <math>99</math>. Since <math>a</math> and <math>b</math> are not both nine, the denominator cannot be <math>1</math>. By choosing <math>a</math> and <math>b</math> appropriately, we can make fractions with each of the other denominators. | ||
+ | |||
+ | Thus, the answer is <math>\boxed{5}</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://youtu.be/0_0XpawpVVs | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=x086uFh-i00 | ||
== See Also == | == See Also == | ||
− | + | [[Category:Introductory Number Theory Problems]] | |
{{AMC12 box|year=2002|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2002|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:23, 16 June 2024
Contents
[hide]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 the 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 denominator.
~ Nafer ~ edit by SpeedCuber7 ~ edit by PojoDotCom
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
Solution 4 (Alcumus)
Since , the denominator must be a factor of . The factors of are and . Since and are not both nine, the denominator cannot be . By choosing and appropriately, we can make fractions with each of the other denominators.
Thus, the answer is .
Video Solution
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=x086uFh-i00
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.