Difference between revisions of "2002 AMC 12A Problems/Problem 20"
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== Solution == | == Solution == | ||
+ | |||
+ | === Solution 1 === | ||
The repeating decimal <math>0.\overline{ab}</math> is equal to | The repeating decimal <math>0.\overline{ab}</math> is equal to | ||
<cmath> | <cmath> | ||
− | \frac{ | + | \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{ | + | \frac{10a+b}{99} |
</cmath> | </cmath> | ||
− | When expressed in lowest terms, the denominator of this fraction will always be a divisor of the number <math>99 = 3\cdot 3\cdot 11</math>. This gives us the possibilities <math>\{1,3,9,11,33,99\}</math>. As <math>a</math> and <math>b</math> are not both nine and not both zero, the | + | When expressed in the lowest terms, the denominator of this fraction will always be a divisor of the number <math>99 = 3\cdot 3\cdot 11</math>. This gives us the possibilities <math>\{1,3,9,11,33,99\}</math>. As <math>a</math> and <math>b</math> are not both nine and not both zero, the denominator <math>1</math> can not be achieved, leaving us with <math>\boxed{\mathrm{(C) }5}</math> possible denominators. |
(The other ones are achieved e.g. for <math>ab</math> equal to <math>33</math>, <math>11</math>, <math>9</math>, <math>3</math>, and <math>1</math>, respectively.) | (The other ones are achieved e.g. for <math>ab</math> equal to <math>33</math>, <math>11</math>, <math>9</math>, <math>3</math>, and <math>1</math>, respectively.) | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | 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 denominator. <math>\boxed{(C)}</math> | ||
+ | |||
+ | ~ Nafer | ||
+ | ~ edit by SpeedCuber7 | ||
+ | ~ edit by PojoDotCom | ||
+ | |||
+ | === 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 | ||
+ | === 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}} |
Latest revision as of 14:23, 16 June 2024
Contents
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.