Difference between revisions of "1988 AIME Problems/Problem 5"

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== Solution ==
 
== Solution ==
<math>10^{99} = 2^{99}5^{99}</math>, so it has <math>(99 + 1)(99 + 1) = 10000</math> factors. Out of these, we only want those factors of <math>10^{99}</math> which are divisible by <math>10^{88}</math>; it is easy to draw a [[bijection]] to the number of factors that <math>10^{11} = 2^{11}5^{11}</math> has, which is <math>(11 + 1)(11 + 1) = 144</math>. Our probability is <math>\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}</math>, and <math>m + n = 634</math>.
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<math>10^{99} = 2^{99}5^{99}</math>, so it has <math>(99 + 1)(99 + 1) = 10000</math> factors. Out of these, we only want those factors of <math>10^{99}</math> which are divisible by <math>10^{88}</math>; it is easy to draw a [[bijection]] to the number of factors that <math>10^{11} = 2^{11}5^{11}</math> has, which is <math>(11 + 1)(11 + 1) = 144</math>. Our probability is <math>\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}</math>, and <math>m + n = \boxed{634}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 22:21, 6 November 2016

Problem

Let $m/n$, in lowest terms, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$. Find $m + n$.

Solution

$10^{99} = 2^{99}5^{99}$, so it has $(99 + 1)(99 + 1) = 10000$ factors. Out of these, we only want those factors of $10^{99}$ which are divisible by $10^{88}$; it is easy to draw a bijection to the number of factors that $10^{11} = 2^{11}5^{11}$ has, which is $(11 + 1)(11 + 1) = 144$. Our probability is $\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}$, and $m + n = \boxed{634}$.

See also

1988 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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