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>. | + | <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 21:21, 6 November 2016
Problem
Let 
, in lowest terms, be the probability that a randomly chosen positive divisor of 
is an integer multiple of 
. Find 
.
Solution
, so it has 
factors. Out of these, we only want those factors of which are divisible by ; it is easy to draw a bijection to the number of factors that 
has, which is 
. Our probability is 
, and 
.
See also
| 1988 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
