Difference between revisions of "1991 AIME Problems/Problem 5"
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== Problem == | == Problem == | ||
| − | Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will be <math>20_{}^{}!</math> the resulting product? | + | Given a [[rational number]], write it as a [[fraction]] in lowest terms and calculate the product of the resulting [[numerator]] and [[denominator]]. For how many rational numbers between 0 and 1 will be <math>20_{}^{}!</math> the resulting [[product]]? |
== Solution == | == Solution == | ||
| − | {{solution}} | + | If the fraction is in the form <math>\frac{a}{b}</math>, then <math>a < b</math> and <math>gcd(a,b) = 1</math>. There are 8 [[prime number]]s less than 20 (<math>\displaystyle 2, 3, 5, 7, 11, 13, 17, 19</math>), and each can only be a factor of one of <math>a</math> or <math>b</math>. There are <math>2^8</math> ways of selecting some [[combination]] of numbers for <math>a</math>; however, since <math>a<b</math>, only half of them will be between <math>0 < \frac{a}{b} < 1</math>. Therefore, the solution is <math>\frac{2^8}{2} = 128</math>. |
== See also == | == See also == | ||
{{AIME box|year=1991|num-b=4|num-a=6}} | {{AIME box|year=1991|num-b=4|num-a=6}} | ||
Revision as of 17:45, 11 March 2007
Problem
Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will be 
the resulting product?
Solution
If the fraction is in the form 
, then 
and 
. There are 8 prime numbers less than 20 (
), and each can only be a factor of one of 
or 
. There are 
ways of selecting some combination of numbers for ; however, since 
, only half of them will be between 
. Therefore, the solution is 
.
See also
| 1991 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 | ||
