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 | + | 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 <math>0</math> and <math>1</math> will <math>20_{}^{}!</math> be 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>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} = \boxed{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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:00, 4 July 2022
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 
and 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
