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Difference between revisions of "1991 AIME Problems/Problem 5"

 
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== Problem ==
 
== Problem ==
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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 ==
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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 ==
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* [[1991 AIME Problems]]
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{{AIME box|year=1991|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 19:00, 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 $0$ and $1$ will $20_{}^{}!$ be the resulting product?

Solution

If the fraction is in the form $\frac{a}{b}$, then $a < b$ and $\gcd(a,b) = 1$. There are 8 prime numbers less than 20 ($2, 3, 5, 7, 11, 13, 17, 19$), and each can only be a factor of one of $a$ or $b$. There are $2^8$ ways of selecting some combination of numbers for $a$; however, since $a<b$, only half of them will be between $0 < \frac{a}{b} < 1$. Therefore, the solution is $\frac{2^8}{2} = \boxed{128}$.

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

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