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

Revision as of 20:53, 22 February 2011

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 ( $\displaystyle 2, 3, 5, 7, 11, 13, 17, 19$ ), 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 $\frac{2^8}{2} = 128$ .

@@ Line 1: / Line 1: @@
 == 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 <math>20_{}^{}!</math> be the resulting [[product]]?
 == Solution ==

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

Page

Toolbox

Search

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

Revision as of 20:53, 22 February 2011

Problem

Solution

See also