Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Fixed a word order mistake)
Line 3: Line 3:
  
 
== 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>.  
+
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} = 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}}

Revision as of 18:20, 4 July 2013

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 $\frac{2^8}{2} = 128$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_5&oldid=54966"