1991 AIME Problems/Problem 5

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 $20_{}^{}!$ the resulting product?

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