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 $20_{}^{}!$ 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 (ProblemsAnswer KeyResources)
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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