Putnam 2009, Problem B-1

Show that every positive rational number can be written as a quotient of products of factorials of (not necessarily distinct) primes. For example,
\[
\frac{10}{9} = \frac{2! \cdot 5!}{3! \cdot 3! \cdot 3!}.
\]

Solution:
Every positive rational number can be uniquely written in lowest terms as \( a/b \) for \( a,b \) positive integers. We prove the statement in the problem by induction on the largest prime dividing either \( a \) or \( b \) (where this is considered to be 1 if \( a = b = 1 \)). For the base case, we can write \( 1/1 = 2!/2! \). For a general \( a/b \), let \( p \) be the largest prime dividing either \( a \) or \( b \); then \( a/b = p^k a'/b' \) for some \( k \neq 0 \) and positive integers \( a',b' \) whose largest prime factors are strictly less than \( p \). We now have
\[
\frac{a}{b} = (p!)^k \frac{a'}{(p-1)!^k b'}, \quad \text{and all prime factors of } a' \text{ and } (p-1)!^k b' \text{ are strictly less than } p.
\]
By the induction assumption, \( \frac{a'}{(p-1)!^k b'} \) can be written as a quotient of products of prime factorials, and so \( a/b = (p!)^k \frac{a'}{(p-1)!^k b'} \) can as well. This completes the induction.

**Remark.** Noam Elkies points out that the representations are unique up to rearranging and canceling common factors.