Problem 220

Solution:

First we factor $12! = 2^{10} 3^{5} 5^{2} 7^{1} 11^{1}$, and note that $2, 5, 11 \equiv -1 \pmod{3}$ while $7 \equiv 1 \pmod{3}$. The desired divisors are precisely $2^{a} 5^{b} 7^{c} 11^{d}$ with $0 \leq a \leq 10$, $0 \leq b \leq 2$, $0 \leq c \leq 1$, $0 \leq d \leq 1$, and $a + b + d$ even. But then for any choice of $a, b$, exactly one $d \in \{0, 1\}$ makes $a + b + d$ even, so we have exactly one $1 \pmod{3}$-divisor for every triple $(a, b, c)$ satisfying the inequality constraints. This gives a total of $(10+1)(2+1)(1+1) = 66$.