HMMT November

First we factor $12!=2^{10} 3^{5} 5^{2} 7^{1} 11^{1}$, and note that $2,5,11 \equiv-1(\bmod 3)$ while $7 \equiv 1$ $(\bmod 3)$. 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 $1(\bmod 3)$-divisor for every triple $(a, b, c)$ satisfying the inequality constraints. This gives a total of $(10+1)(2+1)(1+1)=66$.