2010 AMC 10A Problem 24

The number of trailing zeroes in $90!$ is $\left\lfloor\dfrac{90}{5}\right\rfloor+\left\lfloor\dfrac{90}{25}\right\rfloor=21.$ Let $N=\dfrac{90!}{10^{21}}.$ There are still more than two factors of $2$ left after removing $10^{21},$ so $N\equiv0 \pmod4.$ Let $A$ be the product of factors of $90!$ not divisible by $5,$ and let $B$ be the product of the factors divisible by $5.$ Grouping residues modulo $25$ gives $A\equiv1\pmod{25}$ and $\dfrac{B}{5^{21}}\equiv-1\pmod{25}.$ Therefore $\dfrac{90!}{5^{21}}\equiv-1\pmod{25}.$ Since $2^{21}\equiv2\pmod{25},$ $N=\dfrac{90!}{5^{21}\cdot2^{21}}$ $\equiv-13$ $\equiv12\pmod{25}.$ The number congruent to $0\pmod4$ and $12\pmod{25}$ is $12\pmod{100},$ so the last two nonzero digits form $12.$ Thus, A is the correct answer.