smc

A number in base $b$ that ends in exactly $k$ zeros will be divisible by $b^k$, but not by $b^{k+1}$. Thus, we want to find the highest $k$ for which $12^k|10!$. There are $4$ factors of $3$: $3,6,9$, and an extra factor from $9$. There are $8$ factors of $2$: $2,4,6,8,10$, an extra factor from $4,8$, and a third extra factor from $8$. So, $2^8\cdot 3^4=4^4\cdot 3^4=12^4$ will divide $10!$. Thus, the answer is $\boxed{4}$.