imc

Since $10!$ is divisible by $100$, any factorial greater than $10!$ is also divisible by $100$. The last two digits of all factorials greater than $10!$ are $00$, so the last two digits of $10!+11!+...+2006!$ are $00$. () So all that is needed is the tens digit of the sum $7!+8!+9!$ $7!+8!+9!=5040+40320+362880=408240$ So the tens digit is $\boxed{\text{(C) }4}$. () A slightly faster method would be to take the $(\mathrm{mod}100)$ residue of $7!+8!+9!.$ Since $7!=5040,$ we can rewrite the sum as $$5040+8\cdot 5040+72\cdot 5040\equiv 40+8\cdot 40+72\cdot 40=40+320+2880\equiv 40(\mathrm{mod}100).$$ Since the last two digits of the sum is $40$, the tens digit is $\boxed{4}$.