Quicker way to solve 10! congruent to x (mod 11) I am new to modular arithmetic and solving congruences and the way I went about this was to write out $10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot2$, then multiply numbers until I get a number greater than $11$, replace it with a smaller number in its congruence class and repeat.  Is there a quicker way to go about this?  The next question I must solve is $100!\equiv x \pmod {101}$.

We can pair numbers, since $$10 \equiv -1 \pmod{11}$$ $$9 \equiv -2 \pmod{11}$$
and so on. So your product is the same as
$$(-1) (-2 \cdot 2)(-3 \cdot 3)(-4 \cdot 4)(-5 \cdot 5) \pmod{11}$$
or after dealing with signs,
$$-2^2 \cdot 3^2 \cdot 4^2 \cdot 5^2 \pmod{11}$$
Now $2^2 \equiv 4$, $3^2 \equiv -2$, $4^2 \equiv 5$ and $5^2 \equiv 3$, so this can be written as
$$-(4)(-2)(5)(3) \pmod{11}$$
or $$8 \cdot 5 \cdot 3 \pmod{11}$$
Now $40 \equiv 7$ and $21 \equiv -1$ leads to
$$\boxed{-1 \pmod{11}}$$
