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 $100!\equiv x \pmod {101}$.
 A: Wilson's Theorem states that for any prime number $n$, you get $$(n-1)! \equiv -1 (mod\ n).$$
$10!$ makes $n = 11$, which is prime, so $10! \equiv -1 (mod\ 11) \equiv 10 (mod\ 11)$
Likewise, $100!$ makes $n = 101$, which is prime, so $100! \equiv -1 (mod\ 101) \equiv 100 (mod\ 101)$.
A: 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
$$\boxed{-1 \pmod{11}}$$
A: If you don't know Wilson's theorem you might stumble upon its idea tackling this problem using elementary facts and being lazy.
The non-zero elements in $\displaystyle \mathbb {Z} /11\mathbb {Z}$ form a group and each element has a unique inverse. OK, listing the elements,
$\tag 1 {\overline {1}},{\overline {2}},{\overline {3}},{\overline {4}},{\overline {5}},{\overline {6}},{\overline {7}},{\overline {8}},{\overline {9}},{\overline {10}}$
and you get a little excited since $10!$ modulo $11$ boils down to multiplying them all together.
So an easy first step for the lazy mathematician is to replace the list with
$\tag 2 {\overline {2}},{\overline {3}},{\overline {4}},{\overline {5}},{\overline {6}},{\overline {7}},{\overline {8}},{\overline {9}},{\overline {-1}}$
Looking for pairs of elements, each the multiplicative inverse of the other, you wonder if you can multiply two representative numbers  together and get $12$. Yup, $2 \times 6 = 12$ and $3 \times 4 = 12$. Great, now looking at
$\tag 3 {\overline {5}},{\overline {7}},{\overline {8}},{\overline {9}},{\overline {-1}}$
How about 'multiplying out' to $23 = 12 + 11$. Nope.
How about 'multiplying out' to $34 = 23 + 11$. Nope.
How about 'multiplying out' to $45 = 34 + 11$. Yup, $5 \times 9 = 45$. New list
$\tag 4 {\overline {7}},{\overline {8}},{\overline {-1}}$
How about 'multiplying out' to $56 = 45 + 11$. Yup, $7 \times 8 = 56$.
ANS: $10! \equiv -1 \pmod{11}$

Note: If $p$ is a prime the solutions to $x^2 \equiv 1 \pmod{p}$ are $x \in \{1,-1\}$. So with $p \ge 5$ the elements in the non-empty set
$\quad \displaystyle \mathbb {Z} /p\mathbb {Z} \setminus \{{\overline {0}},{\overline {1}}, {\overline {-1}}\}$
can be always be '$a\text{-}b$ paired off' with 
$\quad \displaystyle ab \equiv 1 (\text{mod } p) \land a \ne b$
