Determine the least non-negative residue $x$ of the following congruences:
(a) $10! ≡ x \pmod {11}$
(b) $10! + 10! ≡ x \pmod {11}$
(c) $10 (10!) + 8 (10!) ≡ x \pmod {11}$
(d) $5 (10!)^{101} + 3 (10!)^{100} ≡ x \pmod {11}$
We remind ourselves of Wilson's Theorem,
$$ (n − 1)! \equiv −1 \pmod {n} \quad \iff \quad n \text{ is prime} $$
(a) Since 11 is prime, Wilson's Theorem gives us
$$ \begin{align} 10! & \equiv -1 \pmod {11} \\ \\ & \equiv 10 \pmod{11} \end{align}$$
The least non-negative residue is $x = 10$.
(b) Since 11 is prime, Wilson's Theorem gives us
$$ \begin{align} 10! & \equiv -1 \pmod {11} \\ \\ 2 \times 10! & \equiv -2 \pmod{11} \\ \\ 10! + 10! & \equiv 9 \pmod {11} \end{align}$$
The least non-negative residue is $x = 9$.
(c) Since 11 is prime, Wilson's Theorem gives us
$$ \begin{align} 10! & \equiv -1 \pmod {11} \\ \\ 18 \times 10! & \equiv -18 \pmod{11} \\ \\ 10(10!) + 8(10!) & \equiv 4 \pmod {11} \end{align}$$
The least non-negative residue is $x = 4$.
(d) Since 11 is prime, Wilson's Theorem gives us
$$ \begin{align} 10! & \equiv -1 \pmod {11} \\ \\ 10!^{100} & \equiv 1 \pmod{11} \\ \\ 10!^{101} & \equiv -1 \pmod{11} \\ \\ 5(10!)^{101} + 3(10!)^{100} & \equiv 5(-1) + 3(1) \pmod{11} \\ \\ & \equiv -2 \pmod{11} \\ \\ \equiv 9 \end{align}$$
The least non-negative residue is $x = 9$.