!!Residue!!

$\newline$ Let, $21! \equiv a\pmod{23}$ $\newline$ $22\cdot21! \equiv 22\cdot a\pmod{23}$ $\newline$ $22! \equiv 22\cdot a\pmod{23}$

But, from Wilson's theorem, we know that $22! \equiv -1 \pmod {23}$

$\therefore$ $22\cdot a \equiv 22! \equiv -1 \pmod {23}$ $22\cdot a \equiv -1 \pmod {23}$ $22\cdot a \equiv -1+23 \pmod {23}$ $22\cdot a \equiv 22 \pmod {23}$ $a \equiv 1 \pmod {23}$

Hence we get the remainder as 1

By Wilson's theorem , $(n-1)! \equiv -1 \text{ (mod n)}$ , where positive integer $n>1$ is a prime. Since 23 is a prime,

$\begin{aligned} 22 ! & \equiv - 1 \text{ (mod 23)} \\ 22 \times 21! & \equiv 22 \text{ (mod 23)} \\ \implies 21! & \equiv \boxed 1 \text{ (mod 23)} \end{aligned}$