What is the remainder when 22! is divided by 43.

43 is a prime number.
So, by Wilson's theorem, 42! mod 43 = 42 mod 43
=> 42*41*40*39*...*23*22! mod 43 = 42 mod 43
=> (-1)(-2)(-3)...(-20)*22! mod 43 = 42 mod 43
=> 22*21*20!*22! mod 43 = 42 * 22*21 mod 43
=> $$22!^2$$ mod 43 = 11
We have to find 22! mod 43. Let this number be x. x is a natural number.
So, $$x^2$$ mod 43 = 11 --> (1)
From the given options, we see that, $$20^2$$ mod 43 = 13
$$1^2$$ mod 43 = 1
$$23^2$$ mod 43 = 13
Only $$21^2$$ mod 43 = 441 mod 43 = 11, which satisfies equation (1)
So, the answer is option a).