Can somebody tell me if this is true : 100! congruents to 201 mod 101. And what can I do to know that. Thanks

Question

101 is prime, so (100!,101)=1. But does it help?

Answer 1

2 integers a and b are congruent modulo n if their difference is an integer multiple of n. This is essentially saying that if 2 numbers leave the same remainder when divided by n then their difference will not leave a remainder when divided by n

According to Wilson's theorem, (p-1)! = -1 (mod p) if and only if p is prime

let p = 101 - we know that p is prime

Then we know that 100! = -1 (mod 101)

and that -1 mod 101 = 100 mod 101 = 201 mod 101

thus 100! = 201 (mod 101)

Answer 2

Well 201 = 100 (mod 101), so we want to know if 100! is congruent to 100 (mod 101).

Essentially we want to know if there exists an integer x such that:
101x + 1 = 100!

We can rewrite as 101x = 100! - 1. So the question is reduced to whether (100! - 1) is divisible by 101. We know that 100! is not divisible by 101, but I can't think of a clever way to find the result for (100! - 1). Hopefully somebody else can take it from here.

EDIT: Ack, I forgot about Wilson! Thanks Steiner.

Answer 3

It's true:
Use Wilson's theorem: If p is prime, then
(p-1)! = -1(mod p)
Since 101 is prime
100! = 100(mod 101) = 201(mod 101)
by just adding 101 to the rhs.