Show that each number of $100!+1,100!+2,100!+3,...,100!+100$ is composite

Question

I'm working in the following Wilson's theorem excercise:

Show that each number of $100!+1,100!+2,100!+3,...,100!+100$ is composite number.

I'm starting from: $$100! \equiv -1 \pmod {101}$$

My thoughts about this is that I may be able to add the required number each time to that congruence, so for the first case I would have:

$$100! +1\equiv (-1)(+1) \pmod {101}$$ $$100! +1\equiv 0 \pmod {101}$$

Showing that $101k=100!+1$ so $100!+1$ is a composite number and I would basically do the same for each number.

Is that correct? Am I missing something? Any hint, help or correction will be really appreciated.

Answer 1

$101$ is prime so by Wilson's theorem $100!\equiv -1\pmod {101}$ so $100!+1\equiv 0\pmod {101} $ so $101|100!+1$ .

For $1 <k\le 100$ we simply note the $k|100! $ (by definition) and $k|k $ so $k|100!+k $.

Note $N! +1$ might be prime for some $N $ but not if $N+1$ is prime. $3!+1=7$ is prime (but $3+1$ isnt).

Answer 2

This has nothing to do with Wilson's theorem. If $1<k\leq n$ then ${n!\over k}\in \mathbb{N}$ and we have:

$$ n!+k = k \cdot \underbrace{({n!\over k}+1)}_{>1}$$