Prime number characterisation using congruences I want to prove that $n$ is prime. From the Wilson's theorem it follows that $n$ is prime if and only if
$$(n-1)! + 1 \equiv 0 \pmod{n}$$ 
However, in my proof, I reduce the congruences to the following form:
$$24((n-1)! + 1) \equiv 0 \pmod{n}$$
I believe I can divide the congruence by 24 only if 24 is coprime to $n$, so I can safely deduce that if $n$ is coprime to 24, it is also prime. Otherwise I do not know what can I find out from the following congrunce. Numerical calculations suggest that this successfully characterises primes grater than 24, but I fail to see why. Can anyone please explain to me which (probably obvious and simple) step I am missing?
Edit:
Perhaps the question was not clear enough. It can be expressed in shorter form, without the background:
Given the congruence
$$24((n-1)! + 1) \equiv 0 \pmod{n}$$
what can one tell about the primality of $n$?
 A: The proposed congruence $24((n-1)! + 1) \equiv 0 \pmod{n}$ does not hold for all positive integers $n$: the first two counterexamples are $n = 9$ and $n = 10$. 
In fact:
$\bullet$ If $2^4 \mid n$, then $24((n-1)!+1)$ is not divisible by $16$, so the congruence fails.  
$\bullet$ For any odd prime $p$, if $p^2 \mid n$ then $24((n-1)!+1) \equiv p \pmod{p^2}$, so the congruence fails.
$\bullet$ If $n$ is of the form $ap$ for a prime $p \geq 5$ and $a \geq 2$, then 
$24((n-1)!+1)$ is not divisible by $p$, so the congruence fails.  
Can you finish it off from here?
A: Yours is special case $\rm\ A = 24\ $ in the following generalization.
Theorem $\rm\,\ \ N\mid A\,(1\!+\!(N\!-\!1)!)\iff N\mid A\ \ or\ \ N\:$ is prime.
Proof $\,\ (\Rightarrow)\ $ $\rm\:d := gcd(N,A).\,$ By Euclid, $\rm\:d=1\:\Rightarrow\:N\mid(1\!+\!(N\!-\!1)!)\:\Rightarrow\:N\,$ prime by Wilson.
Else  $\rm\:d>1\:$ and $\rm\: N =  n d,\ A =  a d,\:$ for $\rm\:n,a\in \Bbb N.\:$ Cancelling $\rm\:d\:$ yields $\rm\:  n\mid \color{#c00}a\,(1\! +\! (\color{#0a0}{nd\!-\!1)!\,}).\ $
Note $\rm\:d>1\:\Rightarrow\: n\mid\color{#0a0}{(n d\!-\!1)!\,},\ $ so $\rm\  n\mid  \color{#c00}a,\ $ so $\rm\: n d\mid  a d,\:$ i.e. $\rm\:N\mid A.\:$  $\ (\Leftarrow)\ $  is clear, by Wilson.

Remark $\ \ $ The above is the special case $\rm\ f(N)\, =\, 1+(N\!-\!1)!\ $ of the following
Theorem $\ \ $ If $\rm\,\ f\, :\, \Bbb N\to \Bbb N\:$ satisfies $\rm\ d>1\:\Rightarrow\:\color{#c00}{gcd(n,f(nd))= 1}\ $ for all $\rm\:n,d\in\Bbb N,\ $ then
$$\rm N\mid A\,f(N)\ \iff\ N\mid A\ \ or\ \ N\mid f(N)\qquad\qquad$$
Proof $\ \ $ Let $\rm\ d = (N,A),\,\ N = nd,\,\ A = ad.\ $ Canceling the factor $\rm\:d\:$ from $\rm\ N,A\ $ yields
if $\rm\:\ d> 1\ $ then $\rm\ N\mid A\,f(N)\iff \color{#c00}n\mid a\,\color{#c00}{f(nd)\stackrel{Euclid}{\iff}} n\mid a\:\iff nd\mid ad\iff N\mid A.$ 
If $\rm\ d = 1\ $ then $\rm\ N\mid A\,f(N) \stackrel{Euclid}{\iff} N\mid f(N),\ $ by $\rm\ (N,A) = d = 1.$
$ $
A: Here's how my sister solved it
(Not a valid proof though)
$N$ be any composite number. Such that $N$'s factorization of primes is $p_1p_2 \dots p_k$
$\{p_1, \dots p_k \} \subset \{1,2, \dots n-1 \}$, therefore $N | (N-1)!$ and therefore $(N-1)!$ can't be $k (\mod N)$, where $k \neq 0$. The case of $N$ being composite is thus dismissed.
