Just put $n=ab$ where $a,b \in \mathbb Z_+$ and $n \ge (a,b)$ then: $(n-1)!=(n-1)...(n-(n-2))(n-(n-1))(n-n)!=(n-1)...(n-a)(n-b)...2.1=2n(n-1)...(a-1)(b-1)...1$

I think it just means when $ a \ne b$ ,there exist 2 different numbers such that they are the divisors.But when $ a = b$ ,as $ n \ge 4$ ,$ \sqrt n \le \frac{n}{2}$,that means there are at least 2 different numbers below $ n-1$ such that $ a = b\left| k \right.,2k\left( {2k \le n - 1} \right)$.

Let $ n=2^k\times m$, where $ m$ is odd. $ m$ divides $ (n-1)!$ because $ m$ is either prime less than $ n$, or product of primes less than $ n$. So, we need to show that $ 2^{k+1}$ divides $ (n-1)!$ Among the numbers $ 2,3,...,n-1$ there are at least $ \frac {n}{2} -1=2^{(k-1)m}-1$ even numbers so their product is divisible by $ 2^{k+1}$ because $ 2^{(k-1)m}-1>k+1$ but in the case $ k=3,m=1$ it isn't correct, but again $ 2^4$ divides $ 7!$ That's the end of the solution.

shobber wrote: If $ n > 4$ is a composite number, then $ 2n$ divides $ (n - 1)!$. first we can see esaly that $ v_2(n)\le [(n-1)/2]<v_2((n-1)!)$ then $ 2n|(n-1)!\iff n|(n-1)!$. 1) if we can write $ n=ab$ such that $ 1<a<b<n$ $ (n-1)!=1.2.3...a.(a+1)...b(b+1)...n$ then $ n|(n-1)!$ 2) if $ n=p^2$ for some $ p>2$ prime. $ (n-1)!=1.2.3...p.(p+1)...(2p)(2p+1)...n$ then $ n|(n-1)!$

shobber wrote: If $n>4$ is a composite number, then $2n$ divides $(n-1)!$. At first hello I can't english good We know that all composite numbers n>4 can write n=ab when 2=<a,a=<b 2<b 1°We take a=b since n=ab=a²>4 than we have a>2 than we have 2a>4 and n=a²>2a well n>2a>4 since n>a>2 and n>2a>4 conspicuous (n-1)*...*2a*...*a*...*2*1 divides 2a*a=2a²=2n 2°Now we take a<b then we have n-1>n-a>n-b>2 the last inequality is true because n-b=n-n/a=n(1-1/a)>4*1/2=2 [n>4;1-1/a>=1/2] Then (n-1)!=(n-1)(n-2)...(n-a)...(n-b)...2*1 This product is conspicuous that divides with 2(n-a)(n-b)=2n(n-a-b+1) Because of (n-1)! divides with 2n(n-a-b+1) it divides also with 2n

PEN A 30 is similar, but weaker. The second solution is easily patchable: If $n=p^2$ with $p$ a prime then $p>2 \implies p^2>2p \implies p^2-1>p^2-p>p$ so $n=2p^2|(p^2-1)!=(p^2-1)\ldots (p^2-p)\ldots p \ldots 2\cdot 1$ since $p^2-p=p(p-2)$, so we have two $p$'s in our product. Otherwise $n=ab$ with $a,b>1$. So we are required to prove $2ab|(2ab-1)!$ If $a,b \not = 2$ then $(n-1)!=(n-1)\ldots a \ldots b \ldots 2 \cdot 1$ is a multiple of $2ab=2n$. If one of $a,b$ (say $b$) is equal to $2$ then $n=8$ (checkable) or $(n-1)!=(n-1) \ldots a \ldots 4 \ldots 2 \cdot 1$. It's also easy to see $4n|(n-1)!$, but difficult to find $ord_2 \left(\frac{(n-1)!}{n} \right)$

Well, de Polignac's formula gives: $ord_2 \left(\frac{(n-1)!}{n} \right)=\sum_{i=1}^{\infty} \left\lfloor{\frac{n-1}{2^i}}\right\rfloor-ord_2(n)$

Just for clarification: I've recently learned that $ord_p(m)$ also denotes the order of $m$ modulo $p$, instead of the exponent $e$ such that $p^e|m$ but $p^e \nmid m $ (or $p^e||m$). So replace $ord_{2}\left(\frac{(n-1)!}{n}\right)$ with $e_2\left(\frac{(n-1)!}{n}\right)$ if you wish. Also, while Wikipedia calls it de Polignac's formula, AoPS Wiki calls it Legendre's formula. (I was following the free weblication Infinity- first page).

Just use Wilsons's theorem

this is false for n=5

The question is for when n is composite and clearly, 5 is prime

We shall prove a more stronger result that $4n$ divides $(n-1)!$ for any composite number $n>4$ which will cover the case of problem statement. Let $n = p.q$ where $q \ge p \ge 2$. Let us define set $S = \{1, 2, 3 ... n-1\}$ $Case 1: q > p$ First let's note that $p, q \in S$ Now, all multiples of $p$ from $p.1$ to $p.(q-1) \in S$ Since $q > p => q > 2 => q-1 \ge 2$ we have that $p.2 \in S$ Also, since $p \ge 2$ we have that $2 \in S$ So, we have that $2, p.2, q \in S$, in other words, $4.p.q$ divides $(n-1)!$ $Case 2: q = p$ Now, all multiples of $p$ from $p.1$ to $p.(p-1) \in S$ Since $p > 2 => p-1 \ge 2$ we have that $p.2 \in S$ Also, since $n = p^2 > 4 => p > 2$ so we have that $2 \in S$ So, we have that $2, p.1, p.2 \in S$, in other words, $4.p^2$ divides $(n-1)!$ Thus $4n$ divides $(n-1)!$.

Can't you just make prime factorization of n and we have all prime factors in (n-1)! that has n.

why is a 13.5 year old thread getting revived....

fabijan_123 wrote: Can't you just make prime factorization of n and we have all prime factors in (n-1)! that has n. yeah but they can appear in n any number of times

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$n=ab$ where $a,b \in \mathbb Z_+$ then: $(n-1)!=(n-1)...(n-a)(n-b)...2*1=2(n-1)...(ab-a)(ab-b)...1=2(n-1)...a(b-1)b(a-1)$, with the a, b, and 2 for a product of 2n in the factors as desired.

I skimmed the solutions posted above. To me, we should analyze the case when $n=16$ separately. Or do I miss something?!

No because we simply need to use the fact that $n$ is composite $\Rightarrow n=ab$ for some $a,b>1$ and $\gcd{a,b}>1$. Then splitting into factors of $2$ and looking at appropriate parities gives the result.

S.Das93 wrote: No because we simply need to use the fact that $n$ is composite $\Rightarrow n=ab$ for some $a,b>1$ and $\gcd{a,b}>1$. Then splitting into factors of $2$ and looking at appropriate parities gives the result. Could you show a solution a case of which includes $n=16$? I guess, no need to mention that $16=2\times8=4\times4$