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How to prove that if $n>4$ is a composite number, then $(n-1)!$ is a multiple of $n$?

I don't have an idea, where to start. Grateful for a hint.

Matthew Conroy
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Snsn
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2 Answers2

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Hints:

$\;n>4\;$ composite $\;\implies \,\exists\,1<a,b\le n-1\;\;s.t.\;\;ab=n\;$.

Now write carefully the expression for $\;(n-1)!\;\ldots$

DonAntonio
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    The obvious obstacle here is if $a=b$. You may want to add a further hint indicating how this issue is solved (this may clarify to the OP why the restriction to $n>4$ plays a role beyond the fact that "if $n=4$ then the result fails"). – Andrés E. Caicedo Feb 26 '14 at 19:29
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    @AndresCaicedo, shall we let the OP reach his own conclussions and problems and, in case of need, he can write back...My point was the OP could arribe by himself to the interrogation "hey, why the $;n>4;$ condition??", and now you blew it... – DonAntonio Feb 26 '14 at 19:31
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    No, my impression with questions at this level is that these issues are too subtle to be detected if the question needs to be asked in the first place. – Andrés E. Caicedo Feb 26 '14 at 19:35
  • This question seems to be at level of senior high school or even college/university: not that low to demmand from peopl t check carefully what they write. – DonAntonio Feb 26 '14 at 19:47
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    I agree with Andres, this answer has a major gap, and it is pedagogically misleading to give no hint of that. – Bill Dubuque Feb 26 '14 at 20:39
  • @DonAntonio could please tell why n=4 doesn't work like how can we prove rigorously ,in writing it's obvious – Riya Verma Jul 02 '19 at 05:36
  • @RiyaVerma Simply because $;4\nmid 3!=6;$ ... – DonAntonio Jul 02 '19 at 11:38
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Hint $\rm\:\!\ n\, =\, \color{#C00}a\:\!\color{#0A0}b\mid1\!\cdot\! 2\cdots\color{#C00} a\:\color{#0A0}{(a\!+\!1)\, (a\!+\!2) \cdots (a\!+\!b)}\cdots (\color{#0af}{ab\!-\!1})=(n\!-\!1)!\ $ when $\rm\, \color{#0a0}{a\!+\!b} \le \color{#0af}{ab\!-\!1} $

$\rm\color{#0A0}b\,$ divides $\rm\color{#0A0}{green}$ product since a sequence of $\rm\,b\,$ consecutive integers has a multiple of $\rm\,b,\,$ and

$\rm\, \color{#0af}{ab\!-\!1} \ge \color{#0a0}{a\!+\!b},\:$ i.e. $\rm\, (\underbrace{a\!-\!1}_{\large \ge\,1})(\underbrace{\color{#c0d}{b\!-\!1}}_{\large\ge\, 2}) \ge 2\,$ is true: $\rm\,a,b\ge 2,\,$ but $\rm\underbrace{not\ both =2,}_{\large n\,=\,ab\,\ne\, 4}\,$ so $\,\color{#c0f}{\rm one}$ is $\,\ge 3.$

Bill Dubuque
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  • I didn't downvote (I even cannot...), but I don't have idea what this answers says. – Timbuc Feb 26 '14 at 20:47
  • @Timbuc It's a hint to show how to infer that $\rm,n\mid (n-1)!$ when $\rm,n = ab,$ is composite $> 4.,$ If you can tell me precisely where you are stuck I can explain further. Note $\ x\mid y\ $ means $x$ divides $y.\ $ – Bill Dubuque Feb 26 '14 at 20:49
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    Thank you sir. At the first line you mean that $;ab;$ divides all that product $;1\cdot2\cdot\ldots\cdot a(a+1)\cdot\ldots\cdot(ab-1);$? Ah, I see...because $;b;$ divides the greens...right, but why do you write "when $;a+b\le ab-1;$, and what happens if this is not true? – Timbuc Feb 26 '14 at 20:54
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    @Tim The inequality $,a+b\le ab-1,$ implies that the $\rm\color{#0a0}{green}$ sequence of factors is a subsequence of the factorial factors, since its end term $,a+b,$ is no greater than the end term $,ab-1,$ of the factors in the product $,(ab!-!1)! = (n!-!1)!\ $ – Bill Dubuque Feb 26 '14 at 21:06