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 October 5th, 2017, 02:50 PM #1 Senior Member   Joined: Dec 2013 Posts: 1,112 Thanks: 41 Prove or disprove Let n=p*q, where p and q are primes. Prove (or disprove) that: n! is divisible by (p!)^q or by (q!)^p Thanks from zylo Last edited by skipjack; October 6th, 2017 at 04:18 AM.
 October 6th, 2017, 02:56 AM #2 Senior Member   Joined: Dec 2013 Posts: 1,112 Thanks: 41 This one is the most interesting. Everybody knows Wilson's theorem. I'm going to give 2 examples. Example 1: We want to check whether n=13 is prime or not. We compute n-1=13-1=12 12 is even hence 12=2*6 We compute (2!)^6 mod 13 We compute (6!)^2 mod 13 What will happen if 13 is prime? Example 2: We want to check whether n=15 is prime or not. We compute n-1=15-1=14 14 is even hence 14=2*7 We compute (2!)^7 mod 15 We compute (7!)^2 mod 15 What will happen if 15 is not prime? Those 2 examples shows that there is room for a new theorem for primality. Instead of computing (n-1)! mod n we break (n-1) is 2 and we compute separately the 2 parts as described above. If you could write in Latex and rewrite what I wrote, then will come the time to prove it. Last edited by skipjack; October 6th, 2017 at 11:25 AM.
 October 6th, 2017, 03:11 AM #3 Senior Member   Joined: Dec 2013 Posts: 1,112 Thanks: 41 Try this by factorizing n-1 (other than 2*stuff) in 2 factors and see what will happen. Try it for the first 100 natural numbers (from 1 to 100). That is only the first step. There are more to come. I will come back tomorrow because I have something to do today. Good work!
 October 6th, 2017, 10:10 AM #4 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749 Are you having a good time having a discussion with yourself? Did you really tell yourself "good work!" ? Maybe a bit of professional help is called for? Thanks from Joppy
 October 9th, 2017, 04:58 AM #5 Senior Member   Joined: Dec 2013 Posts: 1,112 Thanks: 41 Too hard maybe. 172 views! and no serious answer yet.
October 10th, 2017, 03:02 AM   #6
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Quote:
 Originally Posted by romsek Are you having a good time having a discussion with yourself? Did you really tell yourself "good work!" ? Maybe a bit of professional help is called for?
No one here is professional.
I said it a long time ago.
What you have here in this forum is that some people claim wrongly that they are professional because they "master" technical tricks. That's it.
You do not need high IQ just a memorizing such techniques.

I know personally many dumb guys who became maths teachers.

Last edited by skipjack; October 10th, 2017 at 03:27 AM.

 October 10th, 2017, 03:04 AM #7 Senior Member   Joined: Dec 2013 Posts: 1,112 Thanks: 41 Now! Adios The forum is dying because of dumb guys like Romstek, Varchie, Joppy and so on.
October 10th, 2017, 03:39 AM   #8
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Quote:
 Originally Posted by mobel Now! Adios The forum is dying because of dumb guys like Romstek, Varchie, Joppy and so on.
You pose problems you already know the answer to, then when nobody answers your posts (because, you know, you already have the answer...) you call us dumb and quit for a while, only to return weeks or months later and do the same thing all over again.

This forum isn't dying, it's just not what you want it to be... I can't speak for everyone, but generally speaking, I've found that most people here are not interested in stroking people's egos and are much more interested in helping people out with real study problems that they can't solve. In my particular case, I also have less time to dedicate to learning new mathematics than I used to and I often don't know the answers to the problems you're posing. I'm much more interested in helping people with simple physics, numerical techniques or elementary mathematics problems and I dip in where I can to help out.

If you post something and nobody responds, maybe you should ask yourself why no one is responding rather than just moaning about the forum.

 October 10th, 2017, 09:26 AM #9 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90 (3$\displaystyle \cdot$2)!=6$\displaystyle \cdot$5$\displaystyle \cdot$4$\displaystyle \cdot$3$\displaystyle \cdot$2$\displaystyle \cdot$1 = (6!-3!)(3!) which is divisible by (3!)$\displaystyle ^{2}$. The generalization is obvious. Thanks mobel for very interesting thread.
October 10th, 2017, 12:56 PM   #10
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Quote:
 Originally Posted by zylo (3$\displaystyle \cdot$2)!=6$\displaystyle \cdot$5$\displaystyle \cdot$4$\displaystyle \cdot$3$\displaystyle \cdot$2$\displaystyle \cdot$1 = (6!-3!)(3!)
This is not true: $(3 \cdot 2)! = 720$ while $(6! - 3!)(3!) = 4284$.

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