
#1
Apr408, 12:06 PM

P: 26

I need to prove this for any n natural, n>= 5, n not prime.




#2
Apr408, 12:15 PM

P: 688

Think a bit about the prime factors of n... are they smaller than n? Then think of what (n1)! means.




#3
Apr408, 02:56 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

Dodo's hint is excellent but the crucial point is whether the factors of n are less than n1, not jus n itself!




#4
Apr408, 04:10 PM

Sci Advisor
P: 5,935

(n1)! is divisible by n
If n is not prime, then n=j*k, where j and k are >1, therefore both <n. As a result both are factors in (n1)!, so n divides (n1)!.




#5
Apr408, 04:35 PM

P: 39

Almost there, one last thing to consider is the squares of prime numbers. What happens when n=9=3*3 for example? This is where the n>=5 comes in.




#6
Apr508, 12:56 AM

P: 26

You all have good points and I thought of all of them and the reason I've posted this is because of the square numbers as rodigee said. Well I think if n = k*k then k is one of the numbers between 1 and n1. since n=k*k then k divides (nk) and this is smaller the n1 which means n divides (n1)!.




#7
Apr508, 11:03 AM

P: 39

Sorry, xax but I don't really understand your idea heh. This is how I see it:
[tex]n > 4[/tex] implies that [tex]\sqrt{n} > 2[/tex] so we also have that [tex]\sqrt{n}\sqrt{n} > 2\sqrt{n}[/tex] or simply [tex]n > 2\sqrt{n}[/tex], meaning that [tex]2\sqrt{n}[/tex] and [tex]\sqrt{n}[/tex] both show up in [tex](n1)![/tex] so we're good to go. 



#8
Apr508, 01:24 PM

P: 26

You make perfect sense rodigee and this is a nicer demonstation than mine. Thank you.



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