20070721, 15:49  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Consecutive integers.
Prove that the product of the first n consecutive integers is divisible by their sum, if and only if (n + 1) is not a prime. Mally 
20070721, 19:00  #2 
Nov 2005
2×7×13 Posts 
Positively evil! :)

20070722, 10:36  #3 
Bronze Medalist
Jan 2004
Mumbai,India
100000000100_{2} Posts 

20070722, 12:53  #4 
Einyen
Dec 2003
Denmark
6062_{8} Posts 
n! = k * (1+2+3 ....+n) with n! = 0 (mod k) ==> n! = k*n*(n+1)/2 ==> 2*(n1)! / k = (n+1) so (n+1) is not prime. 
20070723, 01:04  #5 
"William"
May 2003
New Haven
3·787 Posts 
The sum of the first n consecutive integers is n(n+1)/2.
If n+1 is prime this clearly cannot divide n! because the prime n+1 divides the sum and not the product. That takes care of the "only if" part. If n+1 is even, we can split the sum into the product of n and (n+1)/2. These are distinct integers less than or equal to n, and hence occur as two of the distinct terms in n! That leaves the case where n+1 is an odd composite. If n+1 is not the square of an odd prime, then it can be factored into two distinct integers, both of which will appear as individual terms in (n1)!. That leaves only the case where n+1=p[sup]2[/sup]. For this case we will get the two factors of p from p and 2p. This works unless 2p >= n but in this case n = p[sup]2[/sup]1, so this works unless 2p >= p[sup]2[/sup]1 Solving this inequality by the usual methods shows that it is only true for real values of p between 1sqrt(2) and 1+sqrt(2). All the odd primes are above this range, so that finishes the "if" part. 
20070723, 06:47  #6 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
WBlipp's answer mirrors my thoughts exactly.
I considered it too much of a labour to write down. What about the case (n+1)=p^3 etc? David 
20070723, 07:02  #7  
Cranksta Rap Ayatollah
Jul 2003
1201_{8} Posts 
Quote:
[quote][spoiler]If n+1 is not the square of an odd prime, then it can be factored into two distinct integers, both of which will appear as individual terms in (n1)!.[/quote] factor it into p^2 and p, both of these will be less than n1, thus factors of (n1)! [/spoiler] Last fiddled with by Orgasmic Troll on 20070723 at 07:02 

20070723, 08:16  #8 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 

20070723, 08:48  #9  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Spoilers!
Quote:
I completely agree with you Dave though it may annoy the originator of spoilers who would naturally want the use of them. Yeah but to a certain limit I would say. We are not school boys to put the answers at the back of the text book! Roger Penrose in his brilliant book 'Road to Reality' asks a lot of key questions that its not practical to refer for the answer anywhere in his some 800 page book. He has however got a separate website for them! The motto should be 'Don't hide the truth but reveal it' ! Any one put that in Latin ? Mally 

20070723, 09:11  #10  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
before you try the problem, the simple solution is not to look at it. Quote:
I recently downsized my living arrangements and although I abandoned many posessions (including books) in the process, I kept "Road to Reality". I haven't "finished" it yet On the same note, I don't understand why Stephen Hawking's "A brief history of Time" gets cited as the most bought and least read book. I found it quite readable if nothing else (and I am a slow reader). David 

20070723, 17:06  #11 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 

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