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 2007-07-21, 15:49 #1 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22×33×19 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
 2007-07-21, 19:00 #2 nibble4bits     Nov 2005 2×7×13 Posts Positively evil! :)
2007-07-22, 10:36   #3
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

1000000001002 Posts

Quote:
 Originally Posted by nibble4bits Positively evil! :)

Nibble hearing from you after a long time indeed!

I thought so too! Well lets solve it and meet evil with good(e)!

Mally

 2007-07-22, 12:53 #4 ATH Einyen     Dec 2003 Denmark 60628 Posts n! = k * (1+2+3 ....+n) with n! = 0 (mod k) ==> n! = k*n*(n+1)/2 ==> 2*(n-1)! / k = (n+1) so (n+1) is not prime.
 2007-07-23, 01:04 #5 wblipp     "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 (n-1)!. 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 1-sqrt(2) and 1+sqrt(2). All the odd primes are above this range, so that finishes the "if" part.
 2007-07-23, 06:47 #6 davieddy     "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
2007-07-23, 07:02   #7
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

12018 Posts

Quote:
 Originally Posted by davieddy 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

[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 (n-1)!.
[/quote]

factor it into p^2 and p, both of these will be less than n-1, thus factors of (n-1)!

[/spoiler]

Last fiddled with by Orgasmic Troll on 2007-07-23 at 07:02

2007-07-23, 08:16   #8
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by Orgasmic Troll factor it into p^2 and p, both of these will be less than n-1, thus factors of (n-1)!
THX Orgasmictroll.
May I suggest we could abandon the spoilers now?
Although it adds to the mystique, it gets tedious

David

2007-07-23, 08:48   #9
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts
Spoilers!

Quote:
 Originally Posted by davieddy THX Orgasmictroll. May I suggest we could abandon the spoilers now? Although it adds to the mystique, it gets tedious David

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

2007-07-23, 09:11   #10
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by mfgoode 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!
In most cases, if you don't want to see the answer
before you try the problem, the simple solution is
not to look at it.

Quote:
 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! Mally
My version (hardback £30) has >1000 pages.
I recently downsized my living arrangements and although
I abandoned many posessions (including books) in the process,

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

David

2007-07-23, 17:06   #11
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

641 Posts

Quote:
 Originally Posted by davieddy In most cases, if you don't want to see the answer before you try the problem, the simple solution is not to look at it.

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