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Old 2007-07-21, 15:49   #1
mfgoode
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Arrow 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
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Old 2007-07-21, 19:00   #2
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Positively evil! :)
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Old 2007-07-22, 10:36   #3
mfgoode
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Quote:
Originally Posted by nibble4bits View Post
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
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Old 2007-07-22, 12:53   #4
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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.
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Old 2007-07-23, 01:04   #5
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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.
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Old 2007-07-23, 06:47   #6
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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
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Old 2007-07-23, 07:02   #7
Orgasmic Troll
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Quote:
Originally Posted by davieddy View Post
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
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Old 2007-07-23, 08:16   #8
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Quote:
Originally Posted by Orgasmic Troll View Post


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
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Old 2007-07-23, 08:48   #9
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Thumbs up Spoilers!

Quote:
Originally Posted by davieddy View Post
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
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Old 2007-07-23, 09:11   #10
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Quote:
Originally Posted by mfgoode View Post


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 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
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Old 2007-07-23, 17:06   #11
Orgasmic Troll
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Quote:
Originally Posted by davieddy View Post
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.
Don't read this sentence.
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