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Topic: factorial problem (Read 1799 times) |
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Deedlit
Senior Riddler
Posts: 476
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factorial problem
« on: Jun 14th, 2005, 1:26am » |
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Show that [n! / ((n+1)(n+2)) ] is even for nonnegative integers n. ( [] is the greatest integer function.)
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| « Last Edit: Jun 14th, 2005, 1:27am by Deedlit » |
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SWF
Uberpuzzler
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Re: factorial problem
« Reply #1 on: Jun 22nd, 2005, 6:32pm » |
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Simple calculation shows it to be true for n<5. It is convenient to continue in terms of n=2m and n=2m-1 for m>2.
When m>2, (2m)!/(2m+2) and (2m-1)!/(2m) are both integers and
(2m)!/(2m+2) - (2m-1)!/(2m) = (2m+1)*(2m-1)!*(m-1) / (2*m*(m+1))
As their difference is a multiple of 2m+1, both of these integers have the same remainder when divided by 2m+1 Also,
(2m)! - (2m)!/(2m+2) = (2m+1)*(2m)! / (2m+2),
and (2m)! must also have the same remainder when divided by 2m+1. In other words, for a given value of m>2, there are 4 integers A, B, C, and r with 0 <= r < 2m+1 and:
(2m)!/(2m+2) = A*(2m+1) + r
(2m-1)!/(2m) = B*(2m+1) + r
(2m)! = C*(2m+1) + r
When m>2, the left side of all three equalities is an even integer. As (2m)! is a multiple of 2m+1, unless 2m+1 is a prime, r is either zero or 2m+1 is prime. By Wilson's Theorem, when 2m+1 is prime, r = 2m. Either way, r is an even number, and since (2m+1) is odd, A, B, and C are all even. B and A are the integer parts of dividing the above by 2m+1. i.e. the integer portions of
(2m-1)!/(2m)/(2m+1) and (2m)!/(2m+1)/(2m+2)
are even, which takes care of odd and even values of n.
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Barukh
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Re: factorial problem
« Reply #2 on: Jun 23rd, 2005, 10:46am » |
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As always, SWF provides an excellent solution! The idea to consider different quotients for even and odd numbers is so nice... 
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SWF
Uberpuzzler
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Re: factorial problem
« Reply #3 on: Jun 27th, 2005, 8:45pm » |
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Thank you, Barukh. The even/odd thing was something I noticed when numerically trying n=1 through 10, then tried to prove it. I wonder if that is the intended way of showing this, or if there is an easier way- where is Deedlit?
Did anyone even try this problem? It sat around for a while with 0 comments, which is rare.
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| « Last Edit: Jul 1st, 2005, 7:21pm by SWF » |
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Barukh
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Re: factorial problem
« Reply #4 on: Jun 28th, 2005, 2:29am » |
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on Jun 27th, 2005, 8:45pm, SWF wrote:| Did anyone even try this problem? It sat around for a while with 0 comments, which is rare. |
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And what about this? I have an idea, but it's based on a conjecture I cannot prove.
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TenaliRaman
Uberpuzzler
I am no special. I am only passionately curious.
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Re: factorial problem
« Reply #5 on: Jul 2nd, 2005, 1:41am » |
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Thats a beautiful proof SWF! So simple, yet so neat!! 
-- AI
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Self discovery comes when a man measures himself against an obstacle - Antoine de Saint Exupery
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Deedlit
Senior Riddler
Posts: 476
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Re: factorial problem
« Reply #6 on: Aug 2nd, 2005, 8:52pm » |
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I agree, very nice!
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