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Joined: Aug 28 2007, 12:59 pm
Last Active: Jan 20 2012, 3:43 am
Status: Offline
Display Name: Fwop_

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Starcraft I: Not Given.
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[03:04 am]
Mini Moose 2707 -- Either way, rekt, GG. ∎
[03:02 am]
Dem0n -- I see
[03:00 am]
Mini Moose 2707 -- That's probably more concise. Depends where you want to do the heavy lifting of the proof.
[03:00 am]
Mini Moose 2707 -- n * (n+1) * (n+2). If n is even, then for some integer k you can write the original expression as (2k)(2k+1)(2k+2), obvious factor of 2 up front. If n is odd, then express it as (2k+1)(2k+2)(2k+3), factor a 2 from the middle term and put it up front.
[02:58 am]
Mini Moose 2707 -- Actually, you could probably do this whole thing quicker
[02:55 am]
Mini Moose 2707 -- Oh, that x in odd * even should be an n
[02:55 am]
Dem0n -- oh so smart XD
[02:55 am]
Mini Moose 2707 -- More difficult one? odd * odd = (2n + 1)(2m + 1) = 4nm + 2(m+n) + 1, 4nm and 2(m+n) are even, therefore their sum is even, therefore 4nm + 2(m+n) + 1 is odd
[02:53 am]
Mini Moose 2707 -- even * even = 2n * 2m = 2 * (2nm), clearly divisible by 2
[02:53 am]
Mini Moose 2707 -- odd * even = (2n + 1) * 2m = 2 * [(2x + 1) * m], clearly divisible by 2
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