Can someone tell me what this is called?

  • Thread starter GopherWT
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I stumbled on something while working on a program earlier; I'm not so arrogant to think it's never been noticed before, and I'm not sure it's even remotely useful, but I'm curious now and, assuming this has been explored pretty thoroughly, wanted to read more about it. Googling equations doesn't really do a lot of good though, so I'm hoping someone can tell me what this property is called?

Anyway, the thing I noticed, while doing some calculations on permutations, that for any x>=2, x*(x-1) is always even. This seemed logical enough when I thought about it, because whether n is odd or even, x-1 will be the opposite, and odd*even=even. But the property extends beyond two terms; for x>=3, x*(x-1)*(x-2) is always divisible by 6; adding (x-3) makes it divisible by 24, and so on.

I know logically why these are true, based on how I was coming up with the equations in the first place, but mathematically it seems a bit unusual. If anyone recognizes this and can point me to more info, or just tell me what it's called, I'd be appreciative.
 

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  • #2
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I stumbled on something while working on a program earlier; I'm not so arrogant to think it's never been noticed before, and I'm not sure it's even remotely useful, but I'm curious now and, assuming this has been explored pretty thoroughly, wanted to read more about it. Googling equations doesn't really do a lot of good though, so I'm hoping someone can tell me what this property is called?

Anyway, the thing I noticed, while doing some calculations on permutations, that for any x>=2, x*(x-1) is always even.
You will usually see this product as n * (n - 1), with the stated or tacit assumption that the numbers involved are integers. If x is a nonintegral real, it doesn't make any sense to talk about its evenness or oddness.
This seemed logical enough when I thought about it, because whether n is odd or even, x-1 will be the opposite, and odd*even=even. But the property extends beyond two terms; for x>=3, x*(x-1)*(x-2) is always divisible by 6; adding (x-3) makes it divisible by 24, and so on.
Again, the numbers involved are integers. With three successive integers, you have either two odds and an even (making an even product) or two evens and an odd (also making an even product). For any three integers in succession, one of them has to be a multiple of 3. Add that to the fact that at least one of the integers is even, and that gives you a product that has a factor of 6 in it.
I know logically why these are true, based on how I was coming up with the equations in the first place, but mathematically it seems a bit unusual. If anyone recognizes this and can point me to more info, or just tell me what it's called, I'd be appreciative.
These are pretty basic number theory concepts, so a search for that term should lead you to more information or textbooks and such.
 
  • #3
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:facepalm:

That is terribly obvious now that you said it. series of n successive integers by definition contains a multiple if n. Thank you :)
 
  • #4
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And for your observation that n(n+1)(n+2)(n+3) is divisible by 24, we have one of the numbers having to be a multiple of 4, plus one other even number, plus one of the numbers has to be a multiple of 3. That gives us 4*2*3 = 24.

In case that "one other even number" bit isn't obvious, no matter which of the four is the number that is divisible by 4, there is another number that is two higher or two lower that must at least be even. If n is divisible by 4, then n+2 is even. If n+1 is divisible by 4, then n+3 is even. If n+2 is divisible by 4, then n is even. If n+3 is divisible by 4, then n+1 is even.
 
  • #5
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Thanks, but the first reply really did make it all click together for me. I know just enough about number theory to know that this question was in the area of number theory, so I apologize that I don't know the correct terminology/notation, but my logic skills are quite solid. Had you not answered so promptly, I'm certain I would've had the Aha moment eventually, but lord knows how long that would've taken me - maybe another minute, maybe a week.

Thanks again for the quick and helpful response :)
 
  • #6
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A product of n consecutive integers is divisible by n!. Here is a quick proof.

Suppose the n consecutive integers are [tex]a, a+1, \dots, a+n-1[/tex]. Then
[tex]\frac{a (a+1) \cdots (a+n-1)}{n!} = \binom{a}{n}[/tex]
(a binomial coefficient), which is an integer.
 
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