Consecutive Integer proofs

Quote:

Original post by Gink How exactly do you prove that the product of n consecutive integers is always divisible by n? (say 2 for example) Here is how far I got Let a/b be 2 consec integers. a b = a + 1 so, a * ( a + 1) = 2n a^2 + a = 2n a(a+1) = 2n What am I missing here?

Well, you're on the right track. Since a and a+1 are consecutive, then one of them is even. Any even number times any other number is also even (factorization theorem). The same idea holds for any n. a(a+1)...(a+n-1)=n*k since one of the a-terms is divisible by n, so the product will be too.

a(a + 1) has to be divisible by 2 because either a is divisible by 2 or a+1 is divisible by
2.

Basically we can write any integer as 2*q + r (by definition of integer division), where r=0 or 1. So then a = 2*q + r. Then a+1 = 2*q + r + 1

So then, if r = 0, the a is divisible by 2. If r=1 then a+1 is divisible by 2. In either case a(a+1) is divisible by 2.

To prove this in general for all n you need to use induction. Are you familiar with the concept of induction?

So, what exactly would I say to finish this off? Since a or a+1 is divisible by 2, 2n | a*(a+1)?

edit : i'm familiar with induction but i have to prove it for specific numbers

Quote:

Original post by Gink How exactly do you prove that the product of n consecutive integers is always divisible by n? (say 2 for example) Here is how far I got Let a/b be 2 consec integers. a b = a + 1 so, a * ( a + 1) = 2n a^2 + a = 2n a(a+1) = 2n What am I missing here?

Where did the line

a * (a + 1) = 2n

come from? n is 2 in this case, right, so you're saying that a * (a + 1) = 4?

Anyway, you should think about remainders on division by n instead. You have n consecutive integers, each with a different remainder on division by n. See where you go from there.

that line is right, when n is the correct value obviously.

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