Jump to content
  • Advertisement
Sign in to follow this  
Gink

Consecutive Integer proofs

This topic is 4815 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Recommended Posts

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?

Share this post


Link to post
Share on other sites
Advertisement
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.

Share this post


Link to post
Share on other sites
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?

Share this post


Link to post
Share on other sites
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

Share this post


Link to post
Share on other sites
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.

Share this post


Link to post
Share on other sites

This topic is 4815 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Guest
This topic is now closed to further replies.
Sign in to follow this  

  • Advertisement
×

Important Information

By using GameDev.net, you agree to our community Guidelines, Terms of Use, and Privacy Policy.

We are the game development community.

Whether you are an indie, hobbyist, AAA developer, or just trying to learn, GameDev.net is the place for you to learn, share, and connect with the games industry. Learn more About Us or sign up!

Sign me up!