consec. int. proof troubles
Author
Can this be proven without induction? If so, any suggestions?
The product of 5 consec. integers is divisible by 5
I can prove it informally, but can't put it into words...
(n or n+1 or n+2 or n+3 or n+4 must be divisible by 5, and 5k = multiple of 5)
All you have to do to prove that N is divisible by D is to prove that when you divide N by D, you get an integer.
In other words, prove there exists an integer X such that N/D = X.
Or better yet, prove there exists an integer X such that N = DX.
I think I've already made it too easy.
In other words, prove there exists an integer X such that N/D = X.
Or better yet, prove there exists an integer X such that N = DX.
I think I've already made it too easy.
Author
Quote:Original post by jjd
Try writing the integers as: 5n, 5n+1, 5n+2, 5n+3, 5n+4, where n = 0,1,...
-Josh
Why would you write it as that?
Author
Quote:Original post by pinacolada
All you have to do to prove that N is divisible by D is to prove that when you divide N by D, you get an integer.
In other words, prove there exists an integer X such that N/D = X.
Or better yet, prove there exists an integer X such that N = DX.
I think I've already made it too easy.
How so? It cant even be factored so that it can be divided. The only way I could think of is assuming one of them is a multiple of 5 then proving that but i doubt that's acceptable.
Quote:Original post by GinkQuote:Original post by jjd
Try writing the integers as: 5n, 5n+1, 5n+2, 5n+3, 5n+4, where n = 0,1,...
-Josh
Why would you write it as that?
You want to show that the product of 5 consecutive integers is divisible by 5. So the product of the above numbers is 5n(5n+1)(5n+2)(5n+3)(5n+4). Is this divisible by 5? With that 5 out the front, the answer becomes obvious
Now that's nice but how can we make it work for a sequence like 7,8,9,10,11? The product of these number can be manipulated into a similar form
7*8*9*10*11 = (5+2)(5+3)(5+4)(5+5)(5+6)
7*8*9*10*11 = (5*1+2)(5*1+3)(5*1+4)(5*2)(5*2+1)
Perhaps you can see how to proceed from here?
-Josh
Yes. If you can write your number as 5*(whatever), and (whatever) is an integer, then you can use the proof I wrote and be done.
lookup proof by induction
take a base case which is true
assume the case is true for n
prove that it also holds for n+1
or something in that direction
take a base case which is true
assume the case is true for n
prove that it also holds for n+1
or something in that direction
Quote:Original post by Airo
lookup proof by induction
Quote:Original post by Gink
Can this be proven without induction?
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