Prove by induction that the product of every three consecutive even
integers is divisible by 48. ex: 2 * 4 * 6 is divisble by 48

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Request for Question Clarification by calebu2-ga on 25 Jul 2002 09:03 PDT

judgementday - can i repost my comment as an answer to your question?
It proves what you needed it to.

Regards

calebu2-ga

Mr. Day,

Before I do a repeat of my answer - here are a few cool links about
induction (just to keep your interest flowing) :

Induction Principle from Mathworld@Wolfram
http://mathworld.wolfram.com/InductionPrinciple.html

Course notes at USC on Mathematical Induction
http://www.math.sc.edu/~sumner/numbertheory/induction/Induction.html

Search term used on google.com : "proof by induction"

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Prove : For every n, the product of the numbers 2n, 2n + 2 and 2n + 4
is divisible by 48.
 
ie. 48 divides 2n(2n+2)(2n+4) = 8n(n+1)(n+2). 
 
Here's the induction part: 
 
Let n = 1 
 
2n(2n+2)(2n+4) = 2*4*6 = 48. True for n = 1. 
 
Assume it is true for n = k. ie. 48 divides 8k(k+1)(k+2). ie. 6
divides k(k+1)(k+2).
 
Consider n = k+1 : 
 
We need 48 to divide 8(k+1)(k+2)(k+3). 
 
Ie. 6 to divide (k+1)(k+2)(k+3). 
 
Now (k+1)(k+2)(k+3) = k(k+1)(k+2) + 3(k+1)(k+2) 
 
By induction 6 divides k(k+1)(k+2) 
and if 2 does not divide k+1, it must divide k+2 (either k+1 is even
or k+2 is even). Therefore 6 = 3 * 2 must divide 3(k+1)(k+2).
 
Therefore 48 divides 8(k+1)(k+2)(k+3). 
 
Hence, by induction, for every .n, the product of the numbers 2n, 2n +
2 and 2n + 4 is divisible by 48

Good luck with the rest of your proofs!

calebu2-ga

great job. Used correct mathematical induction to solve problem

I think the induction proof goes something like this :

Prove : For every n, the product of the numbers 2n, 2n + 2 and 2n + 4
is divisible by 48.

ie. 48 divides 2n(2n+2)(2n+4) = 8n(n+1)(n+2).

Here's the induction part:

Let n = 1

2n(2n+2)(2n+4) = 2*4*6 = 48. True for n = 1.

Assume it is true for n = k. ie. 48 divides 8k(k+1)(k+2). ie. 6
divides k(k+1)(k+2).

Consider n = k+1 :

We need 48 to divide 8(k+1)(k+2)(k+3).

Ie. 6 to divide (k+1)(k+2)(k+3).

Now (k+1)(k+2)(k+3) = k(k+1)(k+2) + 3(k+1)(k+2)

By induction 6 divides k(k+1)(k+2)
and if 2 does not divide k+1, it must divide k+2 (either k+1 is even
or k+2 is even). Therefore 6 = 3 * 2 must divide 3(k+1)(k+2).

Therefore 48 divides 8(k+1)(k+2)(k+3).

Hence, by induction, for every n, the product of the numbers 2n, 2n +
2 and 2n + 4 is divisible by 48.

I'm sure that could be written more clearly, but that's how an
induction proof goes. In some ways it is obvious that any 3
consecutive even numbers must contain the factors 2, 2, 2, 2 and 3 -
but just observing that does not constitute a proof - and it an isn't
inductive proof. Other than that, a good concise answer.

calebu2-ga

The "Answer" (gosh, I hate even saying those two words with all the
Allen Iverson junk in the news) fails to use induction.  calebu2's
"Comment" does the job using induction.

yes please do. This is a first time posting a question so I did not
know how to contact calebu to give him the 2 bucks (a well earned two
bucks).