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Prove that the product of three consecutive even numbers is a multiple of 8? show This into as much detail as possible!

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Not a descriptive title and the homework tag. What have you tried? –  Tyler Mar 29 '12 at 11:10
    
Struggling, need some help, not homework just revision! –  Leah Mar 29 '12 at 11:11
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By consecutive even numbers you mean something like $2k$, $2(k+1)$, $2(k+2)$? If so, it is very trivial: their product is just $8k(k+1)(k+2)$, visibly a multiple of $8$. Mind that it is actually a multiple of $16$. –  Andrea Mori Mar 29 '12 at 11:14

3 Answers 3

It doesn't even have to be consecutive even numbers. Let $a, b,$ and $c$ be even numbers. We have that $2$ divides $a,b,$ and $c$. So we can factor $2$ from $a,b$ and $c$ each. Therefore $$ abc = (2^3)k = 8k $$ for some $k \in Bbb{Z}$. So the product of three even numbers is divisible by $8$.

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$$2n(2n+2)(2n+4)=2^3n(n+1)(n+2)=2^3n(n(n+3)+2)$$

EDIT(for Didier): If

  • $n$ is even, then we can get another $2$
  • $n$ is odd, then $n(n+3)+2$ is even.

So in both cases we can get another factor of $2$, such that the product is divisble by $16$.

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cheers lad ur top man –  Leah Mar 29 '12 at 11:12
    
Expanding the product is unneeded. –  Did Mar 29 '12 at 11:38
    
@DidierPiau correct(ed) –  draks ... Mar 29 '12 at 11:41
    
Right. Now you can show this is always a multiple of 16. –  Did Mar 29 '12 at 11:44

An even number $n$ is of the form $n=2k$.

Three consucutive even numbers are:

$$2k,2k+2,2k+4$$

And their product is:

$$(2k)(2k+2)(2k+4)=(2)(k)(2)(k+1)(2)(k+2)=8(k)(k+1)(k+2)$$

As we see above have factor of $8$, furthermore, at least on of $k,k+1,k+2$ most be even, so we can also conclude that $(2k)(2k+2)(2k+4)$ have a factor of $16$.

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