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Melllyy

Guest

Algebra

Explain using algebra why the product of three consecutive even numbersis always a multiple of 8. eg 2x6x8=96

Hint: let the first even number be 2n

Helpp pleaseee

JohnnyReinB

Member

Registered: 2007-10-08

Posts: 453

Re: Algebra

Well:

Even numbers are multiples of two
Two cubed is 8

so, it'll always have a factor of eight.

Sorry, got no time to elaborate.

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Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Algebra

Three consecutive even numbers multiplied together:

(2n-2)2n(2n+2)

Expanding the brackets we get:
(4n²-4n)(2n+2)
= 8n³ + 8n² - 8n² - 8n
= 8(n³ - n)

Which is always a multiple of 8 (we know n is an integer as 2n is an even integer)

Last edited by Daniel123 (2008-02-15 01:39:54)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Algebra

That can be generalised slightly (and in fact, your example does this already) to be any three even numbers, instead of consecutive ones.

Then they'll be in the form 2a, 2b and 2c, so multiplying them would give 8abc.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Algebra

Of course, we can always generalize it further.  If k divides {a_0, ..., a_n}, then k^n divides a_0 * a_1 * ...  *a_n.

Anyone want to try to get more general?

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Algebra

I can generalise it a smidge:

k^(n+1) also divides a_0 x ... x a_n.

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Algebra