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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
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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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)
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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.
Why did the vector cross the road?
It wanted to be normal.
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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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I can generalise it a smidge:
k^(n+1) also divides a_0 x ... x a_n.
Why did the vector cross the road?
It wanted to be normal.
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