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divisible by 8

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Re: divisible by 8

by logitech » Thu Dec 18, 2008 10:53 pm
Vignesh.4384 wrote:Hi ,

Please help me wiht htis problem
n(n+1)(n+2) can be divided with 8 under 2 conditions:

a) n+1 is a a multiple of 8 ( 7x8x9 or 15x16x17)

b) n is an EVEN number ( 2x3x4 or 6x7x8)

Lets find the numbers for case a

7 8 9
15 16 17
............
95 96 97

8,16,...96 = 12 numbers ( you can divide all numbers by 8 to find out the # of numbers )

1,2,...12 = 12 numbers

Lets find the numbers for case b

2 3 4
4 5 6
......
96 97 98

so 2,4,...96 again if you divide all the numbers by 2

1,2,...48 so 48 numbers


if you add both cases : 12+48 = 60

So the answer is: 60/96 = 5/8
LGTCH
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by 720dreaming » Thu Dec 18, 2008 11:34 pm
Could you explain how you came up with statement 2?

It seems to work, just can't quite wrap my arms around it. Thanks.

Seems to be that there is at least a 2^3 (which divides by 8) if you reduce the string (ie 16*17*18) starting with the even number to primes, I am on the right track?
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by logitech » Thu Dec 18, 2008 11:48 pm
720dreaming wrote:Could you explain how you came up with statement 2?

It seems to work, just can't quite wrap my arms around it. Thanks.

Seems to be that there is at least a 2^3 (which divides by 8) if you reduce the string (ie 16*17*18) starting with the even number to primes, I am on the right track?
I am not sure whether I understand what you are asking. Can you post your solution ?
LGTCH
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by 720dreaming » Fri Dec 19, 2008 12:53 am
Basically what I am asking is how you came up with if n is an even number, then n(n+1)(n+2) is divisible by 8.

I tested it out, seems to work, but wondering how you thought of that.
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by logitech » Fri Dec 19, 2008 1:02 am
720dreaming wrote:Basically what I am asking is how you came up with if n is an even number, then n(n+1)(n+2) is divisible by 8.

I tested it out, seems to work, but wondering how you thought of that.
even numbers : 2n

2n x ( 2n+1 ) x (2n+2 )

2n x ( 2n+1 ) x 2 ( n+1 )

We know that 2n+1 will be always ODD

so we are left with

2n x 2 (n+1)

or

4 n(n+1)

Well if we want to divide this by 8

we need to prove that n(n+1) will always have 2 as a factor

if n = odd

n+1 will be even

if n = even

n will be even

so n(n+1) is divisible with 2

so all that term is divisible by 8

Hope it is clear now ?
LGTCH
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by logitech » Fri Dec 19, 2008 1:06 am
also n(n+1) is consecutive series and N terms of a consecutive set can be divided by N

this is why it is safe to say that n x (n+1) can ve divided by 2
LGTCH
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by 720dreaming » Fri Dec 19, 2008 2:26 am
That makes sense. Thank you.
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by 720dreaming » Fri Dec 19, 2008 2:29 am
That makes sense. Thank you.
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