Hi ,
Please help me wiht htis problem
divisible by 8
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n(n+1)(n+2) can be divided with 8 under 2 conditions:Vignesh.4384 wrote:Hi ,
Please help me wiht htis problem
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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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?
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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I am not sure whether I understand what you are asking. Can you post your solution ?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?
LGTCH
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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.
I tested it out, seems to work, but wondering how you thought of that.
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even numbers : 2n720dreaming 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.
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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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
this is why it is safe to say that n x (n+1) can ve divided by 2
LGTCH
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