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pl solve ths sum...asap

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pl solve ths sum...asap

by kalyand022 » Mon Dec 01, 2008 2:06 am
if N is a positive integer,is (N^3 - N)/4 ?

1>N=2k+1 WHERE K=INTEGER;
2>N^2 + N IS DIVISIBLE BY 6;

ans is A,

but if i take K as 1,2,3 it is satisfyn but what if i take K=0 then
N=1 ans so (N^3 - N)/4 is giving 0....
then how is A the ans?

pl help me out
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by austin » Mon Dec 01, 2008 2:14 am
pardone moi....
Last edited by austin on Mon Dec 01, 2008 2:26 am, edited 1 time in total.
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Re: pl solve ths sum...asap

by logitech » Mon Dec 01, 2008 2:21 am
kalyand022 wrote:if N is a positive integer,is (N^3 - N)/4 ?

1>N=2k+1 WHERE K=INTEGER;
2>N^2 + N IS DIVISIBLE BY 6;

ans is A,

but if i take K as 1,2,3 it is satisfyn but what if i take K=0 then
N=1 ans so (N^3 - N)/4 is giving 0....
then how is A the ans?

pl help me out
(N^3 - N)/4

N(N^2-1)/4

N(N-1)(N+1)/4

so N-1, N and N+1 we have consecutive numbers. This can be divided by 4 in two ways:

1) N is a multiple of 4 3,4,5
2) N-1 and N+1 is divided by 2 2,3,4

Statement 1)

N=2k+1 means N is odd so both N-1 and N+1 are even.

2k, 2k+1, 2k+2 SUF

Statement 2)

N^2 + N IS DIVISIBLE BY 6

N^2+N-6K=0

where K=0,1,2,3,...

You will always have one + and one - solution to this. Since N>0 given in question.

N can be both even or Odd so Insuff

K=2

N^2+N-12=0 --> N=3,-4

K=7

N^2+N-42=0 ---> N=6,-7
LGTCH
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by vittalgmat » Mon Dec 01, 2008 2:37 am
Hi Kalyan,
I can partially answer this.

U are trying to find whether (n^3 -n) is divisible by 4.
n^3 -n = n(n^2 -1) = (n-1)n(n+1) <--------- product of 3 consecutive integers.

So, the product is divisible by 4 if two of the 3 number are even numbers.
ie. if n is ODD, the product is divisible by 4.

Stmt1: n = 2k+1
2k+1 is always an Odd integer. So n is odd integer
So, n-1 and n+1 are even integers. therefore the product of 3 conseq ints is divisible by 4.
Sufficient.



Stmt2: n^2 + n is divisible by 6.
ie n^2 +n = 6P, for some int P.
But n^2 +n = n(n+1) <-------- prod of 2 consequtive ints.

so n(n+1) = 6Q, for some int Q.

At this point we can plugin numbers and prove that this is insufficient.

choose n = 2, so n-1 = 1, n+1 = 3. n(n+1) = 6, which is divisible by 6.
But not divisible by 4.
choose n = 5, so n-1 = 4, n+1=6 n(n+1) divisible by 6 and also 4.
so we get 2 different results. stmt 2 is insufficient.

Hence go with A.

Hope i dint miss anything. Cramya, Logitech ?? Your thoughts.
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by logitech » Mon Dec 01, 2008 2:43 am
vittalgmat wrote:Hi Kalyan,
I can partially answer this.

U are trying to find whether (n^3 -n) is divisible by 4.
n^3 -n = n(n^2 -1) = (n-1)n(n+1) <--------- product of 3 consecutive integers.

So, the product is divisible by 4 if two of the 3 number are even numbers.
ie. if n is ODD, the product is divisible by 4.

Stmt1: n = 2k+1
2k+1 is always an Odd integer. So n is odd integer
So, n-1 and n+1 are even integers. therefore the product of 3 conseq ints is divisible by 4.
Sufficient.



Stmt2: n^2 + n is divisible by 6.
ie n^2 +n = 6P, for some int P.
But n^2 +n = n(n+1) <-------- prod of 2 consequtive ints.

so n(n+1) = 6Q, for some int Q.

At this point we can plugin numbers and prove that this is insufficient.

choose n = 2, so n-1 = 1, n+1 = 3. n(n+1) = 6, which is divisible by 6.
But not divisible by 4.
choose n = 5, so n-1 = 4, n+1=6 n(n+1) divisible by 6 and also 4.
so we get 2 different results. stmt 2 is insufficient.

Hence go with A.

Hope i dint miss anything. Cramya, Logitech ?? Your thoughts.
Beautiful solution. And I am sure you can nail this question in less than 2 minutes since you know the concepts!
LGTCH
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