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
pl solve ths sum...asap
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- logitech
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(N^3 - N)/4kalyand022 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(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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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.
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.
- logitech
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Beautiful solution. And I am sure you can nail this question in less than 2 minutes since you know the concepts!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.
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"