4GMAT_Mumbai wrote:Hi,
Interesting Q ... Thanks !
1. n is an odd number
n^3 - n = (n-1) * n * (n+1) = Product of three consecutive integers
We know that n is an odd number. So, the number before and after n must both be even numbers. So, n^3-n will definitely be divisible by 4. Suff
2. n^2 + n = n (n+1) is divisible by 6
n could be 2. n^3 - n is NOT divisible by 4.
n could be 3. n^3 - n is divisible by 4.
I am getting both 'Yes' and 'No' for an answer for different values of n which satisfy the Statement 2. Hence, Insuff.
My answer would be A ... OA please ... Thanks.
tomada wrote:What if k=0 ? Then, n = 2(0) + 1 = 1.
(1)^3 - 1 = 0, but zero isn't divisible by 4.
How is (A) sufficient?
4GMAT_Mumbai wrote:Hi,
Interesting Q ... Thanks !
1. n is an odd number
n^3 - n = (n-1) * n * (n+1) = Product of three consecutive integers
We know that n is an odd number. So, the number before and after n must both be even numbers. So, n^3-n will definitely be divisible by 4. Suff
2. n^2 + n = n (n+1) is divisible by 6
n could be 2. n^3 - n is NOT divisible by 4.
n could be 3. n^3 - n is divisible by 4.
I am getting both 'Yes' and 'No' for an answer for different values of n which satisfy the Statement 2. Hence, Insuff.
My answer would be A ... OA please ... Thanks.
Night reader wrote:what's the problem Tomada, do you agree with the solution below as well as an answer to your query?
given: n (integer) >0, n(n+1)(n-1)/4
st(1) 2k=n-1 <-- integer Sufficient to answer Yes, because k is an integer and (n-1)/2 is an integer too. So n is odd and alone cannot be divided by 4 BUT one (odd+1)*(another odd-1) gives divisibility by 4. Even with n=1 and k=0 we have 0/4
st(2) n(n+1)/6 Obviously Not Sufficient.
Answer Yes and choice A.
[spoiler]If n is a positive integer, is n^3 - n divisible by 4?
1) n = 2k+1, where k is an integer.
2) n^2 + n is divisible by 6.
[/spoiler]tomada wrote:What if k=0 ? Then, n = 2(0) + 1 = 1.
(1)^3 - 1 = 0, but zero isn't divisible by 4.
How is (A) sufficient?
4GMAT_Mumbai wrote:Hi,
Interesting Q ... Thanks !
1. n is an odd number
n^3 - n = (n-1) * n * (n+1) = Product of three consecutive integers
We know that n is an odd number. So, the number before and after n must both be even numbers. So, n^3-n will definitely be divisible by 4. Suff
2. n^2 + n = n (n+1) is divisible by 6
n could be 2. n^3 - n is NOT divisible by 4.
n could be 3. n^3 - n is divisible by 4.
I am getting both 'Yes' and 'No' for an answer for different values of n which satisfy the Statement 2. Hence, Insuff.
My answer would be A ... OA please ... Thanks.
Night reader wrote:what's the problem Tomada, do you agree with the solution below as well as an answer to your query?
given: n (integer) >0, n(n+1)(n-1)/4
st(1) 2k=n-1 <-- integer Sufficient to answer Yes, because k is an integer and (n-1)/2 is an integer too. So n is odd and alone cannot be divided by 4 BUT one (odd+1)*(another odd-1) gives divisibility by 4. Even with n=1 and k=0 we have 0/4
st(2) n(n+1)/6 Obviously Not Sufficient.
Answer Yes and choice A.
[spoiler]If n is a positive integer, is n^3 - n divisible by 4?
1) n = 2k+1, where k is an integer.
2) n^2 + n is divisible by 6.
[/spoiler]tomada wrote:What if k=0 ? Then, n = 2(0) + 1 = 1.
(1)^3 - 1 = 0, but zero isn't divisible by 4.
How is (A) sufficient?
4GMAT_Mumbai wrote:Hi,
Interesting Q ... Thanks !
1. n is an odd number
n^3 - n = (n-1) * n * (n+1) = Product of three consecutive integers
We know that n is an odd number. So, the number before and after n must both be even numbers. So, n^3-n will definitely be divisible by 4. Suff
2. n^2 + n = n (n+1) is divisible by 6
n could be 2. n^3 - n is NOT divisible by 4.
n could be 3. n^3 - n is divisible by 4.
I am getting both 'Yes' and 'No' for an answer for different values of n which satisfy the Statement 2. Hence, Insuff.
My answer would be A ... OA please ... Thanks.
snotty? why the problem becomes condescending when the word "problem" is put as "question" and not as an "issue"... I myself have repeatedly posted here the desire to set an academic tone. Anyway it's good to know that the only thing you found "touched on" in the post was the word "problem"
tomada wrote:Nightreader, there's no need to be condescending ("what's the problem Tomada"); I posted my comment prior to yours but, even if I hadn't, let's try to keep the snotty attitude outside of the forum. Agreed?
Night reader wrote:what's the problem Tomada, do you agree with the solution below as well as an answer to your query?
given: n (integer) >0, n(n+1)(n-1)/4
st(1) 2k=n-1 <-- integer Sufficient to answer Yes, because k is an integer and (n-1)/2 is an integer too. So n is odd and alone cannot be divided by 4 BUT one (odd+1)*(another odd-1) gives divisibility by 4. Even with n=1 and k=0 we have 0/4
st(2) n(n+1)/6 Obviously Not Sufficient.
Answer Yes and choice A.
[spoiler]If n is a positive integer, is n^3 - n divisible by 4?
1) n = 2k+1, where k is an integer.
2) n^2 + n is divisible by 6.
[/spoiler]tomada wrote:What if k=0 ? Then, n = 2(0) + 1 = 1.
(1)^3 - 1 = 0, but zero isn't divisible by 4.
How is (A) sufficient?
4GMAT_Mumbai wrote:Hi,
Interesting Q ... Thanks !
1. n is an odd number
n^3 - n = (n-1) * n * (n+1) = Product of three consecutive integers
We know that n is an odd number. So, the number before and after n must both be even numbers. So, n^3-n will definitely be divisible by 4. Suff
2. n^2 + n = n (n+1) is divisible by 6
n could be 2. n^3 - n is NOT divisible by 4.
n could be 3. n^3 - n is divisible by 4.
I am getting both 'Yes' and 'No' for an answer for different values of n which satisfy the Statement 2. Hence, Insuff.
My answer would be A ... OA please ... Thanks.
tomada wrote:But, to answer your question, my problem is that, when k=0, n=1. When n=1, n^3-n = 0.
Zero is not divisible by 4. Thus, there is an exception, which occurs when k=0.
Of course, I'm assuming that zero is an integer. If that's a correct assumption, then statement (1) is insufficient.
Night reader wrote:0 is not a divisor (factor). We divide 0 by a number, and Yes 0 is divisible by 4. For this reason, the only means when k turns 0 is n=1, still we have to answer Yes and select choice A.tomada wrote:But, to answer your question, my problem is that, when k=0, n=1. When n=1, n^3-n = 0.
Zero is not divisible by 4. Thus, there is an exception, which occurs when k=0.
Of course, I'm assuming that zero is an integer. If that's a correct assumption, then statement (1) is insufficient.
