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.
If n is a positive integer, is n^3 - n divisible by 4?
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- ithamarsorek
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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.
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.
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
- tomada
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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?
(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.
I'm really old, but I'll never be too old to become more educated.
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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![Smile :)](./images/smilies/smile.png)
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]
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
![Smile :)](./images/smilies/smile.png)
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.
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
- tomada
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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.
I'm really old, but I'll never be too old to become more educated.
- tomada
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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.
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: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.
I'm really old, but I'll never be too old to become more educated.
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![Sad :(](./images/smilies/sad.png)
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.
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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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.
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
- tomada
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Now I see my mistake. I assumed zero was not divisible by 4. I knew that 4 goes into zero "zero times", but didn't see this as indicative of divisibility.
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.
I'm really old, but I'll never be too old to become more educated.
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Remember that 0 is an even integer that is neither positive nor negative. It is divisible by every positive integer.
Don't take offense by nightreader's tone: he probably didn't know that 'what's your problem with ..." sounds harsh to most English-speakers
Don't take offense by nightreader's tone: he probably didn't know that 'what's your problem with ..." sounds harsh to most English-speakers
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This is a very nice thread....
I had similar problems in past : When I had a question..I used to say " I have a doubt" ...
Doubt to a native speaker in his explanation is a critic being found in his speech but actually I meant to ask a question to get clarified on what ever the topic he was explaining...
When a non-native speaker says "do you have a problem" it means "can I help you on a given PROBLEM"![Smile :)](./images/smilies/smile.png)
Tomada are you from united states... I see NJ... I dont thing you have friends in Edison/Edison township
Nor do your software company has branches in bangalore/shanghai/germany
I had similar problems in past : When I had a question..I used to say " I have a doubt" ...
Doubt to a native speaker in his explanation is a critic being found in his speech but actually I meant to ask a question to get clarified on what ever the topic he was explaining...
When a non-native speaker says "do you have a problem" it means "can I help you on a given PROBLEM"
![Smile :)](./images/smilies/smile.png)
Tomada are you from united states... I see NJ... I dont thing you have friends in Edison/Edison township
Nor do your software company has branches in bangalore/shanghai/germany
- tomada
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Hi HSPA,
It's funny you mention Edison, because one of my favorite Indian restaurants (Mogul) is in Edison (or is it Woodbridge). I know it's that area.
It's funny you mention Edison, because one of my favorite Indian restaurants (Mogul) is in Edison (or is it Woodbridge). I know it's that area.
I'm really old, but I'll never be too old to become more educated.