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If n is a positive integer, is n^3 - n divisible by 4?

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by 4GMAT_Mumbai » Wed Mar 02, 2011 7:46 pm
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.
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by tomada » Thu Mar 03, 2011 3:41 pm
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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by Night reader » Thu Mar 03, 2011 3:59 pm
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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by tomada » Thu Mar 03, 2011 4:19 pm
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.
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by tomada » Thu Mar 03, 2011 4:25 pm
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: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.
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by Night reader » Thu Mar 03, 2011 4:28 pm
:( 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.
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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by 721tjm » Thu Mar 03, 2011 4:35 pm
What's the remainder when 0 is divided by 4?
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by Night reader » Thu Mar 03, 2011 4:35 pm
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.
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by tomada » Thu Mar 03, 2011 4:42 pm
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.
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by kevincanspain » Fri Mar 11, 2011 5:12 am
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
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by HSPA » Fri Mar 11, 2011 5:50 am
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" :)

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
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by tomada » Fri Mar 11, 2011 6:23 am
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.
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