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

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by 4meonly » Fri Dec 05, 2008 6:28 am
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

According to OG answer is A

However, I have some doubts.


Main stem: (n^3)-n = n(n^2 -1) = (n-1)(n)(n+1)
Rephrasing main stem (n-1)(n)(n+1) = 4q? Remainder should be =0.

(1)
n=2k+1, where k is an integer, OR n is odd
if k=0, then n=1 and (n-1)(n)(n+1) = 0, 0/4 = 0. This means that (n-1)(n)(n+1) is divisible by 4? Any other odd n gives the product of 3 consequtives divisible by 4.
OG doesn't mention n=1, that's why I am curious
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by bluementor » Fri Dec 05, 2008 8:38 am
4meonly,

I had the same doubt before and it was cleared in the following post:

https://www.beatthegmat.com/og-11-ds-pro ... hlight=147

-BM-
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Re: If n is a positive integer, is (n^3) - n divisible by 4?

by farooq » Wed Oct 28, 2009 12:13 am
4meonly wrote: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

According to OG answer is A

However, I have some doubts.


Main stem: (n^3)-n = n(n^2 -1) = (n-1)(n)(n+1)
Rephrasing main stem (n-1)(n)(n+1) = 4q? Remainder should be =0.

(1)
n=2k+1, where k is an integer, OR n is odd
if k=0, then n=1 and (n-1)(n)(n+1) = 0, 0/4 = 0. This means that (n-1)(n)(n+1) is divisible by 4? Any other odd n gives the product of 3 consequtives divisible by 4.
OG doesn't mention n=1, that's why I am curious
I don't understand this problem completely. I think there are some flaws, especially with the value of k.

A: n=2k+1. Does not specify about the value of K.

SO if I put value of n in the expression n(n-1)(n-2) it gives,

(2k+1)(2K)(2k-1), now if I put k=0, this expression gives "0". Which is not divisible by 4. Therefore answer A is completely INCORRECT.
Regards,
Farooq Farooqui.
London. UK

It is your Attitude, not your Aptitude, that determines your Altitude.
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by NikolayZ » Wed Oct 28, 2009 3:48 pm
Answer is A indeed.
Rephrase the q. (n-1)*n*(n+1). These are consecutive integers.
From stmt(1), we know that n is odd.
If n is odd, then n-1 and n+1 are even, so each of them has at least one 2 in its prime factorizations.
So, the product is definitely divisible by 4.

From stmt2, we know that the product of cons. integers :
n(n+1) is divisible by 6. so the product has at least 2 and 3 in its prime factorization. But we don't know anything else about this one. It could be 30, for example.
30 is divisible by 6, but it is not divisible by 4.

P.s. 0 is the multiple and, consistently, dividend of every number.
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by davo45 » Sun Jul 24, 2011 1:04 am
Hi all,

Perhaps I may be having one of those days where the brain is not fucntioning efficiently, but is there a quicker way to determine that n MUST be odd to ensure N^3 - 1 is ALWAYS divisible by 4?

I understand N, when EVEN, can and cannot be divided by 4, but testing these conditions under exam conditions is a certain way to waste too much time. I appreciate your feedback.
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Re: If n is a positive integer, is (n^3) - n divisible by 4?

by Scott@TargetTestPrep » Wed Nov 25, 2020 6:38 am
4meonly wrote: ↑
Fri Dec 05, 2008 6:28 am
 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

According to OG answer is A

However, I have some doubts.


Main stem: (n^3)-n = n(n^2 -1) = (n-1)(n)(n+1)
Rephrasing main stem (n-1)(n)(n+1) = 4q? Remainder should be =0.

(1)
n=2k+1, where k is an integer, OR n is odd
if k=0, then n=1 and (n-1)(n)(n+1) = 0, 0/4 = 0. This means that (n-1)(n)(n+1) is divisible by 4? Any other odd n gives the product of 3 consequtives divisible by 4.
OG doesn't mention n=1, that's why I am curious
Solution:

We need to determine whether n^3 - n is divisible by 4. Notice that n^3 - n can be factored as n(n^2 - 1) = n(n - 1)(n + 1), which can be expressed as a product of three consecutive integers: (n - 1)(n)(n + 1).

Statement One Alone:

n = 2k + 1, where k is an integer

Statement one tells us that n is an odd integer. Thus, (n - 1) and (n + 1) are both even. We have (n - 1)(n)(n + 1) as even x odd x even, and so each even factor contributes at least one 2 to the product. Therefore, n^3 – n is divisible by 4.

(Note that even if n = 1, resulting in n^3 – n = 0, statement one still holds because zero is divisible by 4.)

Statement one alone is sufficient.

Statement Two Alone:

n^2 + n is divisible by 6

If n = 2, we see that n^2 + n = 4 + 2 = 6 is divisible by 6. However, n(n - 1)(n + 1) = 2(1)(3) = 6 is not divisible by 4.

Statement two alone is not sufficient.

Answer: A

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