If we consider the value of K as 0

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If we consider the value of K as 0

by Paras_0111 » Fri Jul 08, 2016 12:54 am
If n is a positive integer, is (n^3 - n) divisible by 4 ?
(1) n = 2k + 1, where k is an integer.
(2) n2 + n is divisible by 6.

This is a q from OG12 and the answer is A. My question is why can't we consider the value of K as 0 which would make even the first condition NS. Please help. Thank you..

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by MartyMurray » Fri Jul 08, 2016 2:44 am
Paras_0111 wrote:This is a q from OG12 and the answer is A. My question is why can't we consider the value of K as 0 which would make even the first condition NS. Please help. Thank you..
Statement 1: n = 2k + 1

Let's try some values.

k = 0, n = 0 + 1 = 1, n³ - n = 1 - 1 = 0

0 is divisible by 4, or by any other number actually. 0/4 = 0

k = 1, n = 2 + 1 = 3, n³ - n = 27 - 3 = 24

24 is divisible by 4.

So far they are both divisible by 4.

Maybe you thought that 0 isn't divisible by 4?
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by Paras_0111 » Fri Jul 08, 2016 5:10 am
Thank you Marty. That was my mistake. I was assuming 0 isnt divisible by 4. Just a thought. Is it safe to assume that 0 is divisible by the number they are asking for? Because even in answer explanations, no one has considered 0 as a possible value..or perhaps they are already assuming that any number is a factor of 0?

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by MartyMurray » Fri Jul 08, 2016 5:13 am
Paras_0111 wrote:Thank you Marty. That was my mistake. I was assuming 0 isnt divisible by 4. Just a thought. Is it safe to assume that 0 is divisible by the number they are asking for? Because even in answer explanations, no one has considered 0 as a possible value..or perhaps they are already assuming that any number is a factor of 0?
Hmm. Yeah, maybe 0 isn't mentioned because any number is a factor of it.
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by Brent@GMATPrepNow » Fri Jul 08, 2016 5:54 am
If n is a positive integer, is n³ - n divisible by 4 ?

(1) n = 2k + 1, where k is an integer.
(2) n² + n is divisible by 6.
Target question: Is n³ - n divisible by 4?

This is a great candidate for rephrasing the target question.

Aside: Rephrasing the target question can often make data sufficiency questions easier (and faster) to solve. We have a video on this strategy: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

Notice that we can take n³ - n and factor it to get n(n² - 1), which equals n(n-1)(n+1) or (n-1)(n)(n+1)
Now recognize that n-1, n, and n+1 are three consecutive integers. The GMAT often hides this kind of information within given algebraic expressions.

So, at this point, we can rephrase the target question as: Is the product of 3 consecutive integers divisible by 4?

If we dig a little deeper, we can further rephrase the target question to make the question even easier to solve.
To do this, we'll ask, "Under what circumstances is the product of 3 consecutive integers divisible by 4? Well, there are two such circumstances.
Circumstance 1: The first and last integers are even. For example, the product of 2, 3, and 4 will be divisible by 4. In this circumstance, the middle number (n) is odd.
Circumstance 2: The middle integer is divisible by 4. For example, the product of 7, 8, and 9 must be divisible by 4 since the number 8 is already divisible by 4. In this circumstance, the middle number (n) is divisible by 4.

Given these two circumstances, we can rephrase the target question as: Is n either odd or divisible by 4?

At this point, we can check the statements.

Statement 1: n = 2k + 1, where k is an integer
This is a very clever way of telling us that n is odd. In fact, this is the formal definition of an odd number.
Since n is odd, we can now answer the rephrased target question with certainty.
So, statement 1 is SUFFICIENT

Statement 2: n² + n is divisible by 6
Notice that we can take n² + n and rewrite it as (n)(n+1), and we know that n and n+1 are two consecutive integers.
This information yields different possible cases, here are two.
case a: n=2, n+1=3, in which case n is neither odd nor divisible by 4
case b: n=3, n+1=4, in which case n is odd
Since statement 2 yields conflicting answers to our rephrased target question, it is NOT SUFFICIENT.

Answer = A

Cheers,
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by 800_or_bust » Fri Jul 08, 2016 9:35 am
Paras_0111 wrote:If n is a positive integer, is (n^3 - n) divisible by 4 ?
(1) n = 2k + 1, where k is an integer.
(2) n2 + n is divisible by 6.

This is a q from OG12 and the answer is A. My question is why can't we consider the value of K as 0 which would make even the first condition NS. Please help. Thank you..
If k = 0, then n =1 and the expression evaluates to 0. 0 is divisible by 4. In fact, it's divisible by any number.
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by Paras_0111 » Fri Jul 08, 2016 11:50 pm
Thank you everyone for clarifying my doubt. Sincerely appreciate it! :)

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by rocker2112 » Sat Jul 16, 2016 11:05 pm
n3 - n = (n-1)(n)(n+1)

On substituting n = 2k + 1 we get ---> (2k)(2k + 1 )(2k + 2 ) ---> take 2 common ----> (4)(k)(2k+1)(k+1) ----> always divisible by 4

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by Matt@VeritasPrep » Fri Jul 22, 2016 1:58 am
Paras_0111 wrote:Thank you everyone for clarifying my doubt. Sincerely appreciate it! :)
Just in case this makes the logic easier to remember, try thinking of the definition of divisibility. We know that x is divisible by y if

x / y = integer

In our case, we have

0 / 4 = 0

Since 0 (our result) is an integer, 0 (our numerator) must be divisible by 4 (our denominator).