OG12 DS Question 170

[icely]

Hi all, regarding OG12 DS Question 170:

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

Just to confirm my trail of thought regarding Statement 1:

For k = 0 then the n^3-n breakup n(n+1)(n-1) will be 1 * 2 * 0 = 0

That will produce 0/4 = 0 which is an integer making Statement 1 sufficient.

From number properties theory, zero divided by an integer always yields a zero, which is an integer.

My question is whether a divisibility that yields a "zero" as in this example is considered sufficient to get the correct answer?

Thank you.

[adthedaddy]

Please note that in the question, it is explicitly mentioned that "n" is a positive integer.
So n=0 is incorrect. Zero is neither positive nor negative.

'n' can take values 1,2,3,....

[Anurag@Gurome]

n^3 - n = n(n² - 1) = (n - 1)n(n + 1)

(1) n = 2k + 1, where k is an integer implies n = odd, and since n is odd so n - 1 and n + 1 are even.
So, (n - 1)n(n + 1) is divisible by 4; SUFFICIENT.

(2) n² + n is divisible by 6.
If n = 2 then n^3 - n = 6, then n^3-n is NOT divisible by 4.
If n = 3 then n^3 - n = 24, then n^3 - n is divisible by 4
No definite answer; Not sufficient.

The correct answer is A.

[Brent@GMATPrepNow]

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

Target question: Is n^3 - 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 free video on this strategy: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

Notice that we can take n^3 - n and factor it to get n(n^2 - 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: n^2 + n is divisible by 6
Notice that we can take n^2 + 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,
Brent