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.
Divisibility
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n³ - n = n(n² - 1) = n(n - 1)(n + 1) = (n - 1)n(n + 1)ela07mjt wrote: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.
So, n³ - n is nothing but the product of 3 consecutive integer.
Now, if the middle one, i.e. n is odd, (n - 1) and (n + 1) both will be even.
Hence, the product (n - 1)n(n + 1) will be divisible by 4.
Now, if the middle one, i.e. n is even, (n - 1) and (n + 1) both will be odd.
Hence, the product (n - 1)n(n + 1) will be divisible by 4 only if n is divisible by 4.
Statement 1: This means n is an odd integer.
Hence, the product is divisible by 4.
Sufficient
Statement 2: (n² + n) = n(n + 1) is divisible by 6.
Consider the following two cases,
- n = 2 ---> (n³ - n) = 8 - 2 = 6 ---> Not divisible by 4
n = 3 ---> (n³ - n) = 27 - 3 = 24 ---> Divisible by 4
The correct answer is A.
Anju Agarwal
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Zero is divisible by any finite integer n as we can express zero as 0*n.ela07mjt wrote:What if K=0, in that case: n=1 and (n-1)n(n+1)=0*1*2 which is not divisible by 4
Hope that helps.
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Target question: Is n^3 - n divisible by 4?ela07mjt 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.
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