For a positive integer n, is n divisible by 6? (1) n = m3 - m, for any

Statement 1 :
Test n for 3 values of m(1,2,3)

For m = 1
n = \(1^3 - 1 \)= 1 - 1 = 0
Divisible by 6
For m = 2
n = \(2^3 - 2 \)= 8 - 2 = 6
Divisible by 6
For m = 3
n = \(3^3 - 3 \)= 27 - 3 = 24
Divisible by 6
Hence, Statement 1 alone is sufficient.
Statement 2 :
Test n for 3 values of k(1,2,3)

For k = 1
n = \(1^3 + 5 * 1\) = 1 + 5 = 6
Divisible by 6
For k = 2
n = \(2^3 + 2*5 \)= 8 + 10 = 18
Divisible by 6
For k = 3
n = \(3^3 + 5*3\) = 27 + 15 = 42
Divisible by 6
Hence statement 2 alone is sufficient.
So, OA is (D).

The language here makes no sense. It cannot possibly be true that "n = m^3 - m for any positive integer m", since m = 1 and m = 100, say, give you two different values for n, and n can only have one value.

I imagine they mean to say "some" instead of "any". Then m^3 - m = (m-1)(m)(m+1) is the product of three consecutive integers, and the product of c consecutive integers is always divisible by c!, so this product is divisible by 3! = 6. So Statement 1 is sufficient.

Factoring Statement 2, k^3 + 5k = k(k^2 + 5). Notice that exactly one of k and k^2 + 5 will be even, the other odd, so this product is divisible by 2. If k is divisible by 3, then k(k^2 + 5) clearly will be too. If k is not divisible by 3, then we either get a remainder of 1 or of 2 when we divide k by 3. In either case, the remainder when k^2 is divided by 3 will be 1 (you can see that just by plugging k = 1 and k = 2 into k^2 and finding the remainder). If k^2 gives a remainder of 1 when we divide by 3, k^2 + 5 will give us a remainder of zero when we divide by 3, or in other words, k^2 + 5 will be divisible by 3. So no matter the value of k, k(k^2 + 5) will always be a multiple of both 2 and of 3, and thus of 6, and Statement 2 is sufficient.

When discussing Statement 2, I've used some principles of remainder arithmetic ("modular arithmetic", in advanced math) that aren't explicitly tested on the GMAT, though they can occasionally be useful to understand.

n is positive integer.

is n divisible by 6?

1) n = m^3 - m = (m-1)(m)(m+1). Three consecutive integers will have atleast one multiple of 3 and one even number at the very least. So, n is divisible by 6.

Sufficient.

2) n = k^3 + 5k = (k^3 - k) + 6k. From statement 1 we know that (k^3 - k) is always divisible by 6.

Sufficient.

Ans: D

1) n = m (m+1) (m -1) , a product of 3 consecutive integers, so at least one of the integers will be divisible by 2 and another one by 3. So, n is always divisible by 6. Sufficient.

2) n = k (k^2 + 5). When k is 2, n = 2* 9, divisible by 6. When k =3, n = 3 *14, divisible by 6. For any positive integer, one part of the integer will be divisible by 3 and another part by 2. Suffciient.

D is the answer.

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