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