mkbigmoz 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



[spoiler]1. I thought 1 was correct at first, but what if K was 0. Then N = 0+1= 1.

1^3-1 is not divisible by 4. This all depends on if 0 is considered an integer. Ive read various answers on this. Anyone know for sure?[/spoiler]

stmt 1 means it series of odd, so n=1, 3,5...

yes if n=1, n^3 – n=0, and yes 0 is divisble by 4. You can think of it this way - what do you get when you divide 0 by 4? Is there a remainder? No. So yes, 0 is divisible by 4. Infact 0 is divisble by any number.

so even with this case, A is sufficient

Stmt 2 says that product of 2 consecutive integer n and n+1

so, n may be odd or even
If its odd, n^3 – n is divisble by 4 but when n is even, its not ( try with 2)

Choose A