Help with questions regarding factors/divisors

[shanrizvi]

If n is a positive integer, is n^3 - n divisble by 4?

a) n=2k+1, where k is an integer

b) n^2 +n is divisible by 6


I almost always get questions like this wrong! I generally do well on the math section but this is one thing that I just don't get the hang of. What should I do? Any excellent resource that can help me?

[beboppin]

shanrizvi, I'm sorry I can't give you an answer here, but I wanted to post to second your statement. These problems also give me fits and would love to learn about how to approach these. Any resources (maybe MGMAT's number properties book?) that can help us tackle this would be greatly appreciated.

[tohellandback]

IMO A
n^3-n=n(n+1)(n-1)
for any number to be divisible by 4, it must have two 2's

a) n=2k+1. so n is odd
n+1- is even( got out one 2)
n-1- is even(got the 2nd 2)
SUFF

b)n^2+n is divisible by 6
plug in numbers
n^2+n=n(n+1)
n=6, 6*7- not divisible by 4
n=12, 12*13- divisible by 4
NOT SUFF

[ranell]

n^3 - n = n*(n-1)*(n+1)
1) sufficient as n=2k+1 is always odd and thus n^3 - n = n*(n-1)*(n+1) will always be divisible by 4
2) not sufficient
n^2 + n = n*(n+1)
take n=5 and n=2 and you will get two different results

Hence answer is A

[beboppin]

If n is a positive integer, is n^3 - n divisble by 4?

a) n=2k+1, where k is an integer

b) n^2 +n is divisible by 6


I almost always get questions like this wrong! I generally do well on the math section but this is one thing that I just don't get the hang of. What should I do? Any excellent resource that can help me?

How about if k is 0, and I believe 0 is an integer. Then n^3 - n is zero. I guess 0 is considered divisible by any number?

And do you guys agree that these divisibility/number property questions often rely on sequences (i.e. n-1, n, n+1 and such) to get to the answers? I've definitely seen a handful already..

Thanks.