Help with questions regarding factors/divisors

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shanrizvi: Senior | Next Rank: 100 Posts; Posts: 73; Joined: Tue Aug 11, 2009 2:30 am; Thanked: 2 times; GMAT Score:720

by shanrizvi » Mon Aug 17, 2009 1:56 pm

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?

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Source: Beat The GMAT — Data Sufficiency | Mon Aug 17, 2009 1:56 pm

beboppin: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Sat Aug 08, 2009 5:31 pm

Re: Help with questions regarding factors/divisors

by beboppin » Mon Aug 17, 2009 6:21 pm

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.

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Wed Aug 19, 2009 8:35 pm

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

The powers of two are bloody impolite!!

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ranell: Master | Next Rank: 500 Posts; Posts: 224; Joined: Mon May 04, 2009 2:44 pm; Location: Russia, Moscow; Thanked: 10 times; GMAT Score:730

by ranell » Thu Aug 20, 2009 3:38 pm

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

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beboppin: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Sat Aug 08, 2009 5:31 pm

Re: Help with questions regarding factors/divisors

by beboppin » Thu Aug 20, 2009 6:41 pm

shanrizvi wrote: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.

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