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Quick way to do this problem?

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Quick way to do this problem?

by fangtray » Tue Apr 24, 2012 5:01 am
Hello, I took the official prac exam, and got this quesiton right through plugging in numbers.. It took WAAAY too long.

Could someone help me with a fast way of doign this problem?

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
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Source: — Data Sufficiency |

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by Anurag@Gurome » Tue Apr 24, 2012 5:07 am
fangtray wrote:Hello, I took the official prac exam, and got this quesiton right through plugging in numbers.. It took WAAAY too long.

Could someone help me with a fast way of doign this problem?

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
n^3 - n = n(n² - 1) = (n - 1)n(n + 1)
We have to find if the product of 3 consecutive integers is divisible bu 4.

(1) n = 2k + 1, where k is an integer implies n is odd.
As n is odd so both n - 1 and n + 1 are even
Hence (n - 1)n(n + 1) is divisible by 4; SUFFICIENT.

(2) n² + n is divisible by 6
If n = 2 then n^3 - n = 6. Here 6 is not divisible by 4.
If n = 3 then n^3 - n = 24. Here 24 is divisible by 6
No definite answer; NOT sufficient.

The correct answer is A.
Anurag Mairal, Ph.D., MBA
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