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DS OG 11 QUESTION 147

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DS OG 11 QUESTION 147

by treavorhill » Tue Jun 26, 2007 8:41 pm
147. IF N IS A POSITIVE INTEGER, IS N^3-N DIVISIBLE BY 4.

1. N = 2K + 1, WHERE K IS AN INTEGER.

Why is this sufficient?

If K is zero, then N = 1.....ETS is assuming that K cannot equal zero. is zero not an integer?
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Re: DS OG 11 QUESTION 147

by jayhawk2001 » Tue Jun 26, 2007 9:28 pm
treavorhill wrote:147. IF N IS A POSITIVE INTEGER, IS N^3-N DIVISIBLE BY 4.

1. N = 2K + 1, WHERE K IS AN INTEGER.

Why is this sufficient?
n^3 - n = n(n^2 - 1) = n (n+1)(n-1)

If n=2k+1, we have

= (2k+1)(2k+2)(2k)
= 2*2k*(k+1)*(2k+1)
= 4k(k+1)(2k+1)

Since k is an integer, we know the n^3-n is divisible by 4.
Hence sufficient.

By definition every integer (except zero) is a divisor of zero.
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another approach

by discreet » Wed Jun 27, 2007 11:56 pm
n^3-n is a product of 3 consecutive integers....
i.e. It can be expressed as n(n-1)(n+1).

For divisibility by 4,

A product of consecutive integers of the form n^3-n i.e. n(n-1)(n+1) is always divisible by 4,if n is ODD.
Its given that n=2k+1 which will always be odd.Hence sufficient.

Also,using Kaplan's strategy of picking numbers,
If n=2k+1,then by picking simple numbers,it can be easily observed that the product will always be a factor of 4.

eg: K=1 implies n=3 , so the expression is 3*2*4...a factor of 4.
k=2 implies n =5, so the expression is 5*4*6...a factor of 4.

To conclude in general,

A product of consecutive integers of the form n^3-n i.e. n(n-1)(n+1) is always divisible by 4,if n is ODD.
A product of consecutive integers of the form n(n+1)(n+2) or n^3+3n^2+2n is always divisible by 4 if n is EVEN!

I hope this helps...
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