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Helpful Geometry formula sheet:best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

Re: positive integer [#permalink]
21 Oct 2010, 07:15

The option is D, Each statement alone is sufficient Simplifying statement , we get 2x(1-x) Now in option A, we can take 4 as common so the statement becomes 2.4(y+1)(1-4y-4) which is definitely divisible by 4, In option B, we can take 2 common same way, which when multiplied by 2 already in statement gives 4 as common which is definitely divisible by 4 , Hence each statement alone in sufficient

is x^3 - 3x^2 + 2x divisible by 4 [#permalink]
03 Jul 2011, 06:36

if x is a positive integer, is \(x^3 - 3x^2 + 2x\) divisible by 4? 1. x = 4y + 4, where y is an integer

2. x = 2z + 2, where z is an integer

i need some clarification about B

please guide me... _________________

WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post.

Now i got it Thanks Anyway i committed a mistake in the steps.... _________________

WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post.

Re: is x^3 - 3x^2 + 2x divisible by 4 [#permalink]
03 Jul 2011, 10:26

IMO D: x^3 - 3x^2 + 2x = x(x^2-3x+2) = x(x-2)(x-1) Condition 1: x = 4y + 4, where y is an integer x = 4(y+1) If we put x = 4(y+1) in above eq x(x-2)(x-1) it will be divisible by 4 for all values because x = 4(y+1)

2. x = 2z + 2, where z is an integer x = 2(z+1) if we put x in equation "x(x-2)(x-1)" = 2(z+1)(2z)(2z+1) = 4(z+1)z(2z+1) Again this condition is fine. Hence option D

Re: MGMAT DS: Is the polynomial divisible by 4 [#permalink]
26 Sep 2011, 04:10

+ 1 for D.

Time taken: 52 seconds.

However, I did this question on paper, on second thoughts I realised that this can be done with mental calculations. All we need to know here is that can we have 4 a a common multiple of all components of the equation. _________________

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