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by ronnie1985 » Fri Feb 03, 2012 10:17 am
Very tempting to write (A) as answer
But S2. x=2(z+1) = 2k
Exp = x^3-3x^2+2x = 8k^3-3*4k^2+2*2k = 4*m, divisible by 4
(D) is answer.
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by [email protected] » Sat Feb 18, 2012 8:31 pm
If x>0, then is x^3-3x^2+2x divisible by 4?

1. x=4y+4 where y is an integer
2. x=2z+2; were z is an integer


Statement 1 was not a problem... I didnt even had to put in values for statement 1 because any number which is a divisible by 4 when put in any polynomial or a term, has to be divisible by 4.


Statement 2 was a problem because I had to put in values for z and see the answer in the equation...

This was the only thing...



Well there are many times when you actually have to put in values even if you sorted the equations or simplified the equations...
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by kul512 » Tue Feb 21, 2012 4:20 pm
What is the use of the information............x>0.
Does it play any role in the problem solving.
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by fangtray » Sun Apr 01, 2012 6:15 am
venkb wrote:If x>0, then is x^3-3x^2+2x divisible by 4?

1. x=4y+4 where y is an integer
2. x=2z+2; were z is an integer

I saw this quesiton in Manhattan guide, but i am not convinced with its answer.
could someone provide the question with some parenthesis, i'm afraid if i look down i'll accidentally glance over the answer without attempting it first. but curious, how does one do this problem without parenthesis to give you the specific equation?
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by seriousmonkey » Mon Apr 02, 2012 7:07 am
I did this with some basic algebra:

The first statement is true because we can factor out 4 from x=4(y+1) and the expression is (x^3)-(3x^2)+2x as x(x^2-3x+2)

I simplified the expression also by adding x and 1 and subtracting x and 1. So we get:

x^3 - 3x^2 + 2x +x +1 -x -1; this gives x^3-3x^2+3x-1- x+1; this is (x-1)^3 -(x-1) or (x-1)*[(x-1)^2-1];

Substituting the second statement, we get (2z+1)*(4z^2-4z). This is divisible by 4.

Hence the answer is D.
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by somsubhra86 » Tue Apr 24, 2012 5:48 am
I am On A
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by shantanu86 » Tue Apr 24, 2012 7:14 am
venkb wrote:If x>0, then is x^3-3x^2+2x divisible by 4?

1. x=4y+4 where y is an integer
2. x=2z+2; were z is an integer

I saw this quesiton in Manhattan guide, but i am not convinced with its answer.
IMO its D.

Explanation-

eq = x^3-3x^2+2x
=> x(x^2 -3x +2)
=> x(x-1)(x-2)

[A] x= 4(y+1) if x is a multiple of 4 eq will also be a multiple. Therefore, sufficient alone.
x= 2z+2 => eq = 2(z+1)(2z+1)2z .. a multiple of 4.. this this alone is also sufficient.

Also, 0 is a multiple of every number except 0 :)

Hope it helps!!
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by Ganesh hatwar » Wed Jul 25, 2012 11:11 pm
venkb wrote:If x>0, then is x^3-3x^2+2x divisible by 4?

1. x=4y+4 where y is an integer
2. x=2z+2; were z is an integer

I saw this quesiton in Manhattan guide, but i am not convinced with its answer.
D??
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by surmilsehgal » Sun Aug 05, 2012 3:28 am
tha answer is E??
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by cuddytime » Sun Aug 12, 2012 6:25 pm
i like "D" for this answer.

broken down the answer is: x(x-1)(x-2)

you're trying to see if a or b is a multiple of 4 (or has 2 2's when factored out)
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by mehulsayani » Tue Aug 14, 2012 3:53 am
venkb wrote:If x>0, then is x^3-3x^2+2x divisible by 4?

1. x=4y+4 where y is an integer
2. x=2z+2; were z is an integer

I saw this quesiton in Manhattan guide, but i am not convinced with its answer.
Hey here, x>0 and x^3-3x^2+ 2x is the given equation
which can be simplified to x(x^2-3x+2) => x(x-1)(x-2)
which mean its the product of three consecutive numbers.

Consider 1. x= 4y + 4 = 4(y+1) So, x is always divisible by 4 and so is our equation. Therefore, our 1. is sufficient

Consider 2. x= 2z+2 = 2(z+1) So, x is even, we have even*odd*even as our given equation.
which is always divisible by 4,(in case you dont get it, its even(divisible by 2)*odd*even(divisible by 2))

2. is also sufficient.

So, D
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by rajeshsinghgmat » Sun Feb 10, 2013 1:26 am
D is the answer.
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by Java_85 » Wed Sep 04, 2013 5:57 pm
I'm kinda sure it is D.
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by sitam.nitj » Fri Oct 17, 2014 4:59 am
x^3 - 3(x^2) + 2x = x(x-1)(x-2)

Now x(x-1)(x-2) is always divisible by 4 if x is even and x>0 is given in the statement.

1. x = 4y + 4 = 4(y+1) which is always even because ( even * even = Even & Even * odd = Even ) so the expression x^3 - 3(x^2) + 2x is divisible by 4 in first case. (Note - y is an integer but it cannot be less than 0 because x>0)

2. x = 2z + 2 = 2(z+1) which is always divisible by 4 as 2 multiplied by anything is even.

So the answer is D.[/list][/quote]
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by jaspreetsra » Sun Oct 26, 2014 7:10 am
I think that answer is E.
Reason:
If x > 0, then is x^3 - 3(x^2) + 2x divisible by 4?

1. x = 4y + 4, where y is an integer
2. x = 2z + 2, where z is an integer

x^3 - 3(x^2) + 2x = x(x-2)(x-1)
So need x =? to answer this question.

1)Not sufficient
2)Not sufficient

Answer: E
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