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If x is a positive integer, is x^3 - 3x^2 + 2x divisible by

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If x is a positive integer, is x^3 - 3x^2 + 2x divisible by [#permalink] New post 06 Dec 2009, 11:42
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D
E

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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
[Reveal] Spoiler: OA
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Re: MGMAT Problem Divisibility Help, Properties of "0" [#permalink] New post 10 Dec 2009, 08:20
does 0 divide by integer(4) consider as divisible ?
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Re: MGMAT Problem Divisibility Help, Properties of "0" [#permalink] New post 10 Dec 2009, 16:31
tashu wrote:
does 0 divide by integer(4) consider as divisible ?


Yes. 0 divided by anything returns 0. And zero is a non-positive, non-negative integer.
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positive integer [#permalink] New post 21 Oct 2010, 06:34
If x is a positive integer, is x^2-3x^2+2x divisible by 4 ?
1) x= 4y+4, where y is an integer
2) x=2z + 2, where z is an integer
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Re: positive integer [#permalink] New post 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
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Re: positive integer [#permalink] New post 21 Oct 2010, 07:19
x^2-3x^2 + 2x
= -x^2 + 2x
= 2x (1-x)
so if we know x if x is divisible by 2, we know that 2x(1-x) is divisible by 4

1) x = 4y+4 = 4(y+1) -> x is divisible by 4, so enough to identify that 2x(1-x) is divisible by 4

2) x=2z + 2 = 2(z+1) , Again x is divisible by 2, so sufficient,

Ans D
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is x^3 - 3x^2 + 2x divisible by 4 [#permalink] New post 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... :( :(
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Re: is x^3 - 3x^2 + 2x divisible by 4 [#permalink] New post 03 Jul 2011, 06:54
(1) is obviously sufficient, as 4 can be factored out of the expression
x = 4y+4, => x is divisible by 4

and the expression x^3-3x^2+2x is also divisible by 4.

2) x = 2(z+1)

The expression is - x(x^2 + 3x + 2) = x(x+1)(x+2)

So if we simplify this by substituting x = 2(z+1)

then

expression = 2(z+1) (2z+3)(2z+4) = 4 (z+1)(2z+3)(z+2)

which is divisible by 4

Answer - D
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Re: is x^3 - 3x^2 + 2x divisible by 4 [#permalink] New post 03 Jul 2011, 07:11
subhashghosh wrote:
(1) is obviously sufficient, as 4 can be factored out of the expression
x = 4y+4, => x is divisible by 4

and the expression x^3-3x^2+2x is also divisible by 4.

2) x = 2(z+1)

The expression is - x(x^2[highlight]+[/highlight]3x + 2) = [highlight]x(x+1)(x+2)[/highlight] (i think it should be x(x-2)(x-1))

So if we simplify this by substituting x = 2(z+1)

then

expression = 2(z+1) (2z+3)(2z+4) = 4 (z+1)(2z+3)(z+2) [[highlight]2(z+1) (2z) (2z+1) = 4 (z+1)(z)(2z+1)[/highlight]]

which is divisible by 4

Answer - D


Now i got it
Thanks Anyway i committed a mistake in the steps.... :( :(
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Re: is x^3 - 3x^2 + 2x divisible by 4 [#permalink] New post 03 Jul 2011, 07:14
Both sufficient....
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Re: is x^3 - 3x^2 + 2x divisible by 4 [#permalink] New post 03 Jul 2011, 07:16
Yeah, you're right. I'm reading in a hurry these days :(
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Re: is x^3 - 3x^2 + 2x divisible by 4 [#permalink] New post 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
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DS - Number Properties [#permalink] New post 25 Sep 2011, 08:06
Help.

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

Rephrased question: is x(x-1)(x-2) divisible by 4? or is x even?

Both statements say that x is even, and thus answer is D, which is also the OA.

But...

Statement 1 - what if y = -1, and thus x=0?
Statement 2 - what if z = 0, and thus x = 0?

in each case, if x^3 - 3x^2 +2x = 0, is 0 divisible by 4? or more generally, is 0 divisible by any number?
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Re: DS - Number Properties [#permalink] New post 25 Sep 2011, 08:13
syh244 wrote:
Help.

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

Rephrased question: is x(x-1)(x-2) divisible by 4? or is x even?

Both statements say that x is even, and thus answer is D, which is also the OA.

But...

Statement 1 - what if y = -1, and thus x=0?
Statement 2 - what if z = 0, and thus x = 0?

in each case, if x^3 - 3x^2 +2x = 0, is 0 divisible by 4? or more generally, is 0 divisible by any number?



O is a multiple of any number . So even if X = 0 ; its divisible by 4
therefore its D
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Re: DS - Number Properties [#permalink] New post 25 Sep 2011, 08:33
Yes. 0 is divisible by any number.


Another way to solve this ..


x^3-3x^2+2x is divisible by 4?

1. Sufficient

x = 4y+4 = 4(y+1)
As x already has a factor 4, x^2,x^3,2x will all have a factor 4 for sure. Hence x^3-3x^2+2x is divisible by 4.

2. Sufficient

x= 2z+2 = 2(z+1)
x^2 = 4(z+1)^2
x^3 = 8(z+1)^3
2x = 4(z+1)
x^2,x^3 and 2x all have a factor 4. Hence x^3-3x^2+2x is divisible by 4.

Answer is D.
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Re: MGMAT DS: Is the polynomial divisible by 4 [#permalink] New post 25 Sep 2011, 08:40
JimmyWorld wrote:
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


Rephrase:

Is x even?

1) x = 4y + 4= 4(y+1)
x is even.
Sufficient.

2) x = 2z + 2= 2(z+1)
x is even.
Sufficient.

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