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.
Interesting DS question
- ronnie1985
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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...
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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fangtray
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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?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.
- seriousmonkey
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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.
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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IMO its D.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.
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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Ganesh hatwar
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D??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.
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surmilsehgal
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mehulsayani
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Hey here, x>0 and x^3-3x^2+ 2x is the given equationvenkb 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.
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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sitam.nitj
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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]
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]
- jaspreetsra
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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
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












