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If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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12 Sep 2019, 21:30
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Competition Mode Question 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
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Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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Updated on: 12 Sep 2019, 22:57
We are to determine if x^3  3x^2 + 2x divisible by 4, and that x is a positive integer.
X^3  3x^2 + 2x = x(x2)(x1)
From 1, we know x=4y+4 where y is an integer. This statement is sufficient because for every value of y, x is an integral multiple of 4. Hence would be divisible by 4.
From 2, x=2z+2. This also sufficient because we know that x is either 2, a multiple of 4, or 4r+2 where r is an integer. When x=2, we get the function reduced to 0, which is divisible by 4. When x=4r+2, the term (x2) will still make it a multiple of 4 hence it will be divisible by 4.
The answer is D in my view.
Originally posted by eakabuah on 12 Sep 2019, 21:52.
Last edited by eakabuah on 12 Sep 2019, 22:57, edited 1 time in total.



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Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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12 Sep 2019, 22:15
x3−3x2+2x divisible by 4?
(1) x=4y+4, where y is an integer
(4y+4)^33(4y+4)^2+2(4y+4) =4(a) = result will be a multiple of 4
(2) x=2z+2, where z is an integer
(2z+2)^33(2z+2)^2+2(2z+2) = 4(b) = result will be a multiple of 4
SO OA:D, both are sufficient individually.



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Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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Updated on: 13 Sep 2019, 22:35
If x is a positive integer, is \(x^3\)−3\(x^2\)+2x divisible by 4? \(\frac{x(x^23x+2)}{4}\)=? \(\frac{x(x1)(x2)}{4}\)=? (1) x=4y+4, where y is an integer If y=0, x=4 then \(\frac{4(41)(42)}{4}\)=?.........YES If y=1, x=8 then \(\frac{8(81)(82)}{4}\)=?.........YES If y=5, x=24 then \(\frac{24(241)(242)}{4}\)=?.........YES Always yes. SUFFICIENT! (2) x=2z+2, where z is an integer If z=1, x=4 then \(\frac{4(41)(42)}{4}\)=?.........YES If z=2, x=6 then \(\frac{6(61)(62)}{4}\)=?.........YES Always yes. SUFFICIENT! IMO answer is option D Posted from my mobile device
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Originally posted by EncounterGMAT on 12 Sep 2019, 22:51.
Last edited by EncounterGMAT on 13 Sep 2019, 22:35, edited 1 time in total.



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Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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12 Sep 2019, 23:07
its always important to simplify the questions statement; we get is x(x1)(x2) divisible by 4?1 a) x=4y+4 => 4(y+1) when we substitute into 1 we get 4(y+1)(x1)(x2) the 4 taken out common is divisible therefore yes! a is sufficient b)x= 2z+2 => 2(z+1) substituting we get 2(z+1)(2z+1)(2z) => 4*z*(z+1)*(2z+1).... yes b is sufficient therefore "D" is the answer
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Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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12 Sep 2019, 23:17
\(x^33x^2+2x=x(x^23x+2)=x(x1)(x2)\)
\(x(x1)(x2)\) is the product of three consecutive integers. This product will surely be divisible by \(4\) when two of three consecutive integers are even
This can only be possible when \(x\) and \((x2)\) are even.
So all we need to know is whether \(x\) is even
Statements (1) and (2) each independently state that \(x\) is even (Because even+even=even). So we have our answer
(1) and (2) are each independently sufficient
Answer is (D)
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Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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13 Sep 2019, 01:44
given eqn x^3−3x^2+2x can be re written as x(x^23z+2) #1 x=4y+4, where y is an integer test with y=even and odd integer ; value of x will always be a multiple of 4 since x=4(y+1) sufficient #2 x=2z+2, where z is an integer test with z=1,z=2 we get even integer value of x which is >2 ; hence x(x^23z+2) will be divisible by 4 ; 2^2 sufficient IMO D
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



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Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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13 Sep 2019, 02:47
If x is a positive integer, is \(x^3 − 3x^2 + 2x\) divisible by 4? \(x^3 − 3x^2 + 2x = x(x  2)(x  1)\) So if x is a multiple of 4 then \(x^3 − 3x^2 + 2x\) is divisible by 4. (1) \(x = 4y + 4\), where y is an integer Since \(x = 4(y + 1)\) where y ≥ 0 then \(x^3 − 3x^2 + 2x\) is divisible by 4 always. SUFFICIENT. (2) \(x = 2z + 2\), where z is an integer \(x = 2(z + 1)\) where y ≥ 0 If z is odd then x = 4k where k > 0 integer, then \(x^3 − 3x^2 + 2x\) is divisible by 4 always Or if z is even then x = 6, 10, 14 then \(x^3 − 3x^2 + 2x\) is not divisible by 4. INSUFFICIENT. Answer (A).
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Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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13 Sep 2019, 03:34
Quote: 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 \(x^3−3x^2+2x…x(x^23x+2)\) (1) x=4y+4, where y is an integer: sufic. \(x=4(y+1)…x=multiple.4…\frac{m4(x^23x+2)}{4}=x^23x+2=integer\) (2) x=2z+2, where z is an integer: sufic. \(x=2z+2…x(x^23x+2)=(2z+2)((2z+2)^23(2z+2)+2)=(2z+2)(4z^2+4+8z6z6+2);\) \(…=(2z+2)(4z^2+2z)=8z^3+4z^2+8z^2+4z=8z^3+12z^2+4z=4(2z^3+3z^2+z);\) \(…=\frac{4(2z^3+3z^2+z)}{4}=(2z^3+3z^2+z)=integer\) Answer (D)



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Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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13 Sep 2019, 04:38
IMO it's D (1) x=4y+4 (2) x=2z+2 Both are sufficient. a) Because x^3  3x^2 + 2x could be written as x*(x^23x+2) b) from both statements we know that x is even so it has a 2 in it. c) Similarly (x^23x+2) would be even and have a 2 in it. d) The total term would be divisible by 4(2*2)
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Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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13 Sep 2019, 08:48
If x is a positive integer, is \(x^3−3x^2+2x\) divisible by 4?
(1) x=4y+4, where y is an integer x = 4(y+1) and the expression is a multiple of x, hence it is also a multiple of 4. Sufficient.
(2) x=2z+2, where z is an integer x = 2(z+1) Evaluating 3 parts of the expression separately: \(x^3 = 2^3 (z+1)^3\) multiple of 8 \(3x^2 = 3*2^2*(z+1)^2\) multiple of 4 2x = 2*2 multiple of 4 Sufficient. Both 1 and 2 can independently answer the question. So D.




Re: If x is a positive integer, is x^3  3x^2 + 2x divisible by 4?
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