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Math Expert V
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If x is a positive integer, is x^3 - 3x^2 + 2x divisible by 4?  [#permalink]

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Question Stats: 57% (01:44) correct 43% (01:35) wrong based on 60 sessions

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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

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Re: If x is a positive integer, is x^3 - 3x^2 + 2x divisible by 4?  [#permalink]

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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(x-2)(x-1)

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 (x-2) 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, 20:52.
Last edited by eakabuah on 12 Sep 2019, 21:57, 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?  [#permalink]

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1
x3−3x2+2x divisible by 4?

(1) x=4y+4, where y is an integer

(4y+4)^3-3(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)^3-3(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?  [#permalink]

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If x is a positive integer, is $$x^3$$−3$$x^2$$+2x divisible by 4?
$$\frac{x(x^2-3x+2)}{4}$$=?

$$\frac{x(x-1)(x-2)}{4}$$=?

(1) x=4y+4, where y is an integer
If y=0, x=4 then $$\frac{4(4-1)(4-2)}{4}$$=?.........YES
If y=1, x=8 then $$\frac{8(8-1)(8-2)}{4}$$=?.........YES
If y=5, x=24 then $$\frac{24(24-1)(24-2)}{4}$$=?.........YES
Always yes. SUFFICIENT!

(2) x=2z+2, where z is an integer
If z=1, x=4 then $$\frac{4(4-1)(4-2)}{4}$$=?.........YES
If z=2, x=6 then $$\frac{6(6-1)(6-2)}{4}$$=?.........YES
Always yes. SUFFICIENT!

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Originally posted by EncounterGMAT on 12 Sep 2019, 21:51.
Last edited by EncounterGMAT on 13 Sep 2019, 21:35, 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?  [#permalink]

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its always important to simplify the questions statement; we get

is x(x-1)(x-2) divisible by 4?----1

a) x=4y+4 => 4(y+1)
when we substitute into 1 we get 4(y+1)(x-1)(x-2) 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

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Re: If x is a positive integer, is x^3 - 3x^2 + 2x divisible by 4?  [#permalink]

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$$x^3-3x^2+2x=x(x^2-3x+2)=x(x-1)(x-2)$$

$$x(x-1)(x-2)$$ 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 $$(x-2)$$ 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

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Re: If x is a positive integer, is x^3 - 3x^2 + 2x divisible by 4?  [#permalink]

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given eqn
x^3−3x^2+2x can be re written as x(x^2-3z+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^2-3z+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?  [#permalink]

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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.

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Re: If x is a positive integer, is x^3 - 3x^2 + 2x divisible by 4?  [#permalink]

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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

$$x^3−3x^2+2x…x(x^2-3x+2)$$

(1) x=4y+4, where y is an integer: sufic.
$$x=4(y+1)…x=multiple.4…\frac{m4(x^2-3x+2)}{4}=x^2-3x+2=integer$$

(2) x=2z+2, where z is an integer: sufic.
$$x=2z+2…x(x^2-3x+2)=(2z+2)((2z+2)^2-3(2z+2)+2)=(2z+2)(4z^2+4+8z-6z-6+2);$$
$$…=(2z+2)(4z^2+2z)=8z^3+4z^2+8z^2+4z=8z^3+12z^2+4z=4(2z^3+3z^2+z);$$
$$…=\frac{4(2z^3+3z^2+z)}{4}=(2z^3+3z^2+z)=integer$$

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Re: If x is a positive integer, is x^3 - 3x^2 + 2x divisible by 4?  [#permalink]

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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^2-3x+2)
b) from both statements we know that x is even so it has a 2 in it.
c) Similarly (x^2-3x+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?  [#permalink]

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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
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?   [#permalink] 13 Sep 2019, 07:48
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