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# If x, y and z are positive integers, then x^2y^3z^4 must be divisible

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Joined: 02 Sep 2009
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If x, y and z are positive integers, then x^2y^3z^4 must be divisible  [#permalink]

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06 Dec 2016, 07:33
1
00:00

Difficulty:

25% (medium)

Question Stats:

70% (01:34) correct 30% (01:49) wrong based on 50 sessions

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If x, y and z are positive integers, then $$x^2y^3z^4$$ must be divisible by which of the following?

I. x^2 + y^3 + z^4
II. xy + xz
III. xyz + z

A. None
B. I only
C. I and II only
D. I and III only
E. II and III only

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Re: If x, y and z are positive integers, then x^2y^3z^4 must be divisible  [#permalink]

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06 Dec 2016, 08:06
Bunuel wrote:
If x, y and z are positive integers, then $$x^2y^3z^4$$ must be divisible by which of the following?

I. x^2 + y^3 + z^4
II. xy + xz
III. xyz + z

A. None
B. I only
C. I and II only
D. I and III only
E. II and III only

Set $$x=1, y=2, z=3 \implies x^2y^3z^4=1^22^33^4=648$$
$$x^2+y^3+z^4=1^2+2^3+3^4=1+8+81=90$$
Since 648 is not divisible by 90, (I) is not true.

$$xy+xz=1 \times 2 +1\times 3=5$$
Since 648 is not divisible by 5, (II) is not true.

Now set $$x=3, y=2, z=1 \implies x^2 y^3 z^4=3^2 \times 2^3 \times 1^4=72$$
$$xyz+z=2\times 3 \times 1+1=7$$
Since 72 is not divisible by 7, (III) is not true.

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Re: If x, y and z are positive integers, then x^2y^3z^4 must be divisible  [#permalink]

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06 Apr 2019, 03:12
Bunuel wrote:
If x, y and z are positive integers, then $$x^2y^3z^4$$ must be divisible by which of the following?

I. x^2 + y^3 + z^4
II. xy + xz
III. xyz + z

A. None
B. I only
C. I and II only
D. I and III only
E. II and III only

Assign values for a, b, c.

positive integers.

a = 1
b = 1
c = 1.

none of the options are correct.

Re: If x, y and z are positive integers, then x^2y^3z^4 must be divisible   [#permalink] 06 Apr 2019, 03:12
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