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# A tricky one.

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Joined: 19 Sep 2011
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A tricky one. [#permalink]  02 Oct 2011, 02:26
Hi everybody.
I really don't get it, can you explain me pls?

If x^2=y+5, y= z-2 and z=2x, is x^3+y^2+z divisible by 7?
1. x>0
2. y=4
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Manager
Joined: 03 Mar 2011
Posts: 94
Location: United States
Schools: Erasmus (S)
GMAT 1: 730 Q51 V37
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Re: A tricky one. [#permalink]  02 Oct 2011, 03:18
1
KUDOS
$$x^2=2x+3$$
$$x^2-2x-3=0$$
$$x^2-2x+1-4=0$$
$$(x-1)^2-4=0$$
$$(x-1)^2=4$$
$$x-1=2$$ or $$x-1=-2$$
So, $$x=3$$ or $$x=-1$$
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Manager
Joined: 03 Mar 2011
Posts: 94
Location: United States
Schools: Erasmus (S)
GMAT 1: 730 Q51 V37
GPA: 3.9
Followers: 1

Kudos [?]: 100 [0], given: 12

Re: A tricky one. [#permalink]  02 Oct 2011, 02:54
We have:
$$x^2=y+5$$
$$y=z-2$$
$$z=2x$$

$$z=2x$$
$$y=2x-2$$
$$x^2=2x+3$$

$$x^2-2x-3=0$$
$$x=3$$ or $$x=-1$$

If $$x=3$$, then $$y=2x-2=4$$ and $$z=2x=6$$
Therefore $$x^3+y^2+z=27+16+6=49$$

If $$x=-1$$, then $$y=2x-2=-4$$ and $$z=2x=-2$$
Therefore $$x^3+y^2+z=-3+16-2=13$$

If (1) is true, then the only possible x is 3, and the result is divisible by 7.
If (2) is true, then the only possible x is 3, and the result is again divisible by 7.

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Re: A tricky one. [#permalink]  02 Oct 2011, 03:08
Fine, thank you, but I still don't understand how this:
bagrettin wrote:

$$z=2x$$
$$y=2x-2$$
$$x^2=2x+3$$

Turned into this:
bagrettin wrote:
$$x^2-2x-3=0$$
$$x=3$$ or $$x=-1$$

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Re: A tricky one. [#permalink]  02 Oct 2011, 03:22
brilliant, thanks a lot.
_________________

Power is given only to those who dare to lower themselves and pick it up. Only one thing matters, one thing; to be able to dare!

Manager
Joined: 03 Mar 2011
Posts: 94
Location: United States
Schools: Erasmus (S)
GMAT 1: 730 Q51 V37
GPA: 3.9
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Kudos [?]: 100 [0], given: 12

Re: A tricky one. [#permalink]  02 Oct 2011, 03:23
No problem) Thank you!
You are always welcome) Best regards!!!
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Re: A tricky one.   [#permalink] 02 Oct 2011, 03:23
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