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Math Expert V
Joined: 02 Sep 2009
Posts: 57297
M26-35  [#permalink]

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5
19 00:00

Difficulty:   95% (hard)

Question Stats: 25% (02:03) correct 75% (02:37) wrong based on 163 sessions

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What is the value of $$x^2+y^3$$?

(1) $$x^6+y^9=0$$

(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 57297
Re M26-35  [#permalink]

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3
Official Solution:

(1) $$x^6+y^9=0$$.

$$(x^2)^3=(-y^3)^3;$$

$$x^2=-y^3$$;

$$x^2+y^3=0$$. Sufficient.

(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$.

$$3^{3x^2}=\frac{3}{3^{3y^2}*3}$$;

$$3^{3x^2}*3^{3y^2}=1$$;

$$3^{3x^2+3y^2}=1$$;

Since the power of 3 must be zero in order for the above equation to hold true then $$3x^2+3y^2=0$$, so $$x^2+y^2=0$$. The sum of two non-negative values is zero means that both $$x$$ and $$y$$ must be zero therefore $$x=y=0$$ and $$x^2+y^3=0$$. Sufficient.

Answer: D
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Re: M26-35  [#permalink]

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Hi Bunuel,

I keep making mistakes on exponent manipulation. It just doesn't immediately jump out to me that x^6 + y^9 = 0 can be manipulated into something that solves the problem. Do you happen to have a list of questions that test similar concepts that I can practice with? I tried using the "find similar topics" function but many of the problems don't involve the skills I'm having trouble with.
Math Expert V
Joined: 02 Sep 2009
Posts: 57297
Re: M26-35  [#permalink]

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1
1
OutOfTheHills wrote:
Hi Bunuel,

I keep making mistakes on exponent manipulation. It just doesn't immediately jump out to me that x^6 + y^9 = 0 can be manipulated into something that solves the problem. Do you happen to have a list of questions that test similar concepts that I can practice with? I tried using the "find similar topics" function but many of the problems don't involve the skills I'm having trouble with.

Theory on Exponents: math-number-theory-88376.html

All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39
All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60

Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

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Re: M26-35  [#permalink]

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For statement two, can you explain how you got from step 1 to step 2 and to step 3?
Math Expert V
Joined: 02 Sep 2009
Posts: 57297
Re: M26-35  [#permalink]

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1
intogamer wrote:
For statement two, can you explain how you got from step 1 to step 2 and to step 3?

$$x^6+y^9=0$$;

Re-arrange: $$x^6=-y^9$$;

$$(x^2)^3=(-y^3)^3;$$ (since (x^2)^3=x^6 and (-y^3)^3=-y^9);

Take the third root: $$x^2=-y^3$$;

Re-arrange again: $$x^2+y^3=0$$.

Hope it's clear.
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M26-35  [#permalink]

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Bunuel wrote:
intogamer wrote:
For statement two, can you explain how you got from step 1 to step 2 and to step 3?

$$x^6+y^9=0$$;

Re-arrange: $$x^6=-y^9$$;

$$(x^2)^3=(-y^3)^3;$$ (since (x^2)^3=x^6 and (-y^3)^3=-y^9);

Take the third root: $$x^2=-y^3$$;

Re-arrange again: $$x^2+y^3=0$$.

Hope it's clear.

Sorry if I was unclear Bunuel but I meant

$$27^(x^2) = 3/(3^(((3y)^2)+1))$$

but thank you for the breakout of the first.
Math Expert V
Joined: 02 Sep 2009
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Re: M26-35  [#permalink]

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intogamer wrote:
Bunuel wrote:
intogamer wrote:
For statement two, can you explain how you got from step 1 to step 2 and to step 3?

$$x^6+y^9=0$$;

Re-arrange: $$x^6=-y^9$$;

$$(x^2)^3=(-y^3)^3;$$ (since (x^2)^3=x^6 and (-y^3)^3=-y^9);

Take the third root: $$x^2=-y^3$$;

Re-arrange again: $$x^2+y^3=0$$.

Hope it's clear.

Sorry if I was unclear Bunuel but I meant

$$27^(x^2) = 3/(3^(((3y)^2)+1))$$

but thank you for the breakout of the first.

Notice that $$3^{(3y^2+1)}=3^{(3y^2)}*3^1$$

$$27^{x^2}=\frac{3}{3^{3y^2+1}}$$.

$$3^{3x^2}=\frac{3}{3^{3y^2}*3}$$;

Reduce the right hand side by 3: $$3^{3x^2}=\frac{1}{3^{3y^2}}$$;

Cross-multiply: $$3^{3x^2}*3^{3y^2}=1$$;

$$3^{3x^2+3y^2}=1$$;
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Intern  Joined: 21 Jun 2014
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Re: M26-35  [#permalink]

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Hi Bunuel,

A small doubt :

I had marked B because I felt A is insufficient :

x6 + y9 = 0

which means (x2)3 + (y3)3 = 0 (like you did)

Now using a3 + b3 = (a + b)*(a2 - ab + b2)

So either a + b = 0, or a2 - ab + b2 = 0, i.e. either x2 + y3 = 0 or x4 - x2y3 + y6 = 0

Therefore A is insufficient

Can you please tell me where I'm going wrong regarding statement 1?

Thanks!
Math Expert V
Joined: 02 Sep 2009
Posts: 57297
Re: M26-35  [#permalink]

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akshatrustagi wrote:
Hi Bunuel,

A small doubt :

I had marked B because I felt A is insufficient :

x6 + y9 = 0

which means (x2)3 + (y3)3 = 0 (like you did)

Now using a3 + b3 = (a + b)*(a2 - ab + b2)

So either a + b = 0, or a2 - ab + b2 = 0, i.e. either x2 + y3 = 0 or x4 - x2y3 + y6 = 0

Therefore A is insufficient

Can you please tell me where I'm going wrong regarding statement 1?

Thanks!

The point is that only real roots of x^4 - x^2y^3 + y^6 = 0 is x = y = 0, for which x^2 + y^3 = 0.
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Re: M26-35  [#permalink]

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1
Hi Bunuel,

Thanks! That solves my doubt

However, is there any shortcut to directly come to that conclusion? Or did you actually solve the equation to get the roots x = 0 and y = 0?

I mean is there any shortcut to ignore the x^4 - (x^2)(y^3) + y^6, and directly come to the conclusion that x^6 + y^9 = 0?

Thanks in advance
Math Expert V
Joined: 02 Sep 2009
Posts: 57297
Re: M26-35  [#permalink]

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akshatrustagi wrote:
Hi Bunuel,

Thanks! That solves my doubt

However, is there any shortcut to directly come to that conclusion? Or did you actually solve the equation to get the roots x = 0 and y = 0?

I mean is there any shortcut to ignore the x^4 - (x^2)(y^3) + y^6, and directly come to the conclusion that x^6 + y^9 = 0?

Thanks in advance

Direct way is given in the solution:

(1) $$x^6+y^9=0$$.

$$(x^2)^3=(-y^3)^3;$$

$$x^2=-y^3$$;

$$x^2+y^3=0$$. Sufficient.

You took another path which gave you more complex equation.
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Re: M26-35  [#permalink]

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Yes I understand

Can you tell me in which cases do we avoid that another path?
Intern  Joined: 24 Oct 2016
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Re M26-35  [#permalink]

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does gmat only test real number other than complex number?
Math Expert V
Joined: 02 Sep 2009
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Re: M26-35  [#permalink]

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14razy wrote:
does gmat only test real number other than complex number?

Yes, all numbers on the GMAT are by default real numbers.
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Re: M26-35  [#permalink]

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Good question
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M26-35  [#permalink]

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Bunuel wrote:
intogamer wrote:
For statement two, can you explain how you got from step 1 to step 2 and to step 3?

$$x^6+y^9=0$$;

Re-arrange: $$x^6=-y^9$$;

$$(x^2)^3=(-y^3)^3;$$ (since (x^2)^3=x^6 and (-y^3)^3=-y^9);

Take the third root: $$x^2=-y^3$$;

Re-arrange again: $$x^2+y^3=0$$.

Hope it's clear.

$$a^3+b^3=(a+b)(a^2+b^2-ab)$$
If $$a=x^2$$ and $$b=y^3$$
then
$$x^6+y^9=0$$;
$$(x^2+y^3)(x^4+y^6-x^2y^3)=0$$
So either $$x^2+y^3=0$$ or $$x^4+y^6-x^2y^3=0$$

You are taking only 1 possibility when $$x^2+y^3=0$$
Please explain
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Math Expert V
Joined: 02 Sep 2009
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Re: M26-35  [#permalink]

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Kinshook wrote:
Bunuel wrote:
intogamer wrote:
For statement two, can you explain how you got from step 1 to step 2 and to step 3?

$$x^6+y^9=0$$;

Re-arrange: $$x^6=-y^9$$;

$$(x^2)^3=(-y^3)^3;$$ (since (x^2)^3=x^6 and (-y^3)^3=-y^9);

Take the third root: $$x^2=-y^3$$;

Re-arrange again: $$x^2+y^3=0$$.

Hope it's clear.

$$a^3+b^3=(a+b)(a^2+b^2-ab)$$
If $$a=x^2$$ and $$b=y^3$$
then
$$x^6+y^9=0$$;
$$(x^2+y^3)(x^4+y^6-x^2y^3)=0$$
So either $$x^2+y^3=0$$ or $$x^4+y^6-x^2y^3=0$$

You are taking only 1 possibility when $$x^2+y^3=0$$
Please explain

Please check here: https://gmatclub.com/forum/m26-184474.html#p1712598
_________________ Re: M26-35   [#permalink] 06 Aug 2019, 15:37
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