M26-35

Question

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}}

Bunuel · Accepted Answer

Official Solution:

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

Bunuel · Answer

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

OutOfTheHills · Answer

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.

intogamer · Answer

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

Bunuel · Answer

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.

intogamer · Answer

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.

Bunuel · Answer

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 ;

akshatrustagi · Answer

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!

Bunuel · Answer

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.

akshatrustagi · Answer

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

Bunuel · Answer

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.

14razy · Answer

does gmat only test real number other than complex number?

Bunuel · Answer

Yes, all numbers on the GMAT are by default real numbers.

surendrasaini1 · Answer

Good question

draftpunk · Answer

Hell of a question,  Bunuel . Thanks for posting.

coreyander · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Shub1994 · Answer

I think this is a  high-quality  question and I don't agree with the explanation. The explanation doesnt hold true in first case if y is positive hence cant be concluded.

Bunuel · Answer

You are wrong! Can you please come up with any positive value of y for which  x^6+y^9=0  is true!

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