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M26-35

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M26-35  [#permalink]

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New post 16 Sep 2014, 01:26
5
19
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

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

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Re M26-35  [#permalink]

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New post 16 Sep 2014, 01:26
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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New post 07 Nov 2014, 20:09
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.
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New post 09 Nov 2014, 06:17
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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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New post 15 Feb 2016, 21:02
For statement two, can you explain how you got from step 1 to step 2 and to step 3?
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New post 16 Feb 2016, 00:38
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M26-35  [#permalink]

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New post 16 Feb 2016, 08:13
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.
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New post 16 Feb 2016, 08:30
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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Re: M26-35  [#permalink]

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New post 21 Jul 2016, 03:39
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!
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New post 21 Jul 2016, 03:51
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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New post 21 Jul 2016, 06:14
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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
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New post 21 Jul 2016, 08:18
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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New post 22 Jul 2016, 01:03
Yes I understand

Can you tell me in which cases do we avoid that another path?
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New post 27 Nov 2016, 01:55
does gmat only test real number other than complex number?
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New post 12 May 2019, 02:39
Good question
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M26-35  [#permalink]

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New post 06 Aug 2019, 09:08
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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New post 06 Aug 2019, 15:37
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
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Re: M26-35   [#permalink] 06 Aug 2019, 15:37
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