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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x#y=yx^y for all values x and y, then 2#(0(-2#2)=?

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Intern  Joined: 20 May 2014
Posts: 12
If x#y=yx^y for all values x and y, then 2#(0(-2#2)=?  [#permalink]

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Difficulty:   5% (low)

Question Stats: 93% (01:01) correct 7% (01:15) wrong based on 45 sessions

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If x#y = yx^y for all values x and y, then 2#(0(-2#2)=

A. -64
B. 0
C. 32
D. 64
E. 128

Hey guys i am having a hard time understanding the solution to this question. The correct answer is B. This is the solution given by veritas prep:

This function question may look laborious, but can actually be solved quickly noting that it will require the multiplication of each of the values, and that 0 is one of the values. Accordingly, because 0 multiplied by anything is 0, the answer can only be 0. Taken step-by-step, the result of the innermost parentheses will be 2(-2)2, or 8. That term then becomes the y term of the next sequence, for which x is 0. That term will be 8(0)8, or 0. Then, that becomes the y term of the next set, which would be 0(2)0. The answer, still, is 0, and B is the correct answer.

What i am confused about is how the innermost parenthesis equal 8?

Originally posted by pmklings on 02 Apr 2015, 14:34.
Last edited by Bunuel on 03 Apr 2015, 03:24, edited 1 time in total.
Edited the question and added the OA.
Magoosh GMAT Instructor G
Joined: 28 Dec 2011
Posts: 4476
Re: If x#y=yx^y for all values x and y, then 2#(0(-2#2)=?  [#permalink]

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pmklings wrote:
If x#y=yx^y for all values x and y, then 2#(0(-2#2)=

a. -64
b.0
c. 32
d.64
e. 128

Hey guys i am having a hard time understanding the solution to this question. The correct answer is B. This is the solution given by veritas prep:

This function question may look laborious, but can actually be solved quickly noting that it will require the multiplication of each of the values, and that 0 is one of the values. Accordingly, because 0 multiplied by anything is 0, the answer can only be 0. Taken step-by-step, the result of the innermost parentheses will be 2(-2)2, or 8. That term then becomes the y term of the next sequence, for which x is 0. That term will be 8(0)8, or 0. Then, that becomes the y term of the next set, which would be 0(2)0. The answer, still, is 0, and B is the correct answer.

What i am confused about is how the innermost parenthesis equal 8?

Dear pmklings
I'm happy to help. My friend, I believe you are confused by Order of Operations. This idea, something known as PEMDAS, specifies some quite precise ways to view calculations, and sometimes these are anti-intuitive.

Look at the algebraic expression:
yx^y
In that, what do we do first? Do we multiply y times x, and then raise that product to the power of y? Or do we raise just x to the power of y, and then multiply that power by the first y? Well, the standard Order of Operation procedure tells us that exponents take precedent over multiplication, so we have to raise just the x to the power of y, computing that, before we can even consider the first factor of y.
Thus,
(2)(-2)^2 = 2*[(-2)^2] = 2*4 = 8

You may also find this blog article helpful:
http://magoosh.com/gmat/2013/gmat-quant ... g-symbols/

Mike _________________
Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 9417
Location: United States (CA)
Re: If x#y=yx^y for all values x and y, then 2#(0(-2#2)=?  [#permalink]

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1
pmklings wrote:
If x#y = yx^y for all values x and y, then 2#(0(-2#2)=

A. -64
B. 0
C. 32
D. 64
E. 128

Hey guys i am having a hard time understanding the solution to this question. The correct answer is B. This is the solution given by veritas prep:

This function question may look laborious, but can actually be solved quickly noting that it will require the multiplication of each of the values, and that 0 is one of the values. Accordingly, because 0 multiplied by anything is 0, the answer can only be 0. Taken step-by-step, the result of the innermost parentheses will be 2(-2)2, or 8. That term then becomes the y term of the next sequence, for which x is 0. That term will be 8(0)8, or 0. Then, that becomes the y term of the next set, which would be 0(2)0. The answer, still, is 0, and B is the correct answer.

What i am confused about is how the innermost parenthesis equal 8?

First, we have; (-2#2) = 2(-2)^2 = 2(4) = 8

Next, we have: 0(8) = 0

Lastly, we have 2#0 = 0(2)^0 = 0

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If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: If x#y=yx^y for all values x and y, then 2#(0(-2#2)=?   [#permalink] 23 Jan 2020, 10:39
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