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# M26-37

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Thread Master - Part Time MBA Programs
Joined: 11 Jan 2018
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06 Sep 2018, 09:37
divinka8 wrote:
For statement B, how were we able to conclude that x + y = 2 as opposed to 1? Technically that would make the RHS equal to 25, also an integer? Of course once you start checking the LHS you would realize no combo of x and y would make it equal to 25 but was wondering if there is a clue before that check is conducted. Thank you!

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I was thinking the same. For #2,

(X+Y) = 2, then 5^ 1 = 5
(X+Y) = 1, then 5^ 2 = 25.

But, if you plug in x=1, y=0 or x=0, y=1, the equation of 2^x + 3^y = 25 will not hold. Therefore, X+Y cannot equal to 1.
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Joined: 31 Jul 2017
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07 Sep 2018, 08:16
Even though I got the right answer, I hope not to get such qsns on the real GMAT. It's a great question for practice!
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Joined: 23 May 2018
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30 Sep 2018, 11:04
Bunuel wrote:
Official Solution:

(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$

$$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$;

Equate the powers: $$x-y=\frac{4}{x+y}$$;

$$(x-y)(x+y)=4$$.

Now, since both $$x$$ and $$y$$ are integers (and $$x+y \gt 0$$) then $$x-y=2$$ and $$x+y=2$$ so, $$x=2$$ and $$y=0$$. Therefore, $$(x+y)^{xy}=2^0=1=odd$$, so the answer to the question is No. Sufficient. (Note that $$x-y=1$$ and $$x+y=4$$ is not a valid scenario (solution), since in this case we would have that $$x=2.5$$ and $$y=1.5$$, which is not possible as given that both unknowns must be integers)

(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$. Obviously $$\sqrt[(x+y)]{25}$$ must be an integer (since $$2^x+3^y=integer$$) and as $$x+y=integer$$ then the only solution is $$\sqrt[(x+y)]{25}=\sqrt[2]{25}=5$$ giving $$x+y=2$$. So, $$2^x+3^y=5$$. From that, two scenarios are possible:

A. $$x=2$$ and $$y=0$$ (notice that $$x+y=2$$ holds true): $$2^x+3^y=2^2+3^0=5$$, and in this case: $$(x+y)^{xy}=2^0=1=odd$$;

B. $$x=1$$ and $$y=1$$ (notice that $$x+y=2$$ holds true): $$2^x+3^y=2^1+3^1=5$$, and in this case: $$(x+y)^{xy}=2^1=2=even$$.

Hi Bunuel

Can you please explain how we arrive at the last fraction here? Thanks.

$$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$;
_________________
If you can dream it, you can do it.

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Joined: 02 Sep 2009
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30 Sep 2018, 20:45
MsInvBanker wrote:
Bunuel wrote:
Official Solution:

(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$

$$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$;

Equate the powers: $$x-y=\frac{4}{x+y}$$;

$$(x-y)(x+y)=4$$.

Now, since both $$x$$ and $$y$$ are integers (and $$x+y \gt 0$$) then $$x-y=2$$ and $$x+y=2$$ so, $$x=2$$ and $$y=0$$. Therefore, $$(x+y)^{xy}=2^0=1=odd$$, so the answer to the question is No. Sufficient. (Note that $$x-y=1$$ and $$x+y=4$$ is not a valid scenario (solution), since in this case we would have that $$x=2.5$$ and $$y=1.5$$, which is not possible as given that both unknowns must be integers)

(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$. Obviously $$\sqrt[(x+y)]{25}$$ must be an integer (since $$2^x+3^y=integer$$) and as $$x+y=integer$$ then the only solution is $$\sqrt[(x+y)]{25}=\sqrt[2]{25}=5$$ giving $$x+y=2$$. So, $$2^x+3^y=5$$. From that, two scenarios are possible:

A. $$x=2$$ and $$y=0$$ (notice that $$x+y=2$$ holds true): $$2^x+3^y=2^2+3^0=5$$, and in this case: $$(x+y)^{xy}=2^0=1=odd$$;

B. $$x=1$$ and $$y=1$$ (notice that $$x+y=2$$ holds true): $$2^x+3^y=2^1+3^1=5$$, and in this case: $$(x+y)^{xy}=2^1=2=even$$.

Hi Bunuel

Can you please explain how we arrive at the last fraction here? Thanks.

$$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$;

$$16^{\frac{1}{x+y}}=(2^4)^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$
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30 Sep 2018, 22:06
Thank you Bunuel
_________________
If you can dream it, you can do it.

Practice makes you perfect.

Kudos are appreciated.
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Joined: 16 Apr 2014
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16 Dec 2018, 03:24
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 31 Dec 2017
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01 Aug 2019, 02:44
If xx and yy are non-negative integers and x+y>0 is (x+y)^x*y an even integer?

(1) 2^(x-y) = 2^(4/(x+y))
--->x - y = 4/(x+y)
--->x^2 - y^2 =4 (due to x+y>0, we can eliminate x + y in the denominator)
As 4 is an even integer, x^2 & y^2 must be an even integer. In order to generate an even value with an even exponent, x & y have to be an even integer.--->x & y are even integer such that (x+y)^(x*y) is even--->Sufficient.

(2) 2x+3y = 5^(2/(x+y))
All we know is x+y>0-->2x+3y > 0, min value of 2x+3y is 5 (x=y=1) since x & y are integers, so 5^(2/(x+y)) =5
x & y can be any number greater than 1; however, there's no condition to limit x & y, so we have no idea how x&y can be odd or even.---->Not sufficient.
Re: M26-37   [#permalink] 01 Aug 2019, 02:44

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# M26-37

Moderators: chetan2u, Bunuel