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Math Expert V
Joined: 02 Sep 2009
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16 00:00

Difficulty:   95% (hard)

Question Stats: 46% (02:08) correct 54% (02:14) wrong based on 273 sessions

### HideShow timer Statistics If $$x$$ and $$y$$ are non-negative integers and $$x+y \gt 0$$ is $$(x+y)^{xy}$$ an even integer?

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

(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$

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Math Expert V
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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 $$\sqrt[(x+y)]{25}=\sqrt{25}=5$$ 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: $$(x+y)^{xy}=2^1=2=even$$.

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I think this is a poor-quality question and I agree with explanation. I think it's quite a stretch to expect this question to be solved in 2 minutes.
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I think this is a high-quality question and I agree with explanation. Fantabulous question!
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Is there any simple method to know whether a large number is prime or not ?
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Hi Bunuel...
not able to understand how you got x-y=2 and x+y=2 in statement 1.
Math Expert V
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rohitpoonia wrote:
Hi Bunuel...
not able to understand how you got x-y=2 and x+y=2 in statement 1.

We are given that x and y are non-negative integers so x-y and x+y are also integers. 4 = 2*2 --> x-y=2 and x+y=2.
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GMAT 1: 630 Q47 V29 GMAT 2: 740 Q51 V38 ### Show Tags

Bunuel wrote:
If $$x$$ and $$y$$ are non-negative integers and $$x+y \gt 0$$ is $$(x+y)^{xy}$$ an even integer?

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

(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$

Bunuel

Can you suggest any other approach for the (2)?? Basically to reject that the second equation is sufficient

Thanks
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Luckisnoexcuse
Math Expert V
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mynamegoeson wrote:
Bunuel wrote:
If $$x$$ and $$y$$ are non-negative integers and $$x+y \gt 0$$ is $$(x+y)^{xy}$$ an even integer?

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

(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$

Bunuel

Can you suggest any other approach for the (2)?? Basically to reject that the second equation is sufficient

Thanks

I think identifying that RHS and LSH must be integers, is pretty much it.
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Bunuel wrote:
mynamegoeson wrote:
Bunuel wrote:
If $$x$$ and $$y$$ are non-negative integers and $$x+y \gt 0$$ is $$(x+y)^{xy}$$ an even integer?

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

(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$

Bunuel

Can you suggest any other approach for the (2)?? Basically to reject that the second equation is sufficient

Thanks

I think identifying that RHS and LSH must be integers, is pretty much it.

Thank you,,i think now i understand the difference between Q40s and Q50s _________________
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Luckisnoexcuse
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Hi Bunuel,
Can we conclude from the below mentioned eqn,
(x-y)(x+y)=4
that either (x-y) or (x+y) has to be even? & if (x-y) is even the (x+y) is bound to be even.

Hence (x+y)^xy will be even. But if any of x,y is 0, the result is odd.
Kindly help.
Math Expert V
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ManishKM1 wrote:
Hi Bunuel,
Can we conclude from the below mentioned eqn,
(x-y)(x+y)=4
that either (x-y) or (x+y) has to be even? & if (x-y) is even the (x+y) is bound to be even.

Hence (x+y)^xy will be even. But if any of x,y is 0, the result is odd.
Kindly help.

Yes, since both x and y are integers then for (x - y)(x + y) = even to be true at least one of the multiples must be even. Next, it turns out that if x + y is even, then x - y is even too and vise-versa. After this I lost you. What is your question?
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I mean to say is
Since x+y is even, (x+y)^xy will be even, but if any of x,y is 0, the result is 1. So not sufficient???

Thanks!!
Math Expert V
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ManishKM1 wrote:
I mean to say is
Since x+y is even, (x+y)^xy will be even, but if any of x,y is 0, the result is 1. So not sufficient???

Thanks!!

Yes but we can deduce much more from (x - y)(x + y) = 4, then just that both x + y and x - y are even. The only case for (x - y)(x + y) = 4 to be true is when x = 2 and y = 0 (remember that we are given that both are non-negative integers), thus (x + y)^(xy) = 1 = odd.
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Its clear now. Thanks for your prompt reply. Intern  B
Joined: 24 Jul 2017
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GMAT 1: 710 Q48 V40 WE: Research (Energy and Utilities)

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How does y=0 satisfy the "non-negative integer" part? Zero is neither positive nor negative.
Math Expert V
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JTdaniel wrote:
How does y=0 satisfy the "non-negative integer" part? Zero is neither positive nor negative.

Non-negative integers are integers which are not negative, so, 0, 1, 2, 3, ...
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Hi! Can anyone please explain the second statement graphically?
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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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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. Re: M26-37   [#permalink] 06 Sep 2018, 09:37

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# M26-37

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