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M26-37

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Math Expert
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16 Sep 2014, 00:26
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Question Stats:

65% (02:34) correct 35% (02:05) wrong based on 54 sessions

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

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Math Expert
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16 Sep 2014, 00:26
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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 $$\sqrt[(x+y)]{25}=\sqrt[2]{25}=5$$ 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: $$(x+y)^{xy}=2^1=2=even$$.

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26 Jun 2016, 05:43
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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07 Aug 2016, 01:52
I think this is a high-quality question and I agree with explanation. Fantabulous question!

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31 Oct 2016, 20:39
Is there any simple method to know whether a large number is prime or not ?
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15 Dec 2016, 22:33
Hi Bunuel...
not able to understand how you got x-y=2 and x+y=2 in statement 1.

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16 Dec 2016, 05:12
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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21 Jun 2017, 04:58
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

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21 Jun 2017, 06:05
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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21 Jun 2017, 06:25
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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21 Jul 2017, 03:01
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.

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21 Jul 2017, 03:10
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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21 Jul 2017, 06:09
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!!

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21 Jul 2017, 06:15
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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21 Jul 2017, 06:19

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09 Sep 2017, 09:46
How does y=0 satisfy the "non-negative integer" part? Zero is neither positive nor negative.

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09 Sep 2017, 09:48
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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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