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M26-37

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M26-37 [#permalink]

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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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(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: \((x+y)^{xy}=2^1=2=even\).

Two different answers. Not sufficient.


Answer: A
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Re M26-37 [#permalink]

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New post 26 Jun 2016, 06:43
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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Re M26-37 [#permalink]

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New post 07 Aug 2016, 02:52
I think this is a high-quality question and I agree with explanation. Fantabulous question!

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Re: M26-37 [#permalink]

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New post 31 Oct 2016, 21:39
Is there any simple method to know whether a large number is prime or not ?
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New post 15 Dec 2016, 23:33
Hi Bunuel...
not able to understand how you got x-y=2 and x+y=2 in statement 1.

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Re: M26-37 [#permalink]

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New post 21 Jun 2017, 05:58
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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New post 21 Jun 2017, 07:05
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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M26-37 [#permalink]

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New post 21 Jun 2017, 07:25
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 :lol:
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Re: M26-37 [#permalink]

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New post 21 Jul 2017, 04:01
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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New post 21 Jul 2017, 04:10
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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Re: M26-37 [#permalink]

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New post 21 Jul 2017, 07:09
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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New post 21 Jul 2017, 07:15
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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Re: M26-37 [#permalink]

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New post 21 Jul 2017, 07:19
Its clear now. Thanks for your prompt reply. :)

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Re: M26-37 [#permalink]

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New post 09 Sep 2017, 10:46
How does y=0 satisfy the "non-negative integer" part? Zero is neither positive nor negative.

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Re: M26-37   [#permalink] 09 Sep 2017, 10:48
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