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Re M2637
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16 Sep 2014, 01:26
Official Solution: (1) \(2^{xy}=\sqrt[(x+y)]{16}\) \(2^{xy}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}\); Equate the powers: \(xy=\frac{4}{x+y}\); \((xy)(x+y)=4\). Now, since both \(x\) and \(y\) are integers (and \(x+y \gt 0\)) then \(xy=2\) 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 \(xy=1\) 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 M2637
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26 Jun 2016, 06:43
I think this is a poorquality 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, 02:52
I think this is a highquality question and I agree with explanation. Fantabulous question!



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Re: M2637
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31 Oct 2016, 21:39
Is there any simple method to know whether a large number is prime or not ?
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Re: M2637
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15 Dec 2016, 23:33
Hi Bunuel... not able to understand how you got xy=2 and x+y=2 in statement 1.



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16 Dec 2016, 06:12



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Re: M2637
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21 Jun 2017, 05:58
Bunuel wrote: If \(x\) and \(y\) are nonnegative integers and \(x+y \gt 0\) is \((x+y)^{xy}\) an even integer?
(1) \(2^{xy}=\sqrt[(x+y)]{16}\)
(2) \(2^x+3^y=\sqrt[(x+y)]{25}\) BunuelCan you suggest any other approach for the (2)?? Basically to reject that the second equation is sufficient Thanks
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21 Jun 2017, 07:05



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Bunuel wrote: mynamegoeson wrote: Bunuel wrote: If \(x\) and \(y\) are nonnegative integers and \(x+y \gt 0\) is \((x+y)^{xy}\) an even integer?
(1) \(2^{xy}=\sqrt[(x+y)]{16}\)
(2) \(2^x+3^y=\sqrt[(x+y)]{25}\) BunuelCan 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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Re: M2637
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21 Jul 2017, 04:01
Hi Bunuel, Can we conclude from the below mentioned eqn, (xy)(x+y)=4 that either (xy) or (x+y) has to be even? & if (xy) 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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Re: M2637
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21 Jul 2017, 04:10
ManishKM1 wrote: Hi Bunuel, Can we conclude from the below mentioned eqn, (xy)(x+y)=4 that either (xy) or (x+y) has to be even? & if (xy) 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 viseversa. After this I lost you. What is your question?
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Re: M2637
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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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21 Jul 2017, 07:15



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Re: M2637
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21 Jul 2017, 07:19
Its clear now. Thanks for your prompt reply.



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Re: M2637
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09 Sep 2017, 10:46
How does y=0 satisfy the "nonnegative integer" part? Zero is neither positive nor negative.



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09 Sep 2017, 10:48



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23 Apr 2018, 11:39
Hi! Can anyone please explain the second statement graphically?



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Re: M2637
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06 Sep 2018, 08:01
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!
Posted from my mobile device



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Re: M2637
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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!
Posted from my mobile device 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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