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# NEW!!! Tough and tricky exponents and roots questions

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Intern
Joined: 25 Dec 2018
Posts: 13
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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13 Nov 2019, 09:43
Bunuel wrote:
7. If $$x$$ is a positive integer is $$\sqrt{x}$$ an integer?
(1) $$\sqrt{7*x}$$ is an integer
(2) $$\sqrt{9*x}$$ is not an integer

Must know for the GMAT: if $$x$$ is a positive integer then $$\sqrt{x}$$ is either a positive integer itself or an irrational number. (It can not be some reduced fraction eg 7/3 or 1/2)

Also note that the question basically asks whether $$x$$ is a perfect square.

(1) $$\sqrt{7*x}$$ is an integer --> $$x$$ can not be a perfect square because if it is, for example if $$x=n^2$$ for some positive integer $$n$$ then $$\sqrt{7x}=\sqrt{7n^2}=n\sqrt{7}\neq{integer}$$. Sufficient.

(2) $$\sqrt{9*x}$$ is not an integer --> $$\sqrt{9*x}=3*\sqrt{x}\neq{integer}$$ --> $$\sqrt{x}\neq{integer}$$. Sufficient.

Hi Bunuel!
Can't be x equal to 0 in the first statement?
Then sq root of 7x is still an integer and sq root of x is an integer.
With this logic, we have two different answers, hence insufficient.

Correct me please, where is my logic flaw?
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Joined: 02 Sep 2009
Posts: 59622
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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13 Nov 2019, 09:49
LidiiaShchichko wrote:
Bunuel wrote:
7. If $$x$$ is a positive integer is $$\sqrt{x}$$ an integer?
(1) $$\sqrt{7*x}$$ is an integer
(2) $$\sqrt{9*x}$$ is not an integer

Must know for the GMAT: if $$x$$ is a positive integer then $$\sqrt{x}$$ is either a positive integer itself or an irrational number. (It can not be some reduced fraction eg 7/3 or 1/2)

Also note that the question basically asks whether $$x$$ is a perfect square.

(1) $$\sqrt{7*x}$$ is an integer --> $$x$$ can not be a perfect square because if it is, for example if $$x=n^2$$ for some positive integer $$n$$ then $$\sqrt{7x}=\sqrt{7n^2}=n\sqrt{7}\neq{integer}$$. Sufficient.

(2) $$\sqrt{9*x}$$ is not an integer --> $$\sqrt{9*x}=3*\sqrt{x}\neq{integer}$$ --> $$\sqrt{x}\neq{integer}$$. Sufficient.

Hi Bunuel!
Can't be x equal to 0 in the first statement?
Then sq root of 7x is still an integer and sq root of x is an integer.
With this logic, we have two different answers, hence insufficient.

Correct me please, where is my logic flaw?

We are told that x is a positive integer (check the highlighted part) and 0 is neither positive nor negative integer.
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Joined: 27 Oct 2019
Posts: 54
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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24 Nov 2019, 23:07
Bunuel wrote:
10. If $$x$$ and $$y$$ are non-negative integers and $$x+y>0$$ is $$(x+y)^{xy}$$ an even integer?
(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$
(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$

(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$ --> $$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$ --> equate the powers: $$x-y=\frac{4}{x+y}$$ --> $$(x-y)(x+y)=4$$.

Since both $$x$$ and $$y$$ are integers (and $$x+y>0$$) then $$x-y=2$$ and $$x+y=2$$ --> $$x=2$$ and $$y=0$$ --> $$(x+y)^{xy}=2^0=1=odd$$, so the answer to the question is No. Sufficient. (Note that $$x-y=1$$ and $$x+y=4$$ --> $$x=2.5$$ and $$y=1.5$$ is not a valid scenario (solution) as 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 $$\sqrt[(x+y)]{25}=\sqrt[2]{25}=5$$ --> $$x+y=2$$. So, $$2^x+3^y=5$$ --> 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: $$(x+y)^{xy}=2^1=2=even$$.

Since x+y>0 then how do we get x+y=2 and x-y=2?? Why cant x+y=4??

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Joined: 02 Sep 2009
Posts: 59622
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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24 Nov 2019, 23:54
ssshyam1995 wrote:
Bunuel wrote:
10. If $$x$$ and $$y$$ are non-negative integers and $$x+y>0$$ is $$(x+y)^{xy}$$ an even integer?
(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$
(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$

(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$ --> $$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$ --> equate the powers: $$x-y=\frac{4}{x+y}$$ --> $$(x-y)(x+y)=4$$.

Since both $$x$$ and $$y$$ are integers (and $$x+y>0$$) then $$x-y=2$$ and $$x+y=2$$ --> $$x=2$$ and $$y=0$$ --> $$(x+y)^{xy}=2^0=1=odd$$, so the answer to the question is No. Sufficient. (Note that $$x-y=1$$ and $$x+y=4$$ --> $$x=2.5$$ and $$y=1.5$$ is not a valid scenario (solution) as 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 $$\sqrt[(x+y)]{25}=\sqrt[2]{25}=5$$ --> $$x+y=2$$. So, $$2^x+3^y=5$$ --> 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: $$(x+y)^{xy}=2^1=2=even$$.

Since x+y>0 then how do we get x+y=2 and x-y=2?? Why cant x+y=4??

Posted from my mobile device

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Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 24 Nov 2019, 23:54

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