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NEW!!! Tough and tricky exponents and roots questions

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

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New post 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.

Answer: D.


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?
Thank you in advance.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 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.

Answer: D.


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?
Thank you in advance.


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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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 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: \((x+y)^{xy}=2^1=2=even\).

Two different answers. Not sufficient.

Answer: A.

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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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 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: \((x+y)^{xy}=2^1=2=even\).

Two different answers. Not sufficient.

Answer: A.

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


Please read carefully highlighted part.
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Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 24 Nov 2019, 23:54

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