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M26-37

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Re: M26-37  [#permalink]

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New post 06 Sep 2018, 08:37
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!

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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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Re: M26-37  [#permalink]

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New post 07 Sep 2018, 07:16
Even though I got the right answer, I hope not to get such qsns on the real GMAT. It's a great question for practice!
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Re: M26-37  [#permalink]

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New post 30 Sep 2018, 10:04
Bunuel wrote:
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: \((x+y)^{xy}=2^1=2=even\).

Two different answers. Not sufficient.


Answer: A


Hi Bunuel

Can you please explain how we arrive at the last fraction here? Thanks.

\(2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}\);
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Re: M26-37  [#permalink]

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New post 30 Sep 2018, 19:45
MsInvBanker wrote:
Bunuel wrote:
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: \((x+y)^{xy}=2^1=2=even\).

Two different answers. Not sufficient.


Answer: A


Hi Bunuel

Can you please explain how we arrive at the last fraction here? Thanks.

\(2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}\);



\(16^{\frac{1}{x+y}}=(2^4)^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}\)
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Re: M26-37  [#permalink]

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New post 30 Sep 2018, 21:06
Thank you Bunuel
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Re M26-37  [#permalink]

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New post 16 Dec 2018, 02:24
I think this is a high-quality question and I agree with explanation.
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Re M26-37 &nbs [#permalink] 16 Dec 2018, 02:24

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