8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.

Hello Bunuel,
Nice explanations. I can't figure out why exponents over exponents act different sometimes. When to multiply and when to add them? Here, why isn't (x^2)^3 = x^8 and (-y^3)^3 = -y^27, please explain? As we know, 3^3^3 = 27.

Could you please explicate?

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.

if y = 0 then statement 1 become 2^x = 2^2x , i.e. x = 2x , i.e. x = 0 not 2 , what am i doing wrong here if you plz??

4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.

I could not understand the highlighted part. I also don't understand when to consider both positive and negative solutions and when to consider only positive solution when dealing with roots. Please explain.

4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.

Hi Bunuel,

Can you kindly help me understand why 4th root of x^2 > 0 and not 4th root of x^2 = 4th root of |x| similar to square of x^2 = square root of |x| as shown in the step 2 of your solution to the question @ https://gmatclub.com/forum/what-is-x-2- ... 99157.html?

Thank you

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.

Hi Bunuel

I don’t understand why in statement 2, x+y cannot equal 1. X and Y are non-negatives, so we shouldn’t exclude the possibility of X and Y being 0 and 1 or 1 and 0 respectively. Please clarify. Thanks



Posted from my mobile device

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.

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

For (1) x=7 is not the only possible value to satisfy \(\sqrt{7*x}=integer\). For example, x can also be 7*4, 7*9, 7*16, ... basically any number of a form 7*integer^2.

11. What is the value of \(xy\)?
(1) \(3^x*5^y=75\)
(2) \(3^{(x-1)(y-2)}=1\)

Notice that we are not told that the \(x\) and \(y\) are integers.

(1) \(3^x*5^y=75\) --> if \(x\) and \(y\) are integers then as \(75=3^1*5^2\) then \(x=1\) and \(y=2\) BUT if they are not, then for any value of \(x\) there will exist some non-integer \(y\) to satisfy given expression and vise-versa (for example if \(y=1\) then \(3^x*5^y=3^x*5=75\) --> \(3^x=25\) --> \(x=some \ irrational \ #\approx{2.9}\)). Not sufficient.

(2) \(5^{(x-1)(y-2)}=1\) --> \((x-1)(y-2)=0\) --> either \(x=1\) and \(y\) is ANY number (including 2) or \(y=2\) and \(x\) is ANY number (including 1). Not sufficient.

(1)+(2) If from (2) \(x=1\) then from (1) \(3^x*5^y=3*5^y=75\) --> \(y=2\) and if from (2) \(y=2\) then from (1) \(3^x*5^y=3^x*25=75\) --> \(x=1\). Thus \(x=1\) and \(y=2\). Sufficient.

Answer: C.

Hi Brunel,
one question.

I've noticed that values x=1 and y=2 together make statement 2 undefined... so, shouldn't we say that answer is E?

Why it is undefined?

If x = 1 and y = 2, then \(5^{(x-1)(y-2)}=5^{0*0}=5^0=1\)

My bad, I thought 0*0 were undefined.
Thanks a lot!

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.