THEORY TO TACKLE THE PROBLEMS ABOVE:
For more on number theory check the Number Theory Chapter of Math Book: math-number-theory-88376.html

EXPONENTS

Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number a multiplied n times can be written as a^n, where a represents the base, the number that is multiplied by itself n times and n represents the exponent. The exponent indicates how many times to multiple the base, a, by itself.

Exponents one and zero:
a^0=1 Any nonzero number to the power of 0 is 1.
For example: 5^0=1 and (-3)^0=1
• Note: the case of 0^0 is not tested on the GMAT.

a^1=a Any number to the power 1 is itself.

Powers of zero:
If the exponent is positive, the power of zero is zero: 0^n = 0, where n > 0.

If the exponent is negative, the power of zero (0^n, where n < 0) is undefined, because division by zero is implied.

Powers of one:
1^n=1 The integer powers of one are one.

Negative powers:
a^{-n}=\frac{1}{a^n}

Powers of minus one:
If n is an even integer, then (-1)^n=1.

If n is an odd integer, then (-1)^n =-1.

Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
a^n*b^n=(ab)^n

\frac{a^n}{b^n}=(\frac{a}{b})^n

(a^m)^n=a^{mn}

a^m^n=a^{(m^n)} and not (a^m)^n (if exponentiation is indicated by stacked symbols, the rule is to work from the top down)

Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
a^n*a^m=a^{n+m}

\frac{a^n}{a^m}=a^{n-m}

Fraction as power:
a^{\frac{1}{n}}=\sqrt[n]{a}

a^{\frac{m}{n}}=\sqrt[n]{a^m}


ROOTS

Roots (or radicals) are the "opposite" operation of applying exponents. For instance x^2=16 and square root of 16=4.

General rules:
• \sqrt{x}\sqrt{y}=\sqrt{xy} and \frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}.

• (\sqrt{x})^n=\sqrt{x^n}

• x^{\frac{1}{n}}=\sqrt[n]{x}

• x^{\frac{n}{m}}=\sqrt[m]{x^n}

• {\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}

• \sqrt{x^2}=|x|, when x\leq{0}, then \sqrt{x^2}=-x and when x\geq{0}, then \sqrt{x^2}=x

• When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root.

That is, \sqrt{25}=5, NOT +5 or -5. In contrast, the equation x^2=25 has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

• Odd roots will have the same sign as the base of the root. For example, \sqrt[3]{125} =5 and \sqrt[3]{-64} =-4.
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2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer?
(1) y is an even perfect square and z is an odd perfect cube.
(2) \sqrt{z} is not an integer.


2700 = 2*2*3*3*3*5*5
(1) y is an even perfect square and z is an odd perfect cube.

y = 100 or 4
z = 27
x = 1 or 25

Yes, √x is an integer.
Sufficient

(2) \sqrt{z} is not an integer.

Insuff


Option

There's another possibility here. It is possible that z = 1. In that case, x could be, say, 3^3, and then its square root would not be an integer.
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Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

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Oops.. Thanks for pointing out Ian.

(1) y must be even for the statement to be true (only then will the result be positive). 1/81= 3^(-4) ---> the pairs to make 4 in order for x and y to be integers is 1*4 or 2*2. Therefore the only way to construct it is if y=-2 or y=-4.

if x=-3 and y=-4 --->x^y=\frac{1}{81} ---> x*y=12
if x=-9 and y=-2 --->x^y=\frac{1}{81} ---->x*y=18

Insufficient

(2) x must be odd for the statement to be True (only then will the result be negative).-1/64=-2^6 ---> the pairs to make 6 in order for x and y to be integers are 1*6 or 2*3. Therefore the only way to construct it is if x=-1 or x=-3.

if y=-4 and x=-3 ---> y^x=-\frac{1}{64} --->x*y=12
if y=-64 and x=-1 ---> y^x=-\frac{1}{64} ---> x*y=64

Insufficient.

(1)+(2)---> x*y=12 ---> C

1-D
2-D
3-E
4-D
5-B
6-D
7-D
8-E
9-B
10-D
11-A

please post the OA.

Since in both of the posts above, the answer to this question was given incorrectly, I thought I'd post a quick solution. Using the difference of squares,

x - y = (\sqrt{x})^2 - (\sqrt{y})^2 = (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})

So we can simplify the question by using this factorization in the denominator:

\frac{\sqrt{2x} + \sqrt{2y}}{x-y} = \frac{\sqrt{2} (\sqrt{x} + \sqrt{y})}{(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) } = \frac{\sqrt{2}}{\sqrt{x} - \sqrt{y}}

So if we can find the value of \sqrt{x} - \sqrt{y}, we can answer the question.

Now from Statement 1, we have

\begin{align}
x + y &= 4 + 2\sqrt{xy} \\
x - 2\sqrt{xy} + y &= 4 \\
(\sqrt{x} - \sqrt{y})^2 &= 4 \\
\sqrt{x} - \sqrt{y} &= 2
\end{align}
(here we know the root is 2, and not -2, since x > y). So Statement 1 is sufficient.
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Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

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2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer?
(1) y is an even perfect square and z is an odd perfect cube.
(2) \sqrt{z} is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> xyz=2^2*3^3*5^2.

(1) y is an even perfect square and z is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and z=3^3=odd \ perfect \ square then x must be a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. But if z=1^3=odd \ perfect \ cube then x could be 3^3 which makes \sqrt{x} not an integer. Not sufficient.

(2) \sqrt{z} is not an integer. Clearly insufficient.

(1)+(2) As from (1) \sqrt{z}\neq{integer} then z\neq{1}, therefore it must be 3^3 (from 1) --> x is a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. Sufficient.

Answer: C.
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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.
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6. If x>0 then what is the value of y^x?
(1) \frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}
(2) x\neq{1} and x^y=1

(1) \frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy} --> 4^{(x+y)^2-(x-y)^2}=128^{xy} --> applying a^2-b^2=(a-b)(a+b) we'll get: 4^{4xy}=128^{xy} --> 2^{8xy}=2^{7xy} --> 8xy=7xy --> xy=0, since given that x>0 then y=0 hence y^x=0^x=0. Sufficient.

(2) x\neq{1} and x^y=1 --> since x>0 and x\neq{1} then the only case x^y=1 to hold true is when y=0 --> y^x=0^x=0. Sufficient.

Answer: D.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


What are GMAT Club Tests?
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