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

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Kudos [?]: 17603 [18] , given: 2225

NEW!!! Tough and tricky exponents and roots questions [#permalink]  12 Jan 2012, 02:50
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Expert's post
00:00

Difficulty:

5% (low)

Question Stats:

90% (02:09) correct 10% (00:04) wrong based on 10 sessions
Exponents and roots problems are very common on the GMAT. So, it's extremely important to know how to manipulate them, how to factor out, take roots, multiply, divide, etc. Below are 11 problems to test your skills. Please post your thought process/solutions along with the answers.

I'll post OA's with detailed solutions tomorrow. Good luck.

1. If 357^x*117^y=a, where x and y are positive integers, what is the units digit of a?
(1) 100<y^2<x^2<169
(2) x^2-y^2=23

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029239

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.

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029240

3. If x>y>0 then what is the value of \frac{\sqrt{2x}+\sqrt{2y}}{x-y}?
(1) x+y=4+2\sqrt{xy}
(2) x-y=9

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029241

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}}

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029242

5. If x and y are negative integers, then what is the value of xy?
(1) x^y=\frac{1}{81}
(2) y^x=-\frac{1}{64}

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029243

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

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029244

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

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029245

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}}

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029246

9. If x, y and z are non-zero numbers, what is the value of \frac{x^3+y^3+z^3}{xyz}?
(1) xyz=-6
(2) x+y+z=0

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029247

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}

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029248

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

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029249
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 14:33
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Expert's post
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.

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 14:52
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Expert's post
9. If x, y and z are non-zero numbers, what is the value of \frac{x^3+y^3+z^3}{xyz}?
(1) xyz=-6
(2) x+y+z=0

(1) xyz=-6 --> infinitely many combinations of x, y and z are possible which will give different values of the expression in the stem: try x=y=1 and y=-6 or x=1, y=2, z=-3. Not sufficient.

(2) x+y+z=0 --> x=-(y+z) --> substitute this value of x into the expression in the stem --> \frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}, as x=-(y+z) then: \frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3. Sufficient.

Must know for the GMAT: (x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3 and (x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3.

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 14:54
7
KUDOS
Expert's post
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.

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 13:18
3
KUDOS
Bunuel wrote:

3. If x>y>0 then what is the value of \frac{\sqrt{2x}+\sqrt{2y}}{x-y}?
(1) x+y=4+2\sqrt{xy}
(2) x-y=9

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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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 14:31
3
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Expert's post
1. If 357^x*117^y=a, where x and y are positive integers, what is the units digit of a?
(1) 100<y^2<x^2<169
(2) x^2-y^2=23

(1) 100<y^2<x^2<169 --> since both x and y are positive integers then x^2 and y^2 are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> y=11 and x=12. Sufficient.(As cyclicity of units digit of 7 in integer power is 4, therefore the units digit of 7^{23} is the same as the units digit of 7^3, so 3).

(2) x^2-y^2=23 --> (x-y)(x+y)=23=prime --> since both x and y are positive integers then: x-y=1 and x+y=23 --> y=11 and x=12. Sufficient.

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 14:46
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Expert's post
7. If x is a positive integer is \sqrt{x} an integer?
(1) \sqrt{7*y} 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.

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  12 Jan 2012, 13:22
2
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Bunuel wrote:

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}}

6. If x\neq{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

7. If x is a positive integer is \sqrt{x} an integer?
(1) \sqrt{7*y} is an integer
(2) \sqrt{9*x} is not an integer

9. If \frac{x}{y^{-3}}+\frac{y}{x^{-3}}=\frac{1}{(\sqrt{2}xy)^{-2}}, then what is the value of xy?
(1) x^2=y^2
(2) x^3>y^3

At a quick glance, a few comments - hopefully I haven't made any errors:

The meaning of Q4 would be more clear if the terms were enclosed in brackets (right now the asterisk symbol appears to be part of an exponent): 4. \text{If } xyz\neq{0} \text{ is } \left( x^{-4} \right) \left( \sqrt[3]{y} \right) \left( z^{-2} \right) <0 \text{ ?}

Q6 is a bit problematic. If y turns out to be 0, we need to know that x is positive for the expression in question to be defined. I don't know if, in Statement 2, you had in mind the 'trap' that x might be -1, but for Statement 1 to work, that possibility needs to be ruled out in advance.

In Q7, I imagine you meant to write 'x' instead of 'y' in Statement 1, since there's no other mention of y anywhere.

There seems to be something wrong with Q9. If both statements are true, x and y need to have opposite signs, but that would make the left side of the equation in the stem negative and the right side positive, which is clearly impossible. The question also needs to rule out the possibility that x or y are equal to 0, since that would make terms in the equation in the stem undefined (you can't raise zero to a negative power).
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  13 Jan 2012, 17:29
2
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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

(1) x must have a 7 raised at an odd power as a factor. It could also have any number raised to an even power as a factor ---> therefore \sqrt{x} will never be an integer because of the 7. Sufficient

(2) \sqrt{9*x}=3\sqrt{x}---> since 3 is an integer, \sqrt{x} is an not an integer.

Sufficient

Therefore, D
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 14:34
2
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Expert's post
3. If x>y>0 then what is the value of \frac{\sqrt{2x}+\sqrt{2y}}{x-y}?
(1) x+y=4+2\sqrt{xy}
(2) x-y=9

\frac{\sqrt{2x}+\sqrt{2y}}{x-y} --> factor out \sqrt{2} from the nominator and apply a^2-b^2=(a-b)(a+b) to the expression in the denominator: \frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}. So we should find the value of \sqrt{x}-\sqrt{y}.

(1) x+y=4+2\sqrt{xy} --> x-2\sqrt{xy}+y=4 --> (\sqrt{x}-\sqrt{y})^2=4 --> \sqrt{x}-\sqrt{y}=2 (note that since x-y>0 then the second solution \sqrt{x}-\sqrt{y}=-2 is not valid). Sufficient.

(2) x-y=9. Not sufficient.

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 14:38
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Expert's post
5. If x and y are negative integers, then what is the value of xy?
(1) x^y=\frac{1}{81}
(2) y^x=-\frac{1}{64}

(1) x^y=\frac{1}{81} --> as both x and y are negative integers then x^y=\frac{1}{81}=(-9)^{-2}=(-3)^{-4} --> xy=18 or xy=12. Note that as negative integer (x) in negative integer power (y) gives positive number (1/81) then the power must be negative even number. Not sufficient.

(2) y^x=-\frac{1}{64} --> as the result is negative then x must be negative odd number --> y^x=-\frac{1}{64}=(-4)^{-3}=(-64)^{-1} --> xy=12 or xy=64. Not sufficient.

(1)+(2) Only one pair of negative integers x and y satisfies both statements x=-3 and y=-4 --> xy=12. Sufficient.

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 14:49
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Expert's post
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.

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 14:56
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Expert's post
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.

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 15:04
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Expert's post
Just posted solutions. Kudos points given to everyone with correct solutions. Let me know if I missed someone.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  12 Jan 2012, 13:26
1
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rijul007 wrote:
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

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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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  13 Jan 2012, 16:37
1
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Bunuel wrote:
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}}

if we can find the sign of y we can answer the question.

(1) \sqrt[5]{y}>\sqrt[4]{x^2}>0 since it is[\sqrt[4]{x^2}]=[\sqrt[]{x}]

Sufficient

(2) y^3>\frac{1}{z{^4}}>0 since z^4>0

Sufficient

Therefore, D
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  13 Jan 2012, 17:14
1
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Bunuel wrote:
5. If x and y are negative integers, then what is the value of xy?
(1) x^y=\frac{1}{81}
(2) y^x=-\frac{1}{64}

(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
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  14 Jan 2012, 14:37
1
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Expert's post
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.

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  05 Sep 2013, 11:35
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Expert's post
Acropora wrote:
Bunuel wrote:
1. If 357^x*117^y=a, where x and y are positive integers, what is the units digit of a?
(1) 100<y^2<x^2<169
(2) x^2-y^2=23

(1) 100<y^2<x^2<169 --> since both x and y are positive integers then x^2 and y^2 are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> y=11 and x=12. Sufficient.(As cyclicity of units digit of 7 in integer power is 4, therefore the units digit of 7^{23} is the same as the units digit of 7^3, so 3).

(2) x^2-y^2=23 --> (x-y)(x+y)=23=prime --> since both x and y are positive integers then: x-y=1 and x+y=23 --> y=11 and x=12. Sufficient.

Can anybody clarify for me how X-Y = 1 in the 2nd part of the question?

23 is a prime number, so it can be broken into the product of two positive multiples only in one way 23=1*23. Now, since x and y are positive integers, then x-y<x+y, thus x-y=1 and x+y=23.

Does this make sense?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]  12 Jan 2012, 02:51
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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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Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 12 Jan 2012, 02:51
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