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

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

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12 Jan 2012, 03:50
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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 $$\sqrt{x}$$ an 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]

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01 Jul 2014, 10:47
Bunuel wrote:
gkashyap 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.

Hi Bunuel,

Could you please provide me some explanation for (2) when you say " since both x and y are positive integers then: x-y=1 and x+y=23 --> y=11 and x=12"?

How can we assume that x - y = 1? It could have been any two numbers. Is there a theory that if difference of square of two numbers, then they are consecutive? Could you please provide some reference?

Thanks for the help.

Regards,
Gajendra

$$(x-y)(x+y)=23=prime$$. 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?

Yes, it is clear now. Thanks !!

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

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15 Jul 2014, 11:16
Bunuel 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.

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 (2) $$\sqrt{z}\neq{integer}$$ 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.

Hi Bunuel,
Please explain why statement 2 is insufficient. If z^(1/2) is not an integer then z is not a perfect square, so z can be 27, 27*4, 27*25 or 27*25*4. In all these cases the left-out factors are perfect squares. Therefore x must be a perfect square (4,25,100 or 1)
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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15 Jul 2014, 11:42
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shuvabrata88 wrote:
Bunuel 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.

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 (2) $$\sqrt{z}\neq{integer}$$ 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.

Hi Bunuel,
Please explain why statement 2 is insufficient. If z^(1/2) is not an integer then z is not a perfect square, so z can be 27, 27*4, 27*25 or 27*25*4. In all these cases the left-out factors are perfect squares. Therefore x must be a perfect square (4,25,100 or 1)

What about y in this cases? Also, there are other cases possible. For example, z=3, y=2*3^2*5^2 and x=2.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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23 Jul 2014, 06:39
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.

Hello Bunuel, for the second statement, you seem to indicate that since the difference of the squares of two positive integers is a prime number, the two integers are consecutive on the number scale. Does this always hold?
Thank you!

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

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23 Jul 2014, 09:29
Kconfused 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.

Hello Bunuel, for the second statement, you seem to indicate that since the difference of the squares of two positive integers is a prime number, the two integers are consecutive on the number scale. Does this always hold?
Thank you!

Yes, positive integer solutions of x^2-y^2=prime must be consecutive, because of x-y=1 (x=y+1).
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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20 Aug 2014, 03:17
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Bunuel wrote:
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 (2) $$\sqrt{z}\neq{integer}$$ 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.

Hope, you meant cube in the highlighted part
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If you know what you're worth, then go out and get what you're worth. But you gotta be willing to take the hits, and not pointing fingers saying you ain't where you wanna be because of anybody! Cowards do that and You're better than that!
The path is long, but self-surrender makes it short; the way is difficult, but perfect trust makes it easy.

Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.
Failure establishes only this, that our determination to succeed was not strong enough.
Getting defeated is just a temporary notion, giving it up is what makes it permanent.

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

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20 Aug 2014, 03:22
joshnsit wrote:
Bunuel wrote:
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 (2) $$\sqrt{z}\neq{integer}$$ 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.

Hope, you meant cube in the highlighted part

Yes. Edited. Thank you.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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21 Aug 2014, 03:30
Bunuel wrote:
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.

Hi Bunuel, I have a minor doubt. Using the corollary a^3+b^3=(a+b)(a^2-a*b+b^2) and plugging in a=x^2 and b=y^3 in statement 1,
we can say that (x^2)^3+(y^3)^3=(x^2+y^3)(x^4-x^2y^3+y^6)=0
Since any of expressions (x^2+y^3) or (x^4-x^2y^3+y^6) can be zero. It seems that statement 1 is insufficient by this logic.
Kindly comment.
_________________

If you know what you're worth, then go out and get what you're worth. But you gotta be willing to take the hits, and not pointing fingers saying you ain't where you wanna be because of anybody! Cowards do that and You're better than that!
The path is long, but self-surrender makes it short; the way is difficult, but perfect trust makes it easy.

Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.
Failure establishes only this, that our determination to succeed was not strong enough.
Getting defeated is just a temporary notion, giving it up is what makes it permanent.

http://gmatclub.com/forum/1000-sc-notes-at-one-place-in-one-document-with-best-of-explanations-192961.html

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

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21 Aug 2014, 03:41
Bunuel wrote:
joshnsit wrote:
Bunuel wrote:
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 (2) $$\sqrt{z}\neq{integer}$$ 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.

Hope, you meant cube in the highlighted part

Yes. Edited. Thank you.
Thanks for considering change. You may want to change Quant question M26-30 in gmatclub tests. This question actually brought me here. I couldn't get URL, but hopefully question id will be sufficient to trace.
_________________

If you know what you're worth, then go out and get what you're worth. But you gotta be willing to take the hits, and not pointing fingers saying you ain't where you wanna be because of anybody! Cowards do that and You're better than that!
The path is long, but self-surrender makes it short; the way is difficult, but perfect trust makes it easy.

Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.
Failure establishes only this, that our determination to succeed was not strong enough.
Getting defeated is just a temporary notion, giving it up is what makes it permanent.

http://gmatclub.com/forum/1000-sc-notes-at-one-place-in-one-document-with-best-of-explanations-192961.html

Press +1 Kudos, if you think my post gave u a tiny tip.

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

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21 Aug 2014, 04:14
joshnsit wrote:
Bunuel wrote:
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.

Hi Bunuel, I have a minor doubt. Using the corollary a^3+b^3=(a+b)(a^2-a*b+b^2) and plugging in a=x^2 and b=y^3 in statement 1,
we can say that (x^2)^3+(y^3)^3=(x^2+y^3)(x^4-x^2y^3+y^6)=0
Since any of expressions (x^2+y^3) or (x^4-x^2y^3+y^6) can be zero. It seems that statement 1 is insufficient by this logic.
Kindly comment.

The point is that only real roots of x^4 - x^2y^3 + y^6 = 0 is x = y = 0, for which x^2 + y^3 = 0.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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25 Sep 2014, 13:11
Bunuel wrote:
2. If x, y, and z are positive integers and $$xyz=2,700$$. Is $$\sqrt{x}$$ an 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 $$z=3^3=odd \ perfect \ cube$$ 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 (2) $$\sqrt{z}\neq{integer}$$ 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.

Hi Bunuel,
Using stmt (2), is enough to know that Z#1. That would mean that Z must be the only non square left, and that is 3^3.
Why do we also need stmt (1)?

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

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14 Nov 2014, 03:28
ronr34 wrote:
Bunuel wrote:
2. If x, y, and z are positive integers and $$xyz=2,700$$. Is $$\sqrt{x}$$ an 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 $$z=3^3=odd \ perfect \ cube$$ 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 (2) $$\sqrt{z}\neq{integer}$$ 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.

Hi Bunuel,
Using stmt (2), is enough to know that Z#1. That would mean that Z must be the only non square left, and that is 3^3.
Why do we also need stmt (1)?

Hey man, it's a GMAT mechanic. You don't add up informations until you can't prove that either statement holds alone. In this case z is not 1 but we don't know anything about y and x. x might be 5 to the power of 2 or 3 to the power of 3.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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09 Jan 2015, 05:28
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Bunuel wrote:
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.

Hi Bunuel,
I have a confusion. The problem statement with option no. 1 can be written as follows:
if a^3 + b^3 = 0 (1st option) then what is the value of a + b ? (where a = x^2 and b = y^3).
Now, we know that (a^3 + b^3) = (a+b) * (a^2 - a*b + b^2). (a^3 + b^3) has 2 factors.
And according to the explanation, since, (a^3 + b^3) =0 => a^3 = -(b^3) => a = -b => a + b = 0. So, whenever a^3 + b^3 is 0, a+b will also be 0 !!!!
Can these steps be true always? Please reply... i am really confused.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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09 Jan 2015, 05:39
aniteshgmat1101 wrote:
Bunuel wrote:
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.

Hi Bunuel,
I have a confusion. The problem statement with option no. 1 can be written as follows:
if a^3 + b^3 = 0 (1st option) then what is the value of a + b ? (where a = x^2 and b = y^3).
Now, we know that (a^3 + b^3) = (a+b) * (a^2 - a*b + b^2). (a^3 + b^3) has 2 factors.
And according to the explanation, since, (a^3 + b^3) =0 => a^3 = -(b^3) => a = -b => a + b = 0. So, whenever a^3 + b^3 is 0, a+b will also be 0 !!!!
Can these steps be true always? Please reply... i am really confused.

Yes, if a^3 + b^3 = 0, then a + b = 0.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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20 Jun 2015, 22:05
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.

Bunuel : Hi, it is not written that x can not be equal to y.
so what if x= y= 11
or x=y=12

in that case A will not be sufficient.

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

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20 Jun 2015, 23:40
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MyaimHarvard 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.

Bunuel : Hi, it is not written that x can not be equal to y.
so what if x= y= 11
or x=y=12

in that case A will not be sufficient.

x = y is not possible since we are told that y^2 < x^2.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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12 Jul 2015, 16:28
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) $$y^3>\frac{1}{z{^4}}$$

$$xyz\neq{0}$$ 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.

Hello sir,
It was not mentioned in the question stem whether x, y and z are integers. In this question, if x, y and z were non-integers, wouldn't the procedure be different? Would you please explain?

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

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12 Jul 2015, 16:59
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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) $$y^3>\frac{1}{z{^4}}$$

$$xyz\neq{0}$$ 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.

Hello sir,
It was not mentioned in the question stem whether x, y and z are integers. In this question, if x, y and z were non-integers, wouldn't the procedure be different? Would you please explain?

Thanks! :D

This question has a trick in that it will be true for both integers and non-integers alike:

1. x^2 > 0 no matter x=integer or not. Similarly, $$\sqrt[4]{x^2}$$ will be positive for x=integer or non-integer. Try it out. Let x = 1 , $$\sqrt[4]{1^2} = 1 > 0$$ and if x= 0.5, $$\sqrt[4]{0.5^2} = 0.707 >0.$$

Thus it does not matter whether you take x = integer or a fraction.

Coming back to the question,

Per statement 1, as mentioned above, $$\sqrt[4]{1^2} = 1 > 0$$ ---> $$\sqrt[5]{y} > 0$$ -----> y >0. Thus the statement $$(x^{-4})*(\sqrt[3]{y})*(z^{-2})<0$$ will be false. Thus this statement is sufficient.

Per statement 2, z^4 > 0 whether z =1 or z= 0.5 ----> 1/(z^4) >0 ----> y^3 >0---> y>0 (satisfies for both, y =1 or y = 0.25). Thus the given statement $$(x^{-4})*(\sqrt[3]{y})*(z^{-2})<0$$ will be false. Thus this statement is sufficient.

As both the given statements are sufficient, the correct answer is D.

whenever you are in doubt, always test the given statements/conditions with different values (integers, fractions , etc). This way you will get a better picture.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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12 Jul 2015, 17:03
Bunuel wrote:
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.

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.

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

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13 Jul 2015, 00:56
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Bunuel wrote:
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.

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.

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

$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$ (if exponentiation is indicated by stacked symbols, the rule is to work from the top down)
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Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 13 Jul 2015, 00:56

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