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

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03 Aug 2015, 09:29
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
In question we just have to prove $$\sqrt{x}$$ is an integer or not --> x should be a square.

Now, in statement 1, its y=even perfect square and z=odd perfect cube
so y can be $$2^2 or 4^2$$ (both are even perfect square)
z can be$$3^3, 5^3 or 7^3$$ ( all are perfect odd cube)

when i substitute y=$$4^2$$ and z=$$3^3$$ , my equation is x * $$4^2$$ * $$3^3$$= 2700, which upon simplification gives x = 25/4, which is not an integer upon doing as per question stem i.e $$\sqrt{x}$$
So how to solve using both statement as in statement 2 its just given that $$\sqrt{z}$$ is not an integer --> implying that it is odd. like i have assumed z= $$5^3$$ whose $$\sqrt{z}$$ is not integer.

Upon Using both statement i still cannot prove it to be sufficient to arrive at C.

So i chose E.

can you tell me where am i going wrong?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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03 Aug 2015, 10:26
shail2509 wrote:
hi Bunuel
In question we just have to prove $$\sqrt{x}$$ is an integer or not --> x should be a square.

Now, in statement 1, its y=even perfect square and z=odd perfect cube
so y can be $$2^2 or 4^2$$ (both are even perfect square)
z can be$$3^3, 5^3 or 7^3$$ ( all are perfect odd cube)

when i substitute y=$$4^2$$ and z=$$3^3$$ , my equation is x * $$4^2$$ * $$3^3$$= 2700, which upon simplification gives x = 25/4, which is not an integer upon doing as per question stem i.e $$\sqrt{x}$$
So how to solve using both statement as in statement 2 its just given that $$\sqrt{z}$$ is not an integer --> implying that it is odd. like i have assumed z= $$5^3$$ whose $$\sqrt{z}$$ is not integer.

Upon Using both statement i still cannot prove it to be sufficient to arrive at C.

So i chose E.

can you tell me where am i going wrong?

Where from are you getting that y =4^2 ? For this to be true then the number n should have 2^4 but 2700 has 2^2 only.

$$2700 = 2^2*3^3*5^2$$

Also, another incorrect interpretation is that just because $$\sqrt{z} \neq integer$$ ---> z = odd? How about z = 6. $$\sqrt{6} \neq integer$$.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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19 Feb 2016, 20:55
"(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)"

Hi:

please could you help me understand why when we consider(x-1)(y-2) = 0, do we rule out the possibility of RHS being 1 as 1^1 would also equal 1

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

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19 Feb 2016, 21:15
WilDThiNg wrote:
"(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)"

Hi:

please could you help me understand why when we consider(x-1)(y-2) = 0, do we rule out the possibility of RHS being 1 as 1^1 would also equal 1

Thanks

Where do you see RHS=1? (x-1)(y-2)=0 has a RHS = 0.

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

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19 Feb 2016, 21:28
Engr2012 wrote:
WilDThiNg wrote:
"(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)"

Hi:

please could you help me understand why when we consider(x-1)(y-2) = 0, do we rule out the possibility of RHS being 1 as 1^1 would also equal 1

Thanks

Where do you see RHS=1? (x-1)(y-2)=0 has a RHS = 0.

Apologies for being unclear - what i meant was when we consider (x-1)(y-2)=0, we assume that 1 on the RHS (before simplification) would have the exponent 0, hence (x-1)(y-2)=0; i was just wondering as to the reasoning of not considering the possibility of 1 (before simplification) having the exponent 1, as the result would be 1 in both the cases - leaving us instead with two possibilities (x-1)(y-2)=0 or (x-1)(y-2)=1?

i know i am going wrong with my reasoning somewhere, not sure where.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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19 Feb 2016, 21:45
WilDThiNg wrote:
Engr2012 wrote:
WilDThiNg wrote:
"(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)"

Hi:

please could you help me understand why when we consider(x-1)(y-2) = 0, do we rule out the possibility of RHS being 1 as 1^1 would also equal 1

Thanks

Where do you see RHS=1? (x-1)(y-2)=0 has a RHS = 0.

Apologies for being unclear - what i meant was when we consider (x-1)(y-2)=0, we assume that 1 on the RHS (before simplification) would have the exponent 0, hence (x-1)(y-2)=0; i was just wondering as to the reasoning of not considering the possibility of 1 (before simplification) having the exponent 1, as the result would be 1 in both the cases - leaving us instead with two possibilities (x-1)(y-2)=0 or (x-1)(y-2)=1?

i know i am going wrong with my reasoning somewhere, not sure where.

Now it's clearer. You are correct that 1^1 =1 but you need to realize that LHS has an exponent with base 3.

This can only be possible when 3^(some power) = 3^0 giving you (x-1)(y-2)= 0.

You can compare powers on LHS with those on the RHS only when the bases are the same . If you take 1^1 on the RHS, you can not compare the powers of dissimilar bases.

Hope this helps.
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19 Feb 2016, 21:53
Got it - very helpful, thanks!... Silly me
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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02 Aug 2016, 17:56
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!
st 2... you wrote x-y = 1 and x+y = 23..
how you got x-y = 1
thanks
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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02 Aug 2016, 18:04
Celestial09 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(2) \(x^2-y^2=23$$

(1) $$100 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!
st 2... you wrote x-y = 1 and x+y = 23..
how you got x-y = 1
thanks

It's based on the fact that 23 is a prime number and thus, if you know that 2 positive integers 'a' and 'b', such that

a*b = 23 , this can only mean that,

a*b = 1*23 or 23*1

For this question, thus,

x-y = 1 and x+y = 23.

Now, the next question that might arise is that how can you know which one of the 2 cases above is true.

Carefully note that both x and y are POSITIVE integers and for any 2 pairs of positive integers x and y,

x-y is always < x+ y.

Hope this helps.

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

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02 Aug 2016, 18:13
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) $$x-y=9$$

$$\frac{\sqrt{2x}+\sqrt{2y}}{x-y}$$ --> 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.

hi!
st 2 can we derive (sq root x - sq root y) (sq root x + sq root y) = 9
and hence x and y +ve.. we can deduce B is also sufficient?
thanks
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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02 Aug 2016, 18:15
Engr2012 wrote:
Celestial09 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(2) \(x^2-y^2=23$$

(1) $$100 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!
st 2... you wrote x-y = 1 and x+y = 23..
how you got x-y = 1
thanks

It's based on the fact that 23 is a prime number and thus, if you know that 2 positive integers 'a' and 'b', such that

a*b = 23 , this can only mean that,

a*b = 1*23 or 23*1

For this question, thus,

x-y = 1 and x+y = 23.

Now, the next question that might arise is that how can you know which one of the 2 cases above is true.

Carefully note that both x and y are POSITIVE integers and for any 2 pairs of positive integers x and y,

x-y is always < x+ y.

Hope this helps.

Posted from my mobile device

hi!
thanks for explanation.. i need some more time to understand that.. may be i am lacking some basics.
thanks
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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06 Aug 2016, 12:29
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.

Bunuel,

can you please walk me through the following process:
$$27^{x^2}=\frac{3}{3^{3y^2+1}}$$ --> $$3^{3x^2}=\frac{3}{3^{3y^2}*3}$$
I am having a hard time understanding how did {3^{3y^2+1} become {3^{3y^2}*3}.

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

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14 Oct 2016, 13:21
Bunuel wrote:
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: $$(x+y)^{xy}=2^1=2=even$$.

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

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15 Oct 2016, 02:49
yezz wrote:
Bunuel wrote:
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: $$(x+y)^{xy}=2^1=2=even$$.

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??

If y=0, then we'll have $$2^{x}=\sqrt[x]{16}$$:

$$2^x=2^{\frac{4}{x}}$$;

$$x=\frac{4}{x}$$;

$$x^2=4$$

$$x=2$$ or $$x=-2$$ (discard since x is non-negative).
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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13 Sep 2017, 10:03
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) $$x-y=9$$

$$\frac{\sqrt{2x}+\sqrt{2y}}{x-y}$$ --> 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.

Can someone explain why x-y>0? I understand x&y>0 from the question. But why x-y also??
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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13 Sep 2017, 10:05
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Ace800 wrote:
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) $$x-y=9$$

$$\frac{\sqrt{2x}+\sqrt{2y}}{x-y}$$ --> 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.

Can someone explain why x-y>0? I understand x&y>0 from the question. But why x-y also??

$$x>y>0$$

Subtract y: $$x-y>0>-y$$.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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13 Sep 2017, 11:19
Bunuel wrote:
Ace800 wrote:
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) $$x-y=9$$

$$\frac{\sqrt{2x}+\sqrt{2y}}{x-y}$$ --> 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.

Can someone explain why x-y>0? I understand x&y>0 from the question. But why x-y also??

$$x>y>0$$

Subtract y: $$x-y>0>-y$$.

Oh x>y in the question!! I read the question as x & y are both positive i.e >0. that's why I was confused. But thanks for your prompt reply anyway!I appreciate that
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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27 Sep 2017, 12:09
Bunuel wrote:
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 $$(x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3$$.

hi man

This question is elegantly solved with algebra
Another way that I solved this problem is as under:

x + y + z = 0
-3 + 2 + 1 = 0

Now, it doesn't matter actually which one corresponds to which one, that is which one is x, y, and z. As, all of them are raised to same power, their sum will remain the same
Thus, -3^3 + 2^3 + 1 ^3 = -27 + 8 + 1 = -18, and xyz (-3 * 2 * 1) = -6. So we get -18/-6 = 3
other values for x,y, and z such as -4, 2, and 2 will also work with this system

B

Please say to me whether it is okay ...

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

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13 Nov 2017, 11:06
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.

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

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13 Nov 2017, 21:46
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scrumptious829 wrote:
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.

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.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
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Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 13 Nov 2017, 21:46

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