NEW!!! Tough and tricky exponents and roots questions : GMAT Data Sufficiency (DS) - Page 3
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 10:50

# STARTING SOON:

Open Admission Chat with MBA Experts of Personal MBA Coach - Join Chat Room to Participate.

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7076

Kudos [?]: 93091 [18] , given: 10552

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

### Show Tags

12 Jan 2012, 02:50
18
KUDOS
Expert's post
99
This post was
BOOKMARKED
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
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7076

Kudos [?]: 93091 [0], given: 10552

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

### Show Tags

08 Jul 2013, 23:35
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE
_________________
Intern
Joined: 12 Mar 2013
Posts: 14
Followers: 0

Kudos [?]: 0 [0], given: 14

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

### Show Tags

04 Aug 2013, 14:30
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 (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.

Hi Bunuel,

Is it not true to say that 2^2*5^2 could be called a perfect square too? In that case, from statement 1 and 2 one possibility is that x=2^2*5^2 apart from x=1 and x=5^2.

I know in this case it wont make a difference to the answer but I want to clarify the concept.

Thanks
Intern
Joined: 30 Aug 2013
Posts: 3
GMAT 1: 650 Q47 V31
Followers: 0

Kudos [?]: 0 [0], given: 1

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

### Show Tags

05 Sep 2013, 11:30
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?
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7076

Kudos [?]: 93091 [1] , given: 10552

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

### Show Tags

05 Sep 2013, 11:35
1
KUDOS
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?
_________________
Intern
Joined: 30 Aug 2013
Posts: 3
GMAT 1: 650 Q47 V31
Followers: 0

Kudos [?]: 0 [0], given: 1

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

### Show Tags

05 Sep 2013, 12:45
Bunuel wrote:
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?

Ooo I got it. Man.. I can't believe I didn't realize that. Thanks!
Intern
Joined: 17 Jul 2013
Posts: 3
GMAT 1: 680 Q50 V30
GPA: 3.21
Followers: 0

Kudos [?]: 0 [0], given: 1

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

### Show Tags

06 Sep 2013, 23:38
we are asked only about units digit so the problem simply can be seen as (7^x)*(7^y)
that is nothing but 7^(x+y)
from option a) x,y are 11,12
from option b)(x+y)*(x-y)=23(prime number) thus a+b=23
thus the units digit can be inferred from both the options
Intern
Joined: 21 Sep 2013
Posts: 30
Location: United States
Concentration: Finance, General Management
GMAT Date: 10-25-2013
GPA: 3
WE: Operations (Mutual Funds and Brokerage)
Followers: 0

Kudos [?]: 23 [0], given: 82

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

### Show Tags

02 Oct 2013, 23:26
IanStewart 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) $$x-y=9$$

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.

in the question could not demistify the denominator ,otherwise was on track. thanks for the help.
Intern
Joined: 02 May 2013
Posts: 23
Followers: 0

Kudos [?]: 0 [0], given: 12

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

### Show Tags

07 Oct 2013, 00:11
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 (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.

Dear Mr. Bunuel,
As you say, a perfect cube is an integer that can be written as the cube of some OTHER integer. I am wondering whether 1^3 can not be a perfect cube in accordance with this definition.
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7076

Kudos [?]: 93091 [0], given: 10552

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

### Show Tags

07 Oct 2013, 00:15
smallurgo 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 (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.

Dear Mr. Bunuel,
As you say, a perfect cube is an integer that can be written as the cube of some OTHER integer. I am wondering whether 1^3 can not be a perfect cube in accordance with this definition.

Yes, 1 is a perfect cube and perfect square.
_________________
Intern
Joined: 15 Dec 2012
Posts: 6
Location: United States
GMAT 1: 610 Q34 V41
GMAT 2: 620 Q49 V25
GPA: 3.25
Followers: 0

Kudos [?]: 6 [0], given: 33

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

### Show Tags

22 Oct 2013, 16:28
Bunuel, could you please explain me why did you get this
Bunuel wrote:

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

because I got
x^6+y^9=0 --> (x^2)^3 + (y^3)^3=0 --> (x^2)^3 = - (y^3)^3. The difference is in "-" sing in front of (y^3)^3. So if there is a "-" sign in front of it, does it give us the same result?

Thanks
_________________

Please don't forget to give Kudos if you like my post! :-)

Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7076

Kudos [?]: 93091 [0], given: 10552

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

### Show Tags

22 Oct 2013, 23:03
Natics wrote:
Bunuel, could you please explain me why did you get this
Bunuel wrote:

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

because I got
x^6+y^9=0 --> (x^2)^3 + (y^3)^3=0 --> (x^2)^3 = - (y^3)^3. The difference is in "-" sing in front of (y^3)^3. So if there is a "-" sign in front of it, does it give us the same result?

Thanks

Both give the same result: $$x^2=-y^3$$.
_________________
Manager
Joined: 19 Mar 2012
Posts: 128
Followers: 0

Kudos [?]: 12 [0], given: 270

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

### Show Tags

01 Apr 2014, 09:08
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 (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.

Hi Bunuel
Can you please explain second statement.
How did you figured out "$$\sqrt{z}$$ is not an integer"
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7076

Kudos [?]: 93091 [0], given: 10552

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

### Show Tags

01 Apr 2014, 10:36
282552 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 (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.

Hi Bunuel
Can you please explain second statement.
How did you figured out "$$\sqrt{z}$$ is not an integer"

I did not deduce that myself, it's given in the second statement: "(2) $$\sqrt{z}$$ is not an integer."
_________________
Manager
Joined: 19 Mar 2012
Posts: 128
Followers: 0

Kudos [?]: 12 [0], given: 270

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

### Show Tags

01 Apr 2014, 10:54
Bunuel wrote:
282552 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 (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.

Hi Bunuel
Can you please explain second statement.
How did you figured out "$$\sqrt{z}$$ is not an integer"

I did not deduce that myself, it's given in the second statement: "(2) $$\sqrt{z}$$ is not an integer."

I am sorry I wanted to ask According to second statement why can't z be let's say 3 or 3^3 or anything else.
In that case we would be able to tell that we can have a perfect square for x
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7076

Kudos [?]: 93091 [0], given: 10552

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

### Show Tags

01 Apr 2014, 11:34
Bunuel wrote:
282552 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 (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.

I did not deduce that myself, it's given in the second statement: "(2) $$\sqrt{z}$$ is not an integer."

I am sorry I wanted to ask According to second statement why can't z be let's say 3 or 3^3 or anything else.
In that case we would be able to tell that we can have a perfect square for x

Are you talking about the first statement??? Or when we consider the statements together???

(1) says that $$z$$ is an odd perfect cube, thus it can be only 1 or 3^3.
_________________
Manager
Joined: 04 Jan 2014
Posts: 103
Followers: 0

Kudos [?]: 29 [0], given: 20

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

### Show Tags

28 Apr 2014, 12:10
Hi,

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

when you explained the Stmt 1, I found it confusing.. However I took it in a different way. Let me know if my thinking is right.. For the 1st stmt to be right, x must be 7. So sqrt(7) is not a integer.. Sufficient..

I really missed the 2nd stmt when I did it.. Your way of solving and explaining is really good.. keep us posted with a more and more like these..
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7076

Kudos [?]: 93091 [0], given: 10552

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

### Show Tags

29 Apr 2014, 06:46
sheolokesh wrote:
Hi,

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

when you explained the Stmt 1, I found it confusing.. However I took it in a different way. Let me know if my thinking is right.. For the 1st stmt to be right, x must be 7. So sqrt(7) is not a integer.. Sufficient..

I really missed the 2nd stmt when I did it.. Your way of solving and explaining is really good.. keep us posted with a more and more like these..

For (1) x=7 is not the only possible value to satisfy $$\sqrt{7*x}=integer$$. For example, x can also be 7*4, 7*9, 7*16, ... basically any number of a form 7*integer^2.
_________________
Intern
Joined: 24 May 2013
Posts: 29
Concentration: Operations, General Management
WE: Engineering (Energy and Utilities)
Followers: 2

Kudos [?]: 14 [0], given: 21

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

### Show Tags

19 Jun 2014, 00:06
If x, y, and z are positive integers and xyz=2,700. Is x√ an integer?
(1) y is an even perfect square and z is an odd perfect cube.

(2) z√ is not an integer.

For this question, I agree one of the values can be 1.
But issue is we dont apply this logic to all questions. Somewhere we do this and sometimes we don't. I did a question before in which 1 was not considered.
This is really a trap.
Intern
Joined: 21 May 2013
Posts: 8
Followers: 0

Kudos [?]: 1 [0], given: 7

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

### Show Tags

01 Jul 2014, 09:20
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
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7076

Kudos [?]: 93091 [0], given: 10552

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

### Show Tags

01 Jul 2014, 09:33
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?
_________________
Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 01 Jul 2014, 09:33

Go to page   Previous    1   2   3   4   5    Next  [ 94 posts ]

Similar topics Replies Last post
Similar
Topics:
If the square root of t is a real number, is the square root 3 09 Apr 2013, 15:32
16 Is root{x} a prime number? 23 26 Jan 2012, 05:41
6 Factoring w/ Exponents Question 9 06 Jan 2011, 12:37
23 Series A(n) is such that i*A(i) = j*A(j) for any pair of 19 30 Jun 2010, 09:38
1 Is it Ok to take the roots first?For example, in question 1 5 28 Aug 2010, 12:49
Display posts from previous: Sort by