It is currently 20 Oct 2017, 20:56

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

NEW!!! Tough and tricky exponents and roots questions

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Expert Post
20 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129048 [20], given: 12187

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

Show Tags

New post 12 Jan 2012, 03:50
20
This post received
KUDOS
Expert's post
130
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
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 129048 [20], given: 12187

Intern
Intern
avatar
B
Joined: 21 Jan 2015
Posts: 5

Kudos [?]: 1 [0], given: 8

Concentration: International Business, General Management
GMAT ToolKit User
NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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.

Answer: C.


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?

Kudos [?]: 1 [0], given: 8

Expert Post
Math Forum Moderator
avatar
B
Joined: 20 Mar 2014
Posts: 2675

Kudos [?]: 1725 [0], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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. \(\sqrt{6} \neq integer\).
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1725 [0], given: 792

Manager
Manager
avatar
B
Joined: 29 Dec 2014
Posts: 81

Kudos [?]: 2 [0], given: 997

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

Show Tags

New post 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

Kudos [?]: 2 [0], given: 997

Expert Post
Math Forum Moderator
avatar
B
Joined: 20 Mar 2014
Posts: 2675

Kudos [?]: 1725 [0], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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.

Can you rephrase your question?
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1725 [0], given: 792

Manager
Manager
avatar
B
Joined: 29 Dec 2014
Posts: 81

Kudos [?]: 2 [0], given: 997

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

Show Tags

New post 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.

Can you rephrase your question?


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.

Kudos [?]: 2 [0], given: 997

Expert Post
Math Forum Moderator
avatar
B
Joined: 20 Mar 2014
Posts: 2675

Kudos [?]: 1725 [0], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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.

Can you rephrase your question?


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.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1725 [0], given: 792

Manager
Manager
avatar
B
Joined: 29 Dec 2014
Posts: 81

Kudos [?]: 2 [0], given: 997

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

Show Tags

New post 19 Feb 2016, 21:53
Got it - very helpful, thanks!... Silly me :)

Kudos [?]: 2 [0], given: 997

BSchool Forum Moderator
avatar
B
Status: I Declare War!!!
Joined: 02 Apr 2014
Posts: 258

Kudos [?]: 96 [0], given: 546

Location: United States
Concentration: Finance, Economics
GMAT Date: 03-18-2015
WE: Asset Management (Investment Banking)
GMAT ToolKit User Premium Member
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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.

Answer: D.



hello!
can you please help me with a small doubt..
st 2... you wrote x-y = 1 and x+y = 23..
how you got x-y = 1
thanks

Kudos [?]: 96 [0], given: 546

Expert Post
Math Forum Moderator
avatar
B
Joined: 20 Mar 2014
Posts: 2675

Kudos [?]: 1725 [0], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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.

Answer: D.



hello!
can you please help me with a small doubt..
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.

Thus the answer.

Hope this helps.

Posted from my mobile device
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1725 [0], given: 792

BSchool Forum Moderator
avatar
B
Status: I Declare War!!!
Joined: 02 Apr 2014
Posts: 258

Kudos [?]: 96 [0], given: 546

Location: United States
Concentration: Finance, Economics
GMAT Date: 03-18-2015
WE: Asset Management (Investment Banking)
GMAT ToolKit User Premium Member
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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.

Answer: A.



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

Kudos [?]: 96 [0], given: 546

BSchool Forum Moderator
avatar
B
Status: I Declare War!!!
Joined: 02 Apr 2014
Posts: 258

Kudos [?]: 96 [0], given: 546

Location: United States
Concentration: Finance, Economics
GMAT Date: 03-18-2015
WE: Asset Management (Investment Banking)
GMAT ToolKit User Premium Member
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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.

Answer: D.



hello!
can you please help me with a small doubt..
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.

Thus the answer.

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

Kudos [?]: 96 [0], given: 546

Intern
Intern
avatar
B
Joined: 07 Feb 2016
Posts: 16

Kudos [?]: 2 [0], given: 12

GMAT 1: 710 Q47 V40
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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.

Answer: D.


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.

Kudos [?]: 2 [0], given: 12

SVP
SVP
User avatar
B
Joined: 05 Jul 2006
Posts: 1750

Kudos [?]: 430 [0], given: 49

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

Show Tags

New post 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: \((x+y)^{xy}=2^1=2=even\).

Two different answers. Not sufficient.

Answer: A.



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??

Kudos [?]: 430 [0], given: 49

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129048 [0], given: 12187

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

Show Tags

New post 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: \((x+y)^{xy}=2^1=2=even\).

Two different answers. Not sufficient.

Answer: A.



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).
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 129048 [0], given: 12187

Intern
Intern
avatar
B
Joined: 30 Aug 2017
Posts: 11

Kudos [?]: 0 [0], given: 26

Concentration: Real Estate, Operations
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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.

Answer: A.


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

Kudos [?]: 0 [0], given: 26

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129048 [1], given: 12187

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

Show Tags

New post 13 Sep 2017, 10:05
1
This post received
KUDOS
Expert's post
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.

Answer: A.


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


\(x>y>0\)

Subtract y: \(x-y>0>-y\).
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 129048 [1], given: 12187

Intern
Intern
avatar
B
Joined: 30 Aug 2017
Posts: 11

Kudos [?]: 0 [0], given: 26

Concentration: Real Estate, Operations
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

Show Tags

New post 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.

Answer: A.


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


\(x>y>0\)

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 :thumbup:

Kudos [?]: 0 [0], given: 26

Manager
Manager
User avatar
B
Status: love the club...
Joined: 24 Mar 2015
Posts: 176

Kudos [?]: 14 [0], given: 413

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

Show Tags

New post 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 \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).

Answer: B.


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 ...

thanks in advance, man

Kudos [?]: 14 [0], given: 413

Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 27 Sep 2017, 12:09

Go to page   Previous    1   2   3   4   5   [ 98 posts ] 

Display posts from previous: Sort by

NEW!!! Tough and tricky exponents and roots questions

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.