Last visit was: 06 Sep 2026, 01:39 It is currently 06 Sep 2026, 01:39
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3   
555-605 (Medium)|   Algebra|   Exponents|   Roots|      
avatar
Acropora
avatar
Joined: 30 Aug 2013
Last visit: 04 Aug 2014
Posts: 2
Given Kudos: 1
GMAT 1: 650 Q47 V31
GMAT 1: 650 Q47 V31
Posts: 2
Kudos: 0
NEW!!! Tough and tricky exponents and roots questions 05 Sep 2013, 12:30
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.
Show more

Can anybody clarify for me how X-Y = 1 in the 2nd part of the question?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,157
Own Kudos:
839,426
 [1]
Given Kudos: 111,334
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,157
Kudos: 839,426
 [1]
NEW!!! Tough and tricky exponents and roots questions 05 Sep 2013, 12:35
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Acropora
Bunuel
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.

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

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?
User avatar
ranaazad
User avatar
Joined: 16 Jan 2013
Last visit: 04 Jan 2021
Posts: 66
Own Kudos:
29
Given Kudos: 1,323
Location: Bangladesh
GMAT 1: 490 Q41 V18
GMAT 2: 610 Q45 V28
GPA: 2.75
GMAT 2: 610 Q45 V28
Posts: 66
Kudos: 29
NEW!!! Tough and tricky exponents and roots questions 12 Jul 2015, 16:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.

Answer: D.
Show more


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

Thanks! :D
User avatar
ENGRTOMBA2018
User avatar
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,317
Own Kudos:
3,931
 [2]
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
My Reviews
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
Posts: 2,317
Kudos: 3,931
 [2]
NEW!!! Tough and tricky exponents and roots questions 12 Jul 2015, 16:59
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
ranaazad
Bunuel
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.

Answer: D.



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

Thanks! :D
Show more

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

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

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

Coming back to the question,

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

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

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

whenever you are in doubt, always test the given statements/conditions with different values (integers, fractions , etc). This way you will get a better picture.
User avatar
yezz
User avatar
Retired Moderator
Joined: 05 Jul 2006
Last visit: 26 Apr 2022
Posts: 830
Own Kudos:
1,705
Given Kudos: 49
Posts: 830
Kudos: 1,705
NEW!!! Tough and tricky exponents and roots questions 14 Oct 2016, 13:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.
Show more


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??
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,157
Own Kudos:
839,426
 [1]
Given Kudos: 111,334
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,157
Kudos: 839,426
 [1]
NEW!!! Tough and tricky exponents and roots questions 15 Oct 2016, 02:49
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
yezz
Bunuel
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??
Show more

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).
avatar
scrumptious829
avatar
Joined: 31 Oct 2017
Last visit: 10 Mar 2018
Posts: 4
Own Kudos:
2
Given Kudos: 5
Posts: 4
Kudos: 2
NEW!!! Tough and tricky exponents and roots questions 13 Nov 2017, 11:06
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.

Answer: D.
Show more

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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,157
Own Kudos:
839,426
 [1]
Given Kudos: 111,334
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,157
Kudos: 839,426
 [1]
NEW!!! Tough and tricky exponents and roots questions 13 Nov 2017, 21:46
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
scrumptious829
Bunuel
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.

Answer: D.

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.
Show more

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 \(x=\sqrt{9}=3\).
User avatar
duahsolo
User avatar
Joined: 02 Jun 2015
Last visit: 31 Jul 2023
Posts: 143
Own Kudos:
784
Given Kudos: 1,196
Location: Ghana
Posts: 143
Kudos: 784
NEW!!! Tough and tricky exponents and roots questions 16 Feb 2018, 10:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.

Answer: D.
Show more


Hi Bunuel,

Can you kindly help me understand why 4th root of x^2 > 0 and not 4th root of x^2 = 4th root of |x| similar to square of x^2 = square root of |x| as shown in the step 2 of your solution to the question @ https://gmatclub.com/forum/what-is-x-2- ... 99157.html?

Thank you
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,157
Own Kudos:
839,426
 [1]
Given Kudos: 111,334
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,157
Kudos: 839,426
 [1]
NEW!!! Tough and tricky exponents and roots questions 16 Feb 2018, 11:19
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
duahsolo
Bunuel
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.

Answer: D.


Hi Bunuel,

Can you kindly help me understand why 4th root of x^2 > 0 and not 4th root of x^2 = 4th root of |x| similar to square of x^2 = square root of |x| as shown in the step 2 of your solution to the question @ https://gmatclub.com/forum/what-is-x-2- ... 99157.html?

Thank you
Show more

Even roots from a non-negative number is non-negative, so 0 or positive: \(\sqrt[even]{something}\geq 0\), not matter what that "something" is (provided it's positive or 0 because even roots from negative numbers are not defined for the GMAT) )

Next, \(\sqrt[4]{x^2}=\sqrt{\sqrt{x^2}}=\sqrt{|x|}\), which must be positive, so (1) basically says that \(\sqrt[5]{y}>0\), or that y > 0.

Finally, if it were
avatar
Rurugu
avatar
Joined: 25 Jan 2018
Last visit: 10 Dec 2019
Posts: 4
Given Kudos: 45
Posts: 4
Kudos: 0
NEW!!! Tough and tricky exponents and roots questions 05 May 2019, 08:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel

I don’t understand why in statement 2, x+y cannot equal 1. X and Y are non-negatives, so we shouldn’t exclude the possibility of X and Y being 0 and 1 or 1 and 0 respectively. Please clarify. Thanks

Bunuel
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.
Show more

Posted from my mobile device
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,157
Own Kudos:
839,426
 [1]
Given Kudos: 111,334
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,157
Kudos: 839,426
 [1]
NEW!!! Tough and tricky exponents and roots questions 05 May 2019, 09:18
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Rurugu
Hi Bunuel

I don’t understand why in statement 2, x+y cannot equal 1. X and Y are non-negatives, so we shouldn’t exclude the possibility of X and Y being 0 and 1 or 1 and 0 respectively. Please clarify. Thanks

Bunuel
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.

Posted from my mobile device
Show more

If x = 0 and y = 1, then 2^x + 3^y = 4 and if x = 1 and y = 0, then 2^x + 3^y = 3. While \(\sqrt[1]{25}=25\)
User avatar
pudu
User avatar
Joined: 12 Mar 2023
Last visit: 06 Mar 2024
Posts: 221
Own Kudos:
126
Given Kudos: 16
Location: India
Posts: 221
Kudos: 126
NEW!!! Tough and tricky exponents and roots questions 22 Apr 2023, 21:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.

Answer: D.
Show more

how is it possible that even root has only positive value? Square root is even, but we are taking both positive and negative value. Then why not so for other even root?
please someone help me to understand it
User avatar
pudu
User avatar
Joined: 12 Mar 2023
Last visit: 06 Mar 2024
Posts: 221
Own Kudos:
126
Given Kudos: 16
Location: India
Posts: 221
Kudos: 126
NEW!!! Tough and tricky exponents and roots questions 22 Apr 2023, 21:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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

Must know for the GMAT: if \(x\) is a positive integer then \(\sqrt{x}\) is either a positive integer itself or an irrational number. (It can not be some reduced fraction eg 7/3 or 1/2)

Also note that the question basically asks whether \(x\) is a perfect square.

(1) \(\sqrt{7*x}\) is an integer --> \(x\) can not be a perfect square because if it is, for example if \(x=n^2\) for some positive integer \(n\) then \(\sqrt{7x}=\sqrt{7n^2}=n\sqrt{7}\neq{integer}\). Sufficient.

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


Answer: D.
Show more

hi Bunuel, how D?

if x=7 then sqrt of 49 perfect square, but sqrt x is not. If we take 28, 63 etc for x then both sqrt of 7x and sqrt of x is an integer. So I is not sufficient...
I'm I right?
User avatar
pudu
User avatar
Joined: 12 Mar 2023
Last visit: 06 Mar 2024
Posts: 221
Own Kudos:
126
Given Kudos: 16
Location: India
Posts: 221
Kudos: 126
NEW!!! Tough and tricky exponents and roots questions 22 Apr 2023, 22:02
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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

Must know for the GMAT: if \(x\) is a positive integer then \(\sqrt{x}\) is either a positive integer itself or an irrational number. (It can not be some reduced fraction eg 7/3 or 1/2)

Also note that the question basically asks whether \(x\) is a perfect square.

(1) \(\sqrt{7*x}\) is an integer --> \(x\) can not be a perfect square because if it is, for example if \(x=n^2\) for some positive integer \(n\) then \(\sqrt{7x}=\sqrt{7n^2}=n\sqrt{7}\neq{integer}\). Sufficient.

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

Answer: D.
Show more

how D? if we take x as 7 then sqrt(7X) is an integer, but sqrt(x) is not an integer. If we take x as 28,63 etc then both the square roots are integer. Hence B.

Bunuel, Iam i right?
User avatar
pudu
User avatar
Joined: 12 Mar 2023
Last visit: 06 Mar 2024
Posts: 221
Own Kudos:
126
Given Kudos: 16
Location: India
Posts: 221
Kudos: 126
NEW!!! Tough and tricky exponents and roots questions 22 Apr 2023, 22:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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: \((x+y)^{xy}=2^0=1=odd\);

is it correct to take both x-y and x+y as 2? Is there any such integer is existing?
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.
Show more

in statement 1 we get (x-y)(x+y)=4 but we can't take non integer value. So is it right to take both x-y and x+y as 2? is there any such number is existing where both sum and difference of two numbers as the same? Is it not valid to take statement 1 as insufficient?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,157
Own Kudos:
839,426
Given Kudos: 111,334
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,157
Kudos: 839,426
NEW!!! Tough and tricky exponents and roots questions 23 Apr 2023, 02:23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pudu
Bunuel
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.

Answer: D.

how is it possible that even root has only positive value? Square root is even, but we are taking both positive and negative value. Then why not so for other even root?
please someone help me to understand it
Show more

Mathematically, \(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative 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 \(x=\sqrt{9}=3\).

Hope it helps.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,157
Own Kudos:
839,426
Given Kudos: 111,334
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,157
Kudos: 839,426
NEW!!! Tough and tricky exponents and roots questions 23 Apr 2023, 02:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pudu
Bunuel
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

Must know for the GMAT: if \(x\) is a positive integer then \(\sqrt{x}\) is either a positive integer itself or an irrational number. (It can not be some reduced fraction eg 7/3 or 1/2)

Also note that the question basically asks whether \(x\) is a perfect square.

(1) \(\sqrt{7*x}\) is an integer --> \(x\) can not be a perfect square because if it is, for example if \(x=n^2\) for some positive integer \(n\) then \(\sqrt{7x}=\sqrt{7n^2}=n\sqrt{7}\neq{integer}\). Sufficient.

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


Answer: D.

hi Bunuel, how D?

if x=7 then sqrt of 49 perfect square, but sqrt x is not. If we take 28, 63 etc for x then both sqrt of 7x and sqrt of x is an integer. So I is not sufficient...
I'm I right?
Show more

Since the OA is D, you are obviously not right. How is \(\sqrt{x}\) an integer, if x is 28 or 63?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,157
Own Kudos:
839,426
Given Kudos: 111,334
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,157
Kudos: 839,426
NEW!!! Tough and tricky exponents and roots questions 23 Apr 2023, 02:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pudu
Bunuel
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: \((x+y)^{xy}=2^0=1=odd\);

is it correct to take both x-y and x+y as 2? Is there any such integer is existing?
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.

in statement 1 we get (x-y)(x+y)=4 but we can't take non integer value. So is it right to take both x-y and x+y as 2? is there any such number is existing where both sum and difference of two numbers as the same? Is it not valid to take statement 1 as insufficient?
Show more


4 can be expressed as the product of two integers in the following four ways only: 1 * 4, 2 * 2, (-1) * (-4), or (-2) * (-2).

Since both (x - y) and (x + y) are integers, and x + y > 0, we have x - y = 2 and x + y = 2 (the reason x - y = 1 and x + y = 4 is not possible is explained in the referenced solution).

Both x - y and x + y can equal 2 if x = 2 and y = 0, which is clearly stated in the solution you quoted.
User avatar
EnriqueDandolo
User avatar
Joined: 07 Aug 2024
Last visit: 04 Sep 2026
Posts: 16
Own Kudos:
2
Given Kudos: 17
Posts: 16
Kudos: 2
NEW!!! Tough and tricky exponents and roots questions 08 Mar 2025, 14:19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hey Bunuel, I was wondering if you could please explain step by step in more detail how did you apply the algebraic identity (x^2 - y^2)=(x+y)(x-y) to the fraction

Square root(2)*[Square root(x) + Square root(y)] /(x-y).

Many thanks in advance.

I am confused because after applying the same algebraic identity I got as a result the following:

Square root(2)*[Square root(x) + Square root(y)]*(x+y)/(x^2 - y^2)

Bunuel
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.
Show more
   1   2   3   
Moderators:
Bunuel
Math Expert
113157 posts
FrozenDragon
DI Forum Moderator
409 posts