Find all School-related info fast with the new School-Specific MBA Forum

It is currently 01 Oct 2014, 12:29

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Find a nonzero integer Q (1) Q is a prime root of Q^Q=Q^3

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
User avatar
Joined: 07 Oct 2003
Posts: 358
Location: Manhattan
Followers: 2

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

Find a nonzero integer Q (1) Q is a prime root of Q^Q=Q^3 [#permalink] New post 11 Mar 2005, 16:11
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Find a nonzero integer Q

(1) Q is a prime root of Q^Q=Q^3
(2) Q^2=3^2
VP
VP
avatar
Joined: 18 Nov 2004
Posts: 1447
Followers: 2

Kudos [?]: 16 [0], given: 0

 [#permalink] New post 11 Mar 2005, 17:11
"A"

state 1....Q^2(1-Q) = 0...Q = 1 or 0...but Q is nonzero, so Q = 1...suff

State 2....Q can be #3...insuff
Manager
Manager
avatar
Joined: 13 Oct 2004
Posts: 240
Followers: 1

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

 [#permalink] New post 11 Mar 2005, 19:25
St.1 says Q^Q right and not Q*Q?.
VP
VP
User avatar
Joined: 25 Nov 2004
Posts: 1498
Followers: 6

Kudos [?]: 31 [0], given: 0

 [#permalink] New post 11 Mar 2005, 21:02
from i, q= 1 and 3 but q is prime, so q=3. so suff.
from ii, q=-3 or 3. so insuff.

banerjeea_98 wrote:
"A".
state 1....Q^2(1-Q) = 0...Q = 1 or 0...but Q is nonzero, so Q = 1...suff
State 2....Q can be #3...insuff


Could you pls explain your approach?
VP
VP
avatar
Joined: 18 Nov 2004
Posts: 1447
Followers: 2

Kudos [?]: 16 [0], given: 0

 [#permalink] New post 11 Mar 2005, 21:37
MA wrote:
from i, q= 1 and 3 but q is prime, so q=3. so suff.
from ii, q=-3 or 3. so insuff.

banerjeea_98 wrote:
"A".
state 1....Q^2(1-Q) = 0...Q = 1 or 0...but Q is nonzero, so Q = 1...suff
State 2....Q can be #3...insuff


Could you pls explain your approach?



oops, my mistake, didn't see it is Q^Q and not Q^2....still A is suff.... :oops:
Intern
Intern
User avatar
Joined: 19 Jul 2004
Posts: 43
Followers: 1

Kudos [?]: 3 [0], given: 0

 [#permalink] New post 11 Mar 2005, 23:45
MA wrote:
from i, q= 1 and 3 but q is prime, so q=3. so suff.
from ii, q=-3 or 3. so insuff.



This would be wrong, unless the definition is clear.

What is the meaning of "Q is a prime root of Q^Q=Q^3 "

There can be two meaning

1) Q is prime (which is used by MA)
2) Q is a "prime root", which imply square root (2 prime), cube root (3 prime), fifth root(5 prime) but NOT fourth root (4 not prime), not sixth root (6 not prime)

If second one is true then A is not sufficient.

Assuming the (2) to be true, then both 1 and 3 qualify. Since 1 is a "prime root" of itself.

Note that the condition does not say that "Q is a prime root only of "

Ketan
VP
VP
User avatar
Joined: 25 Nov 2004
Posts: 1498
Followers: 6

Kudos [?]: 31 [0], given: 0

 [#permalink] New post 12 Mar 2005, 07:26
ketanm wrote:
MA wrote:
from i, q= 1 and 3 but q is prime, so q=3. so suff.
from ii, q=-3 or 3. so insuff.

This would be wrong, unless the definition is clear.

What is the meaning of "Q is a prime root of Q^Q=Q^3 "
There can be two meaning
1) Q is prime (which is used by MA)
2) Q is a "prime root", which imply square root (2 prime), cube root (3 prime), fifth root(5 prime) but NOT fourth root (4 not prime), not sixth root (6 not prime)
If second one is true then A is not sufficient.
Assuming the (2) to be true, then both 1 and 3 qualify. Since 1 is a "prime root" of itself. Note that the condition does not say that "Q is a prime root only of "Ketan


I think u r wright. i was unclear and confused with prime root. but the safe side is that answer is A.
Senior Manager
Senior Manager
User avatar
Joined: 07 Oct 2003
Posts: 358
Location: Manhattan
Followers: 2

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

 [#permalink] New post 13 Mar 2005, 19:54
sorry about unclear phrasing. Here's the answer that I got (Not OA, but seems right to me)

(1) Q=1, -1, and 3; of which only 3 is prime. Thus, OK
(2) Q= 3 and -3, not enough
SVP
SVP
User avatar
Joined: 03 Jan 2005
Posts: 2255
Followers: 12

Kudos [?]: 200 [0], given: 0

Re: DS: number qualities [#permalink] New post 13 Mar 2005, 21:12
Q^Q=Q^3
QlnQ=3lnQ
lnQ(Q-3)=0
lnQ=0
Q=1
Q-3=0
Q=3

Hmmm how do you solve this to get all solutions (including -1) without plugging in?

Q^Q-Q^3=0
Q^3*(Q^(Q-3)-1)=0
Q=0
Q^(Q-3)=1
Q-3=0
Q=3

How do you get -1?

Last edited by HongHu on 15 Mar 2005, 19:57, edited 1 time in total.
Senior Manager
Senior Manager
User avatar
Joined: 07 Oct 2003
Posts: 358
Location: Manhattan
Followers: 2

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

Re: DS: number qualities [#permalink] New post 15 Mar 2005, 18:29
HongHu wrote:
Q^Q=Q^3
QlnQ=3lnQ
lnQ(Q-3)=0
lnQ=0
Q=1
Q-3=0
Q=3

Hmmm how do you solve this to get all solutions (including -1) without plugging in?

Q^Q-Q^3=0
Q^3*(Q^(Q-3)-1)=0
Q=0
Q^(Q-3)=1
Q-3=0
Q=1

How do you get -1?


you're right, -1 is not one of the possible answers
SVP
SVP
User avatar
Joined: 03 Jan 2005
Posts: 2255
Followers: 12

Kudos [?]: 200 [0], given: 0

 [#permalink] New post 15 Mar 2005, 20:04
Actually -1 IS one of the possible answers. My question is how do you solve for it. :???
Senior Manager
Senior Manager
avatar
Joined: 15 Mar 2005
Posts: 425
Location: Phoenix
Followers: 1

Kudos [?]: 9 [0], given: 0

 [#permalink] New post 16 Mar 2005, 02:00
In the first equation

Q^Q = Q^3

=> Q^(Q-3) = 1

This is of the type a^b = 1 which has the following solutions
a) a = 1, b = anything
b) a = anything, b = 0
c) a = -1 and b = even

a) gives Q = 1.
b) gives Q = 3.
c) gives Q = -1

Overall, three values of Q, out of which only one is prime.

Hence sufficient.

The second one is more straightforward.

Q^2 = 3^2
Q^2 = 9
Q = +3, -3

No unique solution, hence insufficient.

Hope that helps.
_________________

Who says elephants can't dance?

SVP
SVP
User avatar
Joined: 03 Jan 2005
Posts: 2255
Followers: 12

Kudos [?]: 200 [0], given: 0

 [#permalink] New post 16 Mar 2005, 09:34
kapslock wrote:
This is of the type a^b = 1 which has the following solutions
a) a = 1, b = anything
b) a = anything, b = 0
c) a = -1 and b = even


Yes, I suppose you are right. This can only be solved by treating it as a special case. If we have Q^(Q-3) = 2, then we'll have to go through the steps like what I had before. Just wonder if I would be missing any values by those steps ... (I certainly missed one when the equation is Q^(Q-3) = 1. :( )
  [#permalink] 16 Mar 2005, 09:34
    Similar topics Author Replies Last post
Similar
Topics:
If p,q, r, s and t are nonzero integers, is p/r = s/t ? 1) IgnitedMind 3 12 Sep 2008, 10:31
Is \frac{4Q}{11} a positive integer? 1. Q is a prime number arjtryarjtry 5 09 Sep 2008, 17:53
Is 4Q/11 is positive integer? (1) Q is prime number. (2) 2Q Juaz 13 07 Oct 2007, 13:47
1 Find a nonzero integer Q (1) Q is a prime root of Q^Q=Q^3 yezz 21 12 Oct 2006, 15:48
The number q is a positive integer. Is the square root of q kevincan 5 09 Jul 2006, 03:21
Display posts from previous: Sort by

Find a nonzero integer Q (1) Q is a prime root of Q^Q=Q^3

  Question banks Downloads My Bookmarks Reviews Important topics  


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

Powered by phpBB © phpBB Group and phpBB SEO

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®.