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

Question

Find a nonzero integer Q 

(1) Q is a prime root of Q^Q=Q^3 
(2) Q^2=3^2

banerjeea_98 · Answer

"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

prep_gmat · Answer

St.1 says Q^Q right and not Q*Q?.

MA · Answer

from i, q= 1 and 3 but q is prime, so q=3. so suff.
from ii, q=-3 or 3. so insuff.

 

Could you pls explain your approach?

banerjeea_98 · Answer

oops, my mistake, didn't see it is Q^Q and not Q^2....still A is suff.... :oops:

ketanm · Answer

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

MA · Answer

I think u r wright. i was unclear and confused with prime root. but the safe side is that answer is A.

lastochka · Answer

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

HongHu · Answer

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?

lastochka · Answer

you're right, -1 is not one of the possible answers

HongHu · Answer

Actually -1 IS one of the possible answers. My question is how do you solve for it. :???

kapslock · Answer

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.

HongHu · Answer

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.

)

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

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