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

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Find a nonzero integer Q (1) Q is a prime root of Q^Q=Q^3 [#permalink] New post 09 Jun 2003, 23:07
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Find a nonzero integer Q

(1) Q is a prime root of Q^Q=Q^3
(2) Q^2=3^2
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 [#permalink] New post 11 Jun 2003, 23:28
(1) Q=1, -1, and 3; of which only 3 is prime. Thus, OK
(2) Q= 3 and -3, not enough

Thus, A.
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Re: DS: POWERS [#permalink] New post 21 Feb 2008, 11:37
In stolyar's answer. isn't -3 inadmissible in (ii). Resultantly, only 3 is the prime root in (ii).
so shouldn't the answer be D?
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Re: DS: POWERS [#permalink] New post 21 Feb 2008, 17:02
kanyshkae wrote:
In stolyar's answer. isn't -3 inadmissible in (ii). Resultantly, only 3 is the prime root in (ii).
so shouldn't the answer be D?


Keep in mind that -3 also an integer, negative though.

A is my pick

Q^Q=Q^3
Q^2(Q-1)=0
Q = 0 or 1
Since it's given that Q not= 0, Q=1
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Re: DS: POWERS [#permalink] New post 21 Feb 2008, 23:44
Answer: A

Even roots can have +/- numbers and second condition is not sufficient
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Re: [#permalink] New post 22 Feb 2008, 19:33
stolyar wrote:
(1) Q=1, -1, and 3; of which only 3 is prime. Thus, OK
(2) Q= 3 and -3, not enough

Thus, A.

how come 3 is a root of the equation Q^2=Q^3 ??
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Re: DS: POWERS [#permalink] New post 22 Feb 2008, 20:02
stolyar wrote:
Find a nonzero integer Q

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


I am not able to interpret the statement. "Q is a prime root of Q^Q = Q^3." What does it exactly mean ?
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Re: Re: [#permalink] New post 23 Feb 2008, 05:40
vscid wrote:
stolyar wrote:
(1) Q=1, -1, and 3; of which only 3 is prime. Thus, OK
(2) Q= 3 and -3, not enough

Thus, A.

how come 3 is a root of the equation Q^2=Q^3 ??


The red text should be Q^Q = Q^3
Obviously, 3 is one of the roots of above equation.
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Re: DS: POWERS [#permalink] New post 23 Feb 2008, 05:45
sreehari wrote:
kanyshkae wrote:
In stolyar's answer. isn't -3 inadmissible in (ii). Resultantly, only 3 is the prime root in (ii).
so shouldn't the answer be D?


Keep in mind that -3 also an integer, negative though.

A is my pick

Q^Q=Q^3
Q^2(Q-1)=0Q = 0 or 1
Since it's given that Q not= 0, Q=1


The equation in large text is wrong.
It should be Q^Q - Q^3 = 0
-1, 1 and 3 are the roots of the equation but only 3 is prime.
Stoylar's solution is perfect.
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Re: DS: POWERS [#permalink] New post 23 Feb 2008, 08:03
vshaunak@gmail.com wrote:
sreehari wrote:
kanyshkae wrote:
In stolyar's answer. isn't -3 inadmissible in (ii). Resultantly, only 3 is the prime root in (ii).
so shouldn't the answer be D?


Keep in mind that -3 also an integer, negative though.

A is my pick

Q^Q=Q^3
Q^2(Q-1)=0Q = 0 or 1
Since it's given that Q not= 0, Q=1


The equation in large text is wrong.
It should be Q^Q - Q^3 = 0
-1, 1 and 3 are the roots of the equation but only 3 is prime.
Stoylar's solution is perfect.


yup. got it now!
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Re: DS: POWERS   [#permalink] 23 Feb 2008, 08:03
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