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Find a nonzero integer Q (1) Q is a prime root of Q^Q=Q^3 [#permalink]
 10 Jun 2003, 00:07
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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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SVP
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(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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Intern
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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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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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Director
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Answer: A
Even roots can have +/- numbers and second condition is not sufficient
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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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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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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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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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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. _________________
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