Find a nonzero integer Q (1) Q is a prime root of Q^Q=Q^3 : DS Archive
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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]

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11 Mar 2005, 16:11
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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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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
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11 Mar 2005, 19:25
St.1 says Q^Q right and not Q*Q?.
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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?
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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....
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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
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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.
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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
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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.
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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
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15 Mar 2005, 20:04
Actually -1 IS one of the possible answers. My question is how do you solve for it.
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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.
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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. )
16 Mar 2005, 09:34
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# Find a nonzero integer Q (1) Q is a prime root of Q^Q=Q^3

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