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R^c)(R^d)(R^e)=12^12. If R>0, and c,d and e are each [#permalink]
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25 Jul 2003, 06:53
This topic is locked. If you want to discuss this question please repost it in the respective forum. (R^c)(R^d)(R^e)=12^12.
If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
A) 12
B)10
C)9
D)6
E)1
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(R^c)(R^d)(R^e)=12^12.
If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
A) 12 B)10 C)9 D)6 E)1
My answer is (A) i.e. 12
(R^c)(R^d)(R^e)=12^12
=> r^(c+d+e) = 12^12
=> r^(c+d+e) = (2*sqrt3)^24 (see below how I got this)
so, lowest value of c+d+e = 24
which leads to the smallest value of c as 12.
12^12 = (4*3)^12 = 4^12 * 3^12 = [(2^2)]^12 * [(sqrt3) ^ 2] ^ 12 = 2^24 * (sqrt3)^24 = [2 * sqrt(3)] ^ 24



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If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
You are assuming that R is an interger >0. But in the problem c,d,e are defined as negative intergers. R may not be an integer  isn't it?
If we take, R=2√3; c=12, d=7, e=5
then, c+d+e=24
so, R^(c+d+e) = (2√3)^24 = 12^12
That's why I think answer is A. Feedback pls.



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prakuda2000 wrote: If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
You are assuming that R is an interger >0. But in the problem c,d,e are defined as negative intergers. R may not be an integer  isn't it?
If we take, R=2√3; c=12, d=7, e=5 then, c+d+e=24
so, R^(c+d+e) = (2√3)^24 = 12^12 That's why I think answer is A. Feedback pls.
But how can you 'guess'tify R to be a noninteger
Even if c+d+e = 24 , why C is 12 , why can't it be 22 ?
Per the question C, D and E need to be negetive integers , So each one of them may atmost be 1.
So this gives us C to be 22 ?
I feel (A) should be  22 (may be a typo ).
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Re: Some algebra practice [#permalink]
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26 Jul 2003, 02:48
Lynov Konstantin wrote: (R^c)(R^d)(R^e)=12^12.
If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
A) 12 B)10 C)9 D)6 E)1
Sorry, guys. I made a typing mistake in the question.
The condition should be read as follows:
(R^c)(R^d)(R^e)=R^c+d+e= R^12
If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
A) 12
B)10
C)9
D)6
E)1
Forgive me please and give it another shot.
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This post received KUDOS
R^(a+b+c) = 12^(12)
so a+b+c = 12
Given a,b and c are ve distincive integers
abc can be any of(0,0,12), (0,1,11),(1,1,10),(1,2,9)
Lowest is 11 but that choice is not given. Next lowest is
(1,2,9) where integers are distinct and c = 9



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Re: R^c)(R^d)(R^e)=12^12. If R>0, and c,d and e are each [#permalink]
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04 Nov 2017, 12:13
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Re: R^c)(R^d)(R^e)=12^12. If R>0, and c,d and e are each
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