R^c)(R^d)(R^e)=12^-12. If R>0, and c,d and e are each : Quant Question Archive [LOCKED]
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# R^c)(R^d)(R^e)=12^-12. If R>0, and c,d and e are each

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Manager
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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 re-post 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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KL

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25 Jul 2003, 07:28
(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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25 Jul 2003, 11:15
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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25 Jul 2003, 12:30
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 non-integer
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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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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KL

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23 Jan 2004, 11:07
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
23 Jan 2004, 11:07
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