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f(xyz) = 5^x2^y3^z.

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f(xyz) = 5^x2^y3^z.

by abhi332 » Wed Feb 24, 2010 11:44 am
For a three-digit number xyz, where x, y, and z are the digits of the number, the function f(xyz) = 5^x2^y3^z. If f(abc) = 3 * f(def), what is the value of abc - def?

(A) 1
(B) 2
(C) 3
(D) 9
(E) 27


Ans : A
Last edited by abhi332 on Wed Feb 24, 2010 10:48 pm, edited 1 time in total.
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by firdaus117 » Wed Feb 24, 2010 11:56 am
abhi332 wrote:For a three-digit number xyz, where x, y, and z are the digits of the number, the function f(xyz) = 5x2y3z. If f(abc) = 3 * f(def), what is the value of abc - def?

(A) 1
(B) 2
(C) 3
(D) 9
(E) 27


Ans : A
Tried very hard.Didn't get that.Are you sure the question is right?Source please.
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by abhi332 » Wed Feb 24, 2010 12:04 pm
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by ajith » Wed Feb 24, 2010 9:31 pm
abhi332 wrote:For a three-digit number xyz, where x, y, and z are the digits of the number, the function f(xyz) = 5x2y3z. If f(abc) = 3 * f(def), what is the value of abc - def?

(A) 1
(B) 2
(C) 3
(D) 9
(E) 27


Ans : A
f(xyz) = 5^x*2^y*3^z
f(abc) =(5^a*2^b*3^c) = 3*(5^d*2^e*3^f)

=>(5^a*2^b*3^c) = (5^d*2^e*3^f+1)
a=d; b=e; c=f+1
[spoiler]
Now it is evident that abc-def = 1[/spoiler]

[abhi332, Please post the powers as powers! 5x2y3z and 5^x*2^y*3^z are slightly different, and why is it in the DS forum?]
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by abhi332 » Wed Feb 24, 2010 10:49 pm
Thanks Ajit, I have edited the question. is there any option to move this post to PS section
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by ajith » Wed Feb 24, 2010 11:12 pm
Moved To PS Forum
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