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Director
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Edited: For a threedigit number xyz, where x, y, and z
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Updated on: 08 Oct 2007, 00:12
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Edited:
For a threedigit number xyz, where x, y, and z are the digits of the number, the function f(xyz) = 5^x*2^y*3^z. If f(abc) = 3 * f(def), what is the value of abc  def?
(A) 1
(B) 2
(C) 3
(D) 9
(E) 27
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Originally posted by GK_Gmat on 07 Oct 2007, 23:02.
Last edited by GK_Gmat on 08 Oct 2007, 00:12, edited 1 time in total.



Director
Joined: 22 Aug 2007
Posts: 543

Re: PS: Functions
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07 Oct 2007, 23:40
GK_Gmat wrote: For a threedigit 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
Explain your answer.
though it does not make it easier, still, can you please clarify?
5x2y3z is 5x2y3z, where 5,x,2,y,3,z each are digits?



Director
Joined: 03 May 2007
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Schools: University of Chicago, Wharton School

Re: PS: Functions
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07 Oct 2007, 23:56
GK_Gmat wrote: For a threedigit 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
Explain your answer.
f(xyz) = 5x2y3z
f(abc) = 5a 2b 3c = 3(5d 2b 3f)
f(def) = 5d 2b 3f
but the question ask abut the value of abc  def = ?
to be f(abc) = 5a 2b 3c = 3(5d 2b 3f) true; c has to be 3 times of f and a = d and b = e.
so suppose if def = 541, abc = 543
then the difference = 543  541 = 2



Manager
Joined: 29 Jul 2007
Posts: 178

Is it A?
say if def = 122
f(def) = (5^1)*(2^2)*(3^2)
f(abc) = (5^1)*(2^2)*(3^3) or 3*[(5^1)*(2^2)*(3^2)] or 3* f(def)
then abc = 123
so 123122 = 1, A.



Intern
Joined: 15 Mar 2007
Posts: 43

Re: PS: Functions
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10 Oct 2007, 09:31
GK_Gmat wrote: Edited:
For a threedigit number xyz, where x, y, and z are the digits of the number, the function f(xyz) = 5^x*2^y*3^z. If f(abc) = 3 * f(def), what is the value of abc  def?
(A) 1 (B) 2 (C) 3 (D) 9 (E) 27
Explain your answer.
A.
a=1; b=2; c=3
f(abc) = 540 = 3 * f(def) ==> 3 * 180 ==> f(def) = 180 = 5^1*2^2*3^2
abc  def = 123  122 = 1



Intern
Joined: 03 Mar 2006
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Re: PS: Functions
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11 Oct 2007, 00:27
GK_Gmat wrote: Edited:
For a threedigit number xyz, where x, y, and z are the digits of the number, the function f(xyz) = 5^x*2^y*3^z. If f(abc) = 3 * f(def), what is the value of abc  def?
(A) 1 (B) 2 (C) 3 (D) 9 (E) 27
Explain your answer.
My ans is A.
We know that f(abc)=5^a*2^b*3^c = 3f(def)=3*5^d*.... (similarly)
Therefore, from here we have a=d, b=e, c=d+1
So the diff bwt 2 number abc and def must always be 1. Ans A.
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Joined: 10 Jun 2007
Posts: 1357

Re: PS: Functions
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11 Oct 2007, 05:21
GK_Gmat wrote: Edited:
For a threedigit number xyz, where x, y, and z are the digits of the number, the function f(xyz) = 5^x*2^y*3^z. If f(abc) = 3 * f(def), what is the value of abc  def?
(A) 1 (B) 2 (C) 3 (D) 9 (E) 27
Explain your answer.
A it is
5^a * 2^b * 3^c = 5^d * 2^e * 3^(f+1)
ad=0
be=0
bf=1
abc  def = 100*(ad) + 10*(be) + (bf) = 1



Manager
Joined: 07 Sep 2007
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Re: PS: Functions
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11 Oct 2007, 08:20
GK_Gmat wrote: Edited:
For a threedigit number xyz, where x, y, and z are the digits of the number, the function f(xyz) = 5^x*2^y*3^z. If f(abc) = 3 * f(def), what is the value of abc  def?
(A) 1 (B) 2 (C) 3 (D) 9 (E) 27
Explain your answer.
A.
Great question, this one is fun to think about!
We first realize that 5, 2, and 3 are all prime factors.
We know that for since f(abc) = 3*f(def) that there must be one more "3" in abc than def, and no other changes between abc & def.
Therefore, since the c/f digit corresponds with the 3, we know that c must be 1 greater than f, and that a=d and b=e.



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Re: Edited: For a threedigit number xyz, where x, y, and z
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26 Nov 2018, 16:39
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