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Edited: For a three-digit number xyz, where x, y, and z

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Edited: For a three-digit number xyz, where x, y, and z [#permalink]

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New post 08 Oct 2007, 00:02
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Edited:

For a three-digit 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.

Last edited by GK_Gmat on 08 Oct 2007, 01:12, edited 1 time in total.

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Re: PS: Functions [#permalink]

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New post 08 Oct 2007, 00:40
GK_Gmat 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

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?

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Re: PS: Functions [#permalink]

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New post 08 Oct 2007, 00:56
GK_Gmat 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

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

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 [#permalink]

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New post 10 Oct 2007, 09:56
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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 123-122 = 1, A.

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Re: PS: Functions [#permalink]

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New post 10 Oct 2007, 10:31
GK_Gmat wrote:
Edited:

For a three-digit 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

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Re: PS: Functions [#permalink]

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New post 11 Oct 2007, 01:27
GK_Gmat wrote:
Edited:

For a three-digit 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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Re: PS: Functions [#permalink]

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New post 11 Oct 2007, 06:21
GK_Gmat wrote:
Edited:

For a three-digit 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)
a-d=0
b-e=0
b-f=1
abc - def = 100*(a-d) + 10*(b-e) + (b-f) = 1

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Re: PS: Functions [#permalink]

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New post 11 Oct 2007, 09:20
GK_Gmat wrote:
Edited:

For a three-digit 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 three-digit number xyz, where x, y, and z [#permalink]

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Re: Edited: For a three-digit number xyz, where x, y, and z   [#permalink] 02 Sep 2017, 11:01
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