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For a three-digit number xyz, where x, y, and z are the

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

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For a three-digit number xyz, where x, y, and z are the digits of the number, 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Jul 2014, 02:44, edited 1 time in total.
Edited the question and added the OA.
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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]

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New post 29 Jan 2008, 12:49
JCLEONES wrote:
For a three-digit number xyz, where x, y, and z are the digits of the number,
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


Since f(abc) = 3*f(def), I would assume that f = c - 1 from the function above.

The answer should be (A).
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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]

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New post 29 Jan 2008, 12:53
Incognito,

Could you please detail your work?

Thanks

incognito1 wrote:
JCLEONES wrote:
For a three-digit number xyz, where x, y, and z are the digits of the number,
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


Since f(abc) = 3*f(def), I would assume that f = c - 1 from the function above.

The answer should be (A).
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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]

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New post 29 Jan 2008, 12:57
JCLEONES wrote:
For a three-digit number xyz, where x, y, and z are the digits of the number,
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


f(abc)/f(def)=5^(a-d) 2^(b-e) 3^(c-f) = 3^1 -> a = d, b = e, c = f+1 -> A is the answer
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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]

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New post 29 Jan 2008, 13:03
testgmat wrote:
Incognito,

Could you please detail your work?

Thanks

incognito1 wrote:
JCLEONES wrote:
For a three-digit number xyz, where x, y, and z are the digits of the number,
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


Since f(abc) = 3*f(def), I would assume that f = c - 1 from the function above.

The answer should be (A).


Sure. The function consists purely of powers of 5, 2 and 3. None of which are multiples of each other.

Since f(abc) = 3*f(def), and since the function is entirely made up of multiples of 5, 2 and 3, it would imply that the only difference is a power of 3. This would mean that a = d (since there is no multiple of 5 from the functions output), b = e (no multiple of 2 in the functions output) and c = f + 1 (since there is a multiple of 3 in the two functions output). Since abc and def are 3-digit numbers, and c = f + 1, the difference between abc and def would be 1. Hope that helps!
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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]

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New post 29 Jan 2008, 18:08
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JCLEONES wrote:
For a three-digit number xyz, where x, y, and z are the digits of the number,
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

This is how I solved.
f(abc) = 3 * f(def)
So, 5^a*2^b*3^c = 3*[5^d*2^e*3^f]
So, (5^a*2^b*3^c)/(5^d*2^e*3^f) = 3
So, 5^(a-d)*2^(b-e)*3^(c-f) = 3^1
So,5^(a-d)*2^(b-e)*3^(c-f) = 3^1*5^0*2^0
S0, a-d =0, b-e =0 and c -f =1.
So, a =d, b = e and c = f +1
So abc - def is equal to abc - abf
And since c = f + 1, differrence is 1
Answer is A
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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]

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New post 29 Jan 2008, 20:32
JCLEONES wrote:
For a three-digit number xyz, where x, y, and z are the digits of the number,
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


f(abc)=3*f(def) so the only difference is one more multiple of 3 in f(abc). The only way for this to happens is if z is one more. z is the units digit so it's total value is 1 more.

Answer A
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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]

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I went totally tangent on this one.

thought abc is a*b*c and not a number abc

and wasnt sure why every one was thinking its too obvious.

lesson learnt! :shock:
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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]

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

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JCLEONES wrote:
For a three-digit number xyz, where x, y, and z are the digits of the number, 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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Hope it helps.
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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]

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New post 06 Nov 2015, 20:57
oh yeah..tricky one
f(def) = 5^d*2*e*3^f
a is the same multiplied by 3
we have same prime factorization
abc = def+1
abc-def = 1
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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]

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New post 25 Jan 2017, 21:55
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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New post 23 Feb 2017, 19:10
https://gmatclub.com/forum/the-function ... 00847.html

The answer is "A" - VeritasPrepKarishma gives an incredibly clear explanation in a similar problem linked above.
Re: For a three-digit number xyz, where x, y, and z are the   [#permalink] 23 Feb 2017, 19:10
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