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For a threedigit number xyz, where x, y, and z are the [#permalink]
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Updated on: 12 Jul 2014, 03:44
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For a threedigit 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 abcdef ? (A) 1 (B) 2 (C) 3 (D) 9 (E) 27
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Originally posted by JCLEONES on 29 Jan 2008, 13:44.
Last edited by Bunuel on 12 Jul 2014, 03:44, edited 1 time in total.
Edited the question and added the OA.



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Re: For a threedigit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 13:49
JCLEONES wrote: For a threedigit 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 abcdef ? (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 threedigit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 13:53
Incognito, Could you please detail your work? Thanks incognito1 wrote: JCLEONES wrote: For a threedigit 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 abcdef ? (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 threedigit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 13:57
JCLEONES wrote: For a threedigit 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 abcdef ? (A) 1 (B) 2 (C) 3 (D) 9 (E) 27 f(abc)/f(def)=5^(ad) 2^(be) 3^(cf) = 3^1 > a = d, b = e, c = f+1 > A is the answer



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Re: For a threedigit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 14:03
testgmat wrote: Incognito, Could you please detail your work? Thanks incognito1 wrote: JCLEONES wrote: For a threedigit 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 abcdef ? (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 3digit numbers, and c = f + 1, the difference between abc and def would be 1. Hope that helps!
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Re: For a threedigit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 19:08
JCLEONES wrote: For a threedigit 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 abcdef ? (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^(ad)*2^(be)*3^(cf) = 3^1 So,5^(ad)*2^(be)*3^(cf) = 3^1*5^0*2^0 S0, ad =0, be =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 threedigit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 21:32
JCLEONES wrote: For a threedigit 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 abcdef ? (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 threedigit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 21:56
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!



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Re: For a threedigit number xyz, where x, y, and z are the [#permalink]
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12 Jul 2014, 03:44



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Re: For a threedigit number xyz, where x, y, and z are the [#permalink]
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06 Nov 2015, 21:57
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 abcdef = 1



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Re: For a threedigit number xyz, where x, y, and z are the [#permalink]
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23 Feb 2017, 20:10
https://gmatclub.com/forum/thefunction ... 00847.htmlThe answer is "A"  VeritasPrepKarishma gives an incredibly clear explanation in a similar problem linked above.



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Re: For a threedigit number xyz, where x, y, and z are the [#permalink]
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19 May 2018, 13:38
testgmat wrote: Incognito, Could you please detail your work? Thanks incognito1 wrote: JCLEONES wrote: For a threedigit 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 abcdef ? (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). f(abc) = 5^a * 2^b *3^c f(def) = 5^d * 2^e * 3^f since, f(abc) = 3*f(def) 5^a * 2^b *3^c = 5^d * 2^e * 3^(f+1) so, a=d b=e c=f+1 so, for example, if abcis 123, then def is 124, hence difference is 1.




Re: For a threedigit number xyz, where x, y, and z are the
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