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Re: For a three-digit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 13:49
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.
Re: For a three-digit 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 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.
Re: For a three-digit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 13:57
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
Re: For a three-digit 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 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!
_________________
Re: For a three-digit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 19:08
3
4
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
Re: For a three-digit number xyz, where x, y, and z are the [#permalink]
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29 Jan 2008, 21:32
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.
Re: For a three-digit 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 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.
close
63% (02:00) correct 
