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The function f is defined for each positive three-digit

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The function f is defined for each positive three-digit [#permalink] New post 14 Jun 2008, 05:54
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The function f is defined for each positive three-digit integer n by f(n) = 2x3y5z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80

Please explain also your logic :)
Thank you!
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Re: function [#permalink] New post 14 Jun 2008, 12:03
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quantum wrote:
The function f is defined for each positive three-digit integer n by f(n) = 2x3y5z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80

Please explain also your logic :)
Thank you!


I assume f(n) = (2^x)*(3^y)*(5^z)
since f(m) = 9*f(v) = (3^2)*f(v) -> first and third digits of m and v are the same: x(m) = x(v), z(m) = z(v) and second digit of m is higher than v by 2: y(m) = y(v) + 2 -> m-v = 20 -> D
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Re: function [#permalink] New post 14 Jun 2008, 20:42
Maratikus, How did y(m)=y(v)+2, translate to m-v=20?

Can you help explain?
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Re: function [#permalink] New post 14 Jun 2008, 23:44
Yes OA is D.. I suppose it was a typo in this task and x y z must be exponents...
However I still do not understand how to solve it.
Thank You Maratikus
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Re: function [#permalink] New post 15 Jun 2008, 00:37
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f(m) = 9*f(v)

f(m)/f(v) = 9 = 3^2

2^x(m)*3^y(m)*5^z(m)
------------------------ = 2^0 * 3^2 * 5^0
2^x(v)*3^y(v)*5^z(v)


this will result in:

x(m)-x(v) = 0 , hence first hundred's digit is the same
y(m)-y(v) = 2 , hence tens digit differs by two.
z(m)-z(v) = 0 , hence units digit is the same

Since the tens digit differs by two between the two and the first and last digit remain the same,
the difference between the two numbers is 2*10 i.e 20.

Maratikus, thanks for the correction. Also, referring the digits as x(m), y(m) and so on really helps!
Kudos+1 for Maratikus.
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Re: function [#permalink] New post 15 Jun 2008, 01:39
Same as above. OA is D

It's essential to understand extracting 3^2 and placing in correct numerical sequence. The hardest part about this question is trying to figure out what it is exactly that you're solving for.
Re: function   [#permalink] 15 Jun 2008, 01:39
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