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# The function f is defined for each positive three-digit

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Manager
Joined: 14 May 2008
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The function f is defined for each positive three-digit [#permalink]  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

Thank you!
Director
Joined: 01 Jan 2008
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Re: function [#permalink]  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

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
Intern
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Re: function [#permalink]  14 Jun 2008, 20:42
Maratikus, How did y(m)=y(v)+2, translate to m-v=20?

Can you help explain?
Manager
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Re: function [#permalink]  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
Manager
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Re: function [#permalink]  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]  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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# The function f is defined for each positive three-digit

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