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The function f is defined for each positive three-digit

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Manager
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Joined: 08 Aug 2008
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The function f is defined for each positive three-digit [#permalink]

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New post 05 Nov 2008, 12:01
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The function f is defined for each positive three-digit integer n by f(n) = 2^x3^y5^z, 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), then m-v = ?
A. 8
B. 9
C. 18
D. 20
E. 80

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Manager
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Re: Func [#permalink]

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New post 05 Nov 2008, 12:04
i guessed the answer as D.

since 2,3,5 are all primes, the corresponding digits would remain same for units and hundredths except for tens which would be +2 and thus the diffrence as 20.

However, i would like to know if there is another way to solve it.
Thanks.

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Manager
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Re: Func [#permalink]

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New post 06 Nov 2008, 05:13
prasun84 wrote:
i guessed the answer as D.

since 2,3,5 are all primes, the corresponding digits would remain same for units and hundredths except for tens which would be +2 and thus the diffrence as 20.

However, i would like to know if there is another way to solve it.
Thanks.


I suppose yours is the quickest way

as \(f(m) = 9f(v) so y_{m} - y_{v} = 2, x_{m} = x_{v}, z_{m} = z_{v}\)

\(m - v = 100x_{m} + 10y_{m} + z_{m} - 100x_{v} + 10y_{v} + z_{v} = 10y_{m} - 10y_{v} = 10*2 = 20\)

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Director
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Re: Func [#permalink]

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New post 06 Nov 2008, 05:48
prasun84 wrote:
The function f is defined for each positive three-digit integer n by f(n) = 2^x3^y5^z, 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), then m-v = ?
A. 8
B. 9
C. 18
D. 20
E. 80


2^x * 3^y * 5^z = 3^2 * (2^a * 3^b * 5^c)

2^(x-a) * 3 ^(y-b) * 5^(z-c) = 3^2

x-a =0
y-b = 2
z-c = 0

20

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Re: Func   [#permalink] 06 Nov 2008, 05:48
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The function f is defined for each positive three-digit

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