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Indices and Algebra

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Manager
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Indices and Algebra [#permalink]

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New post 17 Mar 2009, 20:12
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^x*3^y*5^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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Re: Indices and Algebra [#permalink]

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New post 17 Mar 2009, 21:08
milind1979 wrote:
The function f is defined for each positive three-digit integer n by f(n) = 2^x*3^y*5^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


D.20
lets m = abc and v= efg
f(m)= 2^a*3^b*5^c
f(v) = 2^e*3^f*5^g

f(m)= 9*f(v) = 3^2*(2^e*3^f*5^g) = 2^e*3^f+2*5^g

2^a*3^b*5^c = 2^e*3^f+2*5^g

=> units digits of m & v are same, and 100s digit of m & v are same, b = f+2 i.e only the 10s digit is of m is greater than v by 2
so we have a(f+2)c - a f c = 20

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Re: Indices and Algebra [#permalink]

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New post 17 Mar 2009, 21:54
Got D as well.
Took the same route.

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Senior Manager
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Re: Indices and Algebra [#permalink]

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New post 19 Mar 2009, 09:10
Agree with D. But to simplyfy the solution, by reducing the number of variables,

Let V be abc. Then f(v) = \(2^a * 3^b * 5^c\).
Given, f(m) = 9f(v) ==> f(m) = \(2^a\) * \(3^(b+2)\) * \(5^c\)==> m = a(b+2)c.

Note: - Here m and v are expressed in terms face value.

m-v = a(b+2)c - abc. Consider the place value of each digit.

(100a + 10(b+2) + c) - (100a + 10b + c). Solving this we get 20.

So m-v = 20.

Kudos [?]: 376 [0], given: 15

Re: Indices and Algebra   [#permalink] 19 Mar 2009, 09:10
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