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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]

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New post 09 Oct 2008, 05:27
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) = 9 f(v), then m - v =

A. 8
B. 9
C. 18
D. 20
E. 80

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Re: math ---function  [#permalink]

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New post 09 Oct 2008, 06:47
20 :-)
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New post 09 Oct 2008, 07:56
how did u get 20 guys :)
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New post 09 Oct 2008, 09:17
albany09 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) = 9 f(v), then m - v =

A. 8
B. 9
C. 18
D. 20
E. 80


100 <= n <= 999
n = 100x + 10y + z
f(n) = \((2^x)(3^y)(5^z)\)

Assume:
m = 100(a) + 10(b) + (c) ---------- (1)
v = 100(d) + 10(e) +(f) ---------- (2)

f(m) = \((2^a)(3^b)(5^c)\) --------- (3)
f(v) = \((2^d)(3^e)(5^f)\) ---------- (4)

We know that f(m)/f(v) = 9 = \(3^2\)
\((2^a)(3^b)(5^c)\) / \((2^d)(3^e)(5^f)\) = \(3^2\) ---------- From (3) & (4)

So, a must be equal to d
,and c must be equal to f.

m - v = 10(b) - 10(e) = 10(b - e) ----- from equation (1) - (2) ---------- (5)
,And
(3^(b-e)) = \(3^2\)
b - e = 2 ---------- (6)

m - v = 10*2 = 20 ---------- From (5) & (6)

The answer is D.
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New post 09 Oct 2008, 11:22
tnx... wish if theres something more simple..

i hate functions!
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New post 09 Oct 2008, 23:52
spiridon wrote:
tnx... wish if theres something more simple..

i hate functions!



If you look at the problem, it states that f(m) = 9f(v) and since the functions will be in the form of 2^x*3^y*5^z hence, 10s digit of m must be greater than that of v by 2 (as 9 = 3^2).

Thus, if all the other digits remain the same in m and v but 10s digit is greater by 2 in m then m-v should simly be 2*10 = 20.

I hope, this helps.

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Re: math ---function &nbs [#permalink] 09 Oct 2008, 23:52
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