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f(n) = (2^x)*(3^y)*(5^z), where x, y, and z are hundreds',

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Manager
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f(n) = (2^x)*(3^y)*(5^z), where x, y, and z are hundreds', [#permalink] New post 29 Jul 2007, 10:35
f(n) = (2^x)*(3^y)*(5^z), where x, y, and z are hundreds', tens', units' digit of n. If f(u)=9f(v), u-v=?

A. 20
B. 12
C. 14
D. 40
E. 80
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 [#permalink] New post 29 Jul 2007, 18:39
Answer A is correct

f(v)=2^x*3^y*5^z and it is given that x is the hundreds y is the tenth and z is the units digit of v there fore v=100x+10y+z

f(u)=9f(v)=9(2^x*3^y*5^z)=3^2*2^x*3^y*5^z=2^x*3^(y+2)*5^z
u=100x+10(y+2)+z=100x+10y+20+z

u-v=100x+10y+20+z-100x-10y-z=20
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 [#permalink] New post 29 Jul 2007, 18:51
eileen1017 wrote:
Answer A is correct

f(v)=2^x*3^y*5^z and it is given that x is the hundreds y is the tenth and z is the units digit of v there fore v=100x+10y+z

f(u)=9f(v)=9(2^x*3^y*5^z)=3^2*2^x*3^y*5^z=2^x*3^(y+2)*5^z
u=100x+10(y+2)+z=100x+10y+20+z

u-v=100x+10y+20+z-100x-10y-z=20


Wow, thanks! I couldn't get this one.... :?
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 [#permalink] New post 30 Jul 2007, 11:38
I get A..

here is how...

f(n)=2^x + 3^y + 5^z

where x=100, y=10 and z=1

so lets just look at the x, y and z... (100X+10y+z)

f(u)=9f(v)

well 9=3^2

f(u)=9f(v)=>(100x+10(y+2)+z); f(v)=(100x+10y+z)

f(u)-f(v)=(100x+10y+20+z)-(100x+10y+z); all the terms cancel out, excpt for 20...

u-v=20

A it is..
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 [#permalink] New post 30 Jul 2007, 17:48
Great the answer is A thanks.
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 [#permalink] New post 30 Jul 2007, 17:55
eileen1017 wrote:
Answer A is correct

f(v)=2^x*3^y*5^z and it is given that x is the hundreds y is the tenth and z is the units digit of v there fore v=100x+10y+z

f(u)=9f(v)=9(2^x*3^y*5^z)=3^2*2^x*3^y*5^z=2^x*3^(y+2)*5^z
u=100x+10(y+2)+z=100x+10y+20+z

u-v=100x+10y+20+z-100x-10y-z=20


good job eileen1017. got A too.
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Re: FUNC 1 f(n) [#permalink] New post 30 Jul 2007, 22:33
nfa1rhp wrote:
f(n) = (2^x)*(3^y)*(5^z), where x, y, and z are hundreds', tens', units' digit of n. If f(u)=9f(v), u-v=?

A. 20
B. 12
C. 14
D. 40
E. 80


2^x1*3^y1*5^z1 = 3^2 * 2^x1*3^y1*5^z1 = 2^x1*3^(y1+2)*5^z1

So

x1(y1+2)z1 - x1y1z1 = 20.
Re: FUNC 1 f(n)   [#permalink] 30 Jul 2007, 22:33
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f(n) = (2^x)*(3^y)*(5^z), where x, y, and z are hundreds',

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