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

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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), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80

OA is D
Last edited by rakeshd347 on Wed Oct 09, 2013 4:03 am, edited 1 time in total.
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by theCodeToGMAT » Tue Oct 08, 2013 11:24 pm
Is the Answer [spoiler]{D}[/spoiler]????

Steps:
f(m) = 9* f(v)

f(m)/f(v) = 9 = 3^2

That means .. the powers of 2 & 5 are exactly same.

So,
Let consider different numbers with hundreds & units digit same in both numbers and Tens digit difference of 2
131 - 111 == 20
255 - 275 == 20

So, Answer [spoiler] {D}[/spoiler]
R A H U L

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by GMATGuruNY » Wed Oct 09, 2013 3:53 am
rakeshd347 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), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80
For every 3-digit integer xyz, f(xyz) = (2^x)(3^y)(5^z).

f(v):
Let x=1, y=1, and z=1, implying that v=111.
Then:
f(v) = f(111) = 2¹3¹5¹ = 30.

f(m):
f(m) = 9f(v) = 9*30 = 270.
Thus:
f(xyz) = 270
(2^x)(3^y)(5^z) = 270
(2^x)(3^y)(5^z) = 2¹3³5¹
xyz = 131.
Result:
m=131.

Thus:
m-v = 131-111 = 20.

The correct answer is D.
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