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.