Carcass wrote:
The function f is defined for each positive three-digit integer n by \(f(n) = 2^x3^y5^z\) , 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 \(m-v=\) ?
n = xyz
\(f(n) = 2^x3^y5^z\)
Let m = abc and v = efg
then, \(f(m) = 2^a3^b5^c\)
\(f(v) = 2^e3^f5^g\)
We are given, \(f(m)=9*f(v) = 3^2 f(v)\)
i.e. \(2^a3^b5^c = 2^e3^{f+2}5^g\)
Using the property of exponents - When the bases are equal, then the powers are also equal. We have,
a = e
b = f+2
c = g
So, m - v = 020
Hence, the difference is 20