here is my opinion
f(m)= 2^x * 3^y * 5^z
f(v)= 2^x' * 3^y' * 5^z'
=2^(x-x')* 3^(y-y') * 5^ (z-z')=9
so x-x'=0, z-z'=0, y-y'=2
y is the tenth digit of the three digit number. y-y'=2 is the tenth digit of the number (m-v)
x-x'=0 is the hunderd digit, and z-z' is the unit.
so m-v=20 |