How to solve

let 3 digit no be
f(xyz) = (2^x) * (3^y) * (5^z)

lets say no is 123
the
f(123)= 2*(3^2)*(5^3)


Now we have two numbers and one is nine time the other
9 is actually square of three
so
yout digit at second position in one number is greather than the digit at second position in another number
and this greater is actually 2 points (because of 9)

so 2 points at 10th place will make 20.


Let me give an example:
123
f(123)= 2*(3^2)*(5^3)

so now no2 is =
= 9*f(123)
= 2*(3^2)*(5^3) * 9
= 2*(3^(2+2))*(5^3)
= f(143)

the difference of 143 and 123 is 20.

You can solve the above qn for xyz and will get the same answer.


Hope now you understood that what this question si asking.

jazzcat4u 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

The answer is D. Need answers to 2 questions regarding this problem - (the first question is more important to me though):

(1) Can someone please translate what this question is asking. Based on the little that I know of functions, is it safe to assume that I can rephrase this question as asking?

(2) Can someone please show the work involved on how to solve this problem?

Thanks!