Calculate the number of trailing zeros in equation f(n) = f(n-1) * f(n-2) where f(0) and f(1) are any given arbitary numbers This question is doable if you can calculate the number by multiplying f(n-1) and f(n-2).  

Is it possible to do this question if we entirely want to skip multiplying these 2 numbers and still be able to calculate say the number of trailing zeros in f(20).

Why do I want to skip multiplying these two numbers?
Because as this function grows exponentially, so does the value. We might get an integer overflow on some languages. Hence.
Also, our aim is to calculate the number of trailing zeros and not calculate the number itself.

What I tried. To try and calculate trailing zeros and f(n-1) and f(n-2), So that I deduce the number of trailing zeros in f(n) and go on from there. 

But to calculate trailing zeros in f(n-1), I need to know what f(n-1) is? Which is where my whole approach fails.
Assuming f(0) and f(1) are integers, find which powers of two and of five are factors of f(0) and f(1). The powers in f(n) can be calculated like a Fibonacci sequence and the number of trailing zeroes is the smaller of both sequences. 

The Fibonacci series is defined by $fib(0)=0$, $fib(1)=1$, and for $n≥2$ we have $fib(n) = fib(n-2)+fib(n-1)$, so the series starts (0, 1, 1, 2, 3, 5, 8, 13 ...). If we start with different values and use the same recursion formula, we get for example the sequence (1, 0, 1, 1, 2, 3, 5, 8, 13...). If we define $fib(-1)=1$, then the n-th element of this sequence equals $fib(n-1)$. 

The sequence starting with (x, y) and the same recursion formula (x, y, x+y, x+2y, 2x + 3y ...) equals $x \cdot fib(n-1) + y \cdot fib (n)$. 

Let $f(0)$ have the factor $2^x$ and $f(1)$ the factor $2^y$ and no higher power of two, then $f(n)$ has the factor $2^z$, where $z = x \cdot fib(n-1) + y \cdot fib (n)$. 

Likewise, let $f(0)$ have the factor $5^a$ and $f(1)$ the factor $5^b$ and no higher power of five, then $f(n)$ has the factor $5^c$, where $c = a \cdot fib(n-1) + b \cdot fib (n)$. 

The number of trailing zeroes in $f(n)$ is 

    $min(z_n, c_n)$ where 

    $z_n = x \cdot fib(n-1) + y \cdot fib (n)$
    $c_n = a \cdot fib(n-1) + b \cdot fib (n)$