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

Question

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.