Question 5.

Looking at the additional information about the prime numbers should make one realise that they are the key to solving the question. 

f(16000) can be written as $f\left(2^8\times\ 5^4\right)$

Now, we can try to find these individual values:

For any prime p: f(p)=1

$f\left(p^2\right)=f\left(p\right)f\left(p\right)+f\left(p\right)+f\left(p\right)=1+1+1=3$

$f\left(p^3\right)=f\left(p^2\right)f\left(p\right)+f\left(p^2\right)+f\left(p\right)=3+3+1=7$

This way, we can find the function output for any prime number raised to a power. 

We can see that each new exponent is twice the previous output +1, solving this way till prime raised to power 8

$f\left(p^4\right)=7+7+1=15$

$f\left(p^5\right)=15+15+1=31$

$f\left(p^6\right)=31+31+1=63$

$f\left(p^7\right)=63+63+1=127$

$f\left(p^8\right)=127+127+1=255$

Using these values in the original expression of $f\left(2^8\times\ 5^4\right)=f\left(2^8\right)f\left(5^4\right)+f\left(2^8\right)+f\left(5^4\right)$ we get

$f\left(2^8\times\ 5^4\right)=\left(255\times\ 15\right)+255+15=4095$

Therefore, Option A is the correct answer.