, 2018

Since $80 = 2^4 \cdot 5$, its positive divisors are $1, 2, 4, 5, 8, 10, 16, 20, 40, 80$.

For an integer $n$ to share exactly two positive common divisors with $80$, these divisors must be either $1$ and $2$ or $1$ and $5$. ($1$ is a common divisor of any two integers. The second common divisor must be a prime number since any composite divisor will cause there to be at least one more common divisor which is prime.)

Since $1 \leq n \leq 30$ and $n$ is a multiple of $2$ or of $5$, then the possible values of $n$ come from the list $$2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20, 22, 24, 25, 26, 28, 30$$ We remove the multiples of $4$ from this list (since they would share at least the divisors $1,2,4$ with $80$) and the multiples of $10$ from this list (since they would share at least the divisors $1,2,5,10$ with $80$).

This leaves the list $$2, 5, 6, 14, 15, 18, 22, 25, 26$$ The common divisors of any number from this list and 80 are either $1$ and $2$ or $1$ and $5$.

There are $9$ such integers.
We start with $f(50)$ and apply the given rules for the function until we reach $f(1)$: $$\text{f(50) = f(25) (since 50 is even and 1 2 (50)=25 ) = f(24) + 1 (since 25 is odd and 25-1 = 24 ) = f(12) + 1 ( 1 2 (24)=12 ) = f(6) + 1 ( 1 2 (12)=6 ) = f(3) + 1 ( 1 2 (6)=3 ) = (f(2) + 1) + 1 ( 3 -1 = 2 ) = f(1) + 1 + 1 ( 1 2 (2)=1 ) = 1 + 1 + 1 ( f(1)=1 ) = 3}$$ Therefore, $f(50)=3$.