Suppose $f$ is a function that satisfies $f(2)=20$ and $f(2 n)+n f(2)=f(2 n+2)$ for all positive integers $n$. What is the value of $f(10)$ ?
With $n=1$, we have $f(2 \times 1)+1 \times f(2)=f(2 \times 1+2)$ which simplifies to $f(2)+f(2)=f(4)$, so $f(4)=20+20=40$.

With $n=2$, we have $f(2 \times 2)+2 \times f(2)=f(2 \times 2+2)$ which simplifies to $f(4)+2 f(2)=f(6)$.

Since $f(4)=40$ and $f(2)=20$, we have $f(6)=40+2(20)=80$.

Continuing with $n=3$, we have $f(6)+3 f(2)=f(8)$, so $f(8)=80+3(20)=140$.

With $n=4$, we have $f(8)+4 f(2)=f(10)$, so $f(10)=140+4(20)=220$.

ANSWER: 220