Find $f(10)$ in the functional equation $f(f(n))=3n$ 
For each positive integer $n$, define $f(n)$ such that $f(n+1) > f(n)$ and $f(f(n))=3n$. What is the value of $f(10)?$

This question was really hard for me. Since $n$ is a positive integer and $f(f(n)) = 3n$, I deduced that $1<f(1)<f(f(1))$ so $f(1) = 2$ and I couldn't manage to carry on because if i used $2<f(2)<f(f(2))$, I would get $f(f(2)) = 6$ but I wouldn't know how to work out $f(2)$.
If someone could please show me step by step how to get the answer, I would appreciate it as I would like to know how to get the answer, thanks.
 A: You already deduced that $f(1) = 2$. From the $f(f(n))=3n$ condition, we have that $f(f(1))=f(2)=3$, hence $f(2)=3$. Similarly, we have that $f(3)=f(f(2))=3\cdot 2=6$, and $f(6)=f(f(3))=3\cdot 3=9$, and $f(9)=f(f(6))=3 \cdot 6=18$. So until now we have that 
$$f(1)=2\, , \; f(2)=3 \, , \;  f(3)=6 \, ,\; f(6)=9 \, , \; f(9)=18 \, .$$  
Since $f(n)< f(n+1)$, we have that $f(3)=6 < f(4) < f(5) < 9=f(6)$. As $f(n)$ is integer, the only solution is $f(4)=7$ and $f(5)=8$. Now we have that $f(7)=f(f(4))= 3\cdot 4=12$ and $f(12)=f(f(7))=3\cdot 7=21$. Finally, we can write $f(9)=18 < f(10) < f(11) < 21=f(12)$. So now you can deduce what $f(10)$ is.
A: Well, note that
$$f(f(n))=3n \implies f(f(f(n)))=f(3n)=3f(n)$$
So $f(1)=2 \implies f(3)=6=f(f(2)) \implies f(2)=3$ (because $f$ is injective, easily proved from $f(f(n))=3n$). Now, working our way through:
$$f(6)=3f(2)=9$$
Thus $f(3)<f(4)<f(5)<f(6) \implies f(4)=7, f(5)=8$
$$f(9)=3f(3)=18$$
$$f(12)=3f(4)=21$$
$$\implies 18<f(10)<f(11)<21 \implies f(10)=19$$
