$f(k)$ is a strictly increasing function from $Z^+$ to $Z^+$. $f(f(k))=3k.$ Find $f(1)+f(9)+f(99)$. $f(k)$ is a strictly increasing function from $Z^+$ to $Z^+$. 
$f(f(k))=3k.
$ Find $f(1)+f(9)+f(99)$. 
I don't really know where to start with this question; I tried doing $f(9)=f(f(f(f(f(f(1))))))$ but this (unsurprisingly) didn't bring me anywhere. I would appreciate hints but no complete solutions yet, please.
 A: In f(f(k))=3k put k=1 ,f(f(1))=3 since 3 third element in codomain hence f(1) = 1 or 2 or 3.$\\$
Case 1. Let f(1) =3 then f(f(1))=f(3)=3 hence f(x) is not increasing function.$\\$
Case 2. Let f(1)=2 then f(f(x))=f(2)=3 that is one of the possibility.$\\$
Case 3 . Let f(1)=1 then f(f(1))=f(1)=3 hence not possible.
Thus f(1)=2 , f(2)=3 , put f(f(2))=f(3)= 6 and f(f(3))=f(6)=9
Put k=6 we get f(f(6))=f(9)=18 $\\$
for f(99) put k = 33 f(f(33))=99 and at f(9)=18 hence f(f(9))=f(18)=27, put k=18 f(f(18))=f(27)=54. $\\$Hence f(f(27))=f(54)=81 . f(f(54))=f(81)=162
f(f(81))=f(162)=243 $\\$ hence as x varies from 81 to 162 y changes from 162 to 243 for every 1 unit change in x there is 1 unit change in y hence f(99)=179 $\\$
Hence f(1)+f(9)+f(99)= 2 +18+179=199
A: We know $f(f(1)) = 3$ and since $f$ is strictly increasing, that means $f(1) = 2$ and $f(2) = 3$. (You can work out the details for yourself: what happens if $f(1) = 1$, or  $f(1) \ge 3$?) So now $f(3) = f(f(2)) = 6$ and $f(4) < f(5) < f(6) = f(f(3)) = 9$. Therefore $f(4) = 7$ and $f(5) = 8$.
Just continue in this manner until you understand what the pattern is.
