induction, recursive function, discrete mathematics

Question

Please help solve following recursive function. How can I solve $n-10$ for $M(99)$ or $M(98)$ if $n>100$ ? :

Find $M(99), M(100)$, and $M(98)$ when

$$ M(n) = \begin{cases} n-10, & \text{if $n>100$} \\ M(M(n + 11)), & \text{if $n \le 100$} \end{cases}$$ for all positive integers $n$.

Answer 1

$$M(100)=M(M(100+11))$$ $$=M(M(111))$$ $$=M(111-10)$$ $$=M(101)$$ $$=101-10$$ $$=91$$

$$M(99)=M(M(99+11))$$ $$=M(M(110))$$ $$=M(110-10)$$ $$=M(100)$$ $$=91$$

$$M(98)=M(M(98+11))$$ $$=M(M(109))$$ $$=M(109-10)$$ $$=M(99)$$ $$=91$$