$\displaystyle S(n)$ is the number of digits of $\displaystyle n$. Is it possible to calculate:
$\displaystyle S(S(S(S(2012!^{2012!}))))$
Can You help me with this exercise?
...Eek. That is a large number.
You could try "guessing" at values for the answer, then using that guess to put a lower bound and upper bound. For example, if $\displaystyle S(S(k)) = 2$, then $\displaystyle 10^9 \le k \le 10^{99} - 1$.
Don't know if that method'll work but it's the only feasible method I see...problem is you'll have to determine between which two bounds $\displaystyle 2012!^{2012!}$ lies between. Have fun with that.