Sum of digits

Nov 2012
2
0
Austria
\(\displaystyle S(n)\) 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?
 
Jun 2012
616
115
AZ
...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.