$\displaystyle S(n)$ 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?

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- Nov 4th 2012, 04:25 PMmakramerSum of digits
$\displaystyle S(n)$ 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? - Nov 4th 2012, 06:41 PMrichard1234Re: Sum of digits
...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.