Digit sum of digit sum of $2020^{2021}$ Let $S(n)$ be the sum of the digits of a non-negative integer $n$. What is $S\left(S\left(2020^{2021}\right)\right)$?
Other than seeing that this is $7 \pmod 9$ (since every non-negative integer is congruent to its digit sum $\mod 9$), I'm not sure what else can be done.
 A: When reading carefully the problem in the pdf you linked, one realizes that it asks for $S\left(S\left(S\left(2020^{2021}\right)\right)\right)$ rather than for $S\left(S\left(2020^{2021}\right)\right)$. I do not think that there is a reasonable method to evaluate the latter without a computer; notwithstanding, the actual answer is doable.
Observe, first of all, that $2020^{2021}$ has $\lceil 2021\cdot \log_{10}(2020)\rceil=6681$ digits; the last $2021$ are, of course, zeroes, since $10\mid 2020$. Thus $$S\left(2020^{2021}\right)\leqslant (6681-2021)\cdot 9=41940$$ We similarly obtain $$S\left(S\left(2020^{2021}\right)\right)\leqslant 4+4\cdot 9=40$$ At the same time, since $S(x)\equiv x\bmod 9$, we have $$S\left(S\left(2020^{2021}\right)\right)\equiv 2020^{2021}\equiv 4^{2021}\equiv 4^{336\cdot 6+5}\equiv 4^5\equiv 7\mod 9$$ Where I used that $\varphi(9)=6$. Hence $$S\left(S\left(2020^{2021}\right)\right)\in\{7, 16, 25, 34\}$$ Therefore, $S\left(S\left(S\left(2020^{2021}\right)\right)\right)=7.$
