A late 2020 Easter math problem (No calculators, Python, or WolframAlpha!)

Let $n=412^{2020}$ . We first want to find how many digits $n$ has. We could estimate some powers of $412$ and compare these to powers of ten, but we don't actually even need to do that.

Since $412<10^3$ , we have $412^{2020}<10^{6060}$ , so that $n$ has at most $6060$ digits.

The largest possible $d(n)$ would be obtained if all these digits were equal to $9$ ; in other words $d(n)\le9\times 6060=54540$ .

This has $5$ digits. So $d(d(n))\le 9 \times 5=45$ (we could be more careful bounding this, but again we don't need to be in this case).

The largest digit sum of any positive integer less than or equal to $45$ is $d(39)=12$ . Therefore $d(d(d(n)))\le12$ .

Now, note that $d(x) \equiv x \pmod9$ for any positive integer $x$ .

We have $412 \equiv 7 \pmod9$ , and that $7^3 \equiv 1$ .

So $412^{2019}\equiv 1$ (since $2019$ is a multiple of $3$ ).

Finally, $n=412^{2020}\equiv 7$ , so $d(d(d(n))) \equiv 7 \pmod9$ as well. This, together with the fact $d(d(d(n)))\le12$ , means that we must have $d(d(d(n)))=\boxed7$ .