a_507_bc wrote: Show that $S(2^{2^{2 \cdot 2023}})>2023$, where $S(m)$ denotes the digit sum of $m$. Let $2^{2^{4046}}=\sum_{i=0}^{\lfloor2^{4046}\log_{10}(2)\rfloor}a_i10^i$ with $0\leq a_i\leq9$ for all nonegative integers $i<2^{4046}\log_{10}(2)$ and let $s_k:=\sum_{i=0}^{k-1}a_i10^i$ for postive integers $k<2^{4046}\log_{10}(2)+1$. We have $a_0=6$. Let $1\leq n\leq2018$ and note that $4^n<2^{4046}\log_{10}(2)$ and $2^{4^n}\mid2^{2^{4046}}$. So $2^{4^n}\mid s_{4^n}$ and there is a $r$ with $0<r<5^{4^n}$ such that $s_{4^n}=2^{4^n}r$. In particular we have $2^{4^n}\leq s_{4^n}<10^{4^n}$ and $10^{4^{n-1}}< s_{4^n}<10^{4^n}$. But since $s_{4^n}:=\sum_{i=0}^{4^n-1}a_i10^i$ one of the digits $a_i$ with $4^{n-1}\leq i<4^n$ is non-zero. So we have $a_0=6$ and $2018$ different non-zero digits $a_i$ with $i>0$ of $2^{2^{4046}}$. Thus $S\left(2^{2^{2 \cdot 2023}}\right)>2023$.

This problem was proposed by me.