Bye 2018

The number of digits in a number $N$ (that is not a power of 10) is simply $\left\lceil \log_{10} N\right\rceil$ . $\begin{aligned} \frac{A}{101} &= \left\lceil\frac{1}{101}\log_{10}2018^{2018}\right\rceil \\ &= \left\lceil\frac{2018}{101}\log_{10}{(2.018\times 10^3})\right\rceil \\ &= \left\lceil \frac{2018}{101}\left(3 + \log_{10}2.018 \right)\right\rceil \\ &\approx \left\lceil 20(3+\log_{10}2)\right\rceil \\ &\approx 60 + \log_{10}2^{20}\\ &\approx 60 + 6 = 66 \end{aligned}$ Thus, we need $k$ such that $2k+1<66<2k+3$ and $k=\boxed{32}$