11.1. The numbers $2^{2019}$ and $5^{2019}$ are written consecutively. How many digits are written in total?
Answer: 2020.

Solution. Let the number $2^{2019}$ contain $\mathrm{m}$ digits, and the number $5^{2019}$ contain $\mathrm{n}$ digits. Then the following inequalities hold: $10^{\mathrm{m}-1}<2^{2019}<10^{\mathrm{m}}, 10^{\mathrm{n}-1}<5^{2019}<10^{\mathrm{n}}$ (the inequalities are strict because the power of two or five is not equal to the power of ten). Multiplying these inequalities, we get: $10^{\mathrm{m}+\mathrm{n}-2}<10^{2019}<10^{\mathrm{m}+\mathrm{n}}$. From this, it follows that the exponent 2019 is between $\mathrm{m}+\mathrm{n}-2$ and $\mathrm{m}+\mathrm{n}$, so $2019=\mathrm{m}+\mathrm{n}-1$ and $\mathrm{m}+\mathrm{n}=2020$. This means that a total of 2020 digits have been written.