3.59 Impurities constitute $20 \%$ of the total volume of the solution. What is the smallest number of filters through which the solution must be passed so that the final impurity content does not exceed $0.01 \%$, if each filter absorbs $80 \%$ of the impurities? (It is known that $\lg 2 \approx 0.30$.)
3.59 Impurities constitute $\frac{1}{5}$ of the solution. After the first filtration, $\left(\frac{1}{5}\right)^{2}$ impurities remain, and after the $k$-th filtration, $-\left(\frac{1}{5}\right)^{k+1}$ impurities remain. According to the condition, $\left(\frac{1}{5}\right)^{k+1} \leqslant 10^{-4}$; $-(k+1) \lg 5 \leqslant-4$, from which $k \geqslant 4.7$.

Answer: 5 filters.