41. (8th grade) Find the smallest natural number that has the following properties: a) its representation in the decimal system ends with the digit 6; b) if the last digit 6 is erased and this digit 6 is written in front of the remaining digits, the resulting number is four times the original number.
41. The desired number can be written as: $N=\overline{a_{1} a_{2} a_{3} \ldots a_{n-1}}-1$. After rearranging the digit 6, we get the number $K=4 N=6 \bar{a}_{1} a_{2} a_{3} \ldots a_{n-1}$. Since the last digit of the number $N$ is 6, the last digit of the number $K$ is 4, i.e., $a_{n-1}=4$.

Substituting the value $a_{n-1}=4$ into $N$ and multiplying $N$ by 4, we find that $a_{n-2}=8$. Substituting the value $a_{n-2}=8$ into $N$ and multiplying $N$ by 4, we find that $a_{n-3}=3$, and so on, until we get the digit 6 in the number $4 N$. The number 153846 satisfies all the conditions of the problem and is the smallest. The problem has other solutions as well.