Find the smallest natural number $A$, satisfying the following conditions:

a) its notation ends with the digit 6;

b) when the digit 6 is moved from the end of the number to its beginning, the number is increased four times.
The desired number is written in the form $A=\overline{x_{1}} \overline{\overline{1}} \overline{-x} \bar{k} \overline{\bar{k}}$. Then $4 A=\overline{6 x_{1}} \bar{\ldots} \overline{x_{k}} \overline{\text {. }}$. To determine the digits of the number $4 A$, we will sequentially multiply the number $A$ by 4:

$6 \cdot 4=24$. Therefore, $x_{k}=4, 4 x_{k}+2=4 \cdot 4+2=18 ; x_{k-1}=8, 4 x_{k-1}+1=4 \cdot 8+1=33 ; x_{k-2}=3, 4 x_{k-2}+3=4 \cdot 3+3=$ $15 ; x_{k-3}=5, 4 x_{k-3}+1=4 \cdot 5+1=21 ; x_{k-4}=1, 4 x_{k-4}+2=4 \cdot 1+2=6 ; x_{k-5}=6$.

The calculation process ends here, as the digit 6 will be ahead: $4 A=615384, A=153846$.

## Answer

$A=153846$.