1. (POL) Find the smallest natural number $n$ with the following properties: (a) In decimal representation it ends with 6. (b) If we move this digit to the front of the number, we get a number 4 times larger.
1. From the conditions of the problem we have $n=10 x+6$ and $4 n=$ $6 \cdot 10^{m}+x$ for some integer $x$. Eliminating $x$ from these two equations, we get $40 n=6 \cdot 10^{m+1}+n-6 \Rightarrow n=2\left(10^{m+1}-1\right) / 13$. Hence we must find the smallest $m$ such that this fraction is an integer. By inspection, this happens for $m=6$, and for this $m$ we obtain $n=153846$, which indeed satisfies the conditions of the problem.