IMO 1962 - Problem 1

Problem: 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.
Solution: From the conditions of the problem we have $n=10x+6$ and $4n=6.10^{m}+x$ for some integer $x$. Eliminating $x$ from these two equations, we get $40n=6.10^{m+1}+n-6\implies n=\frac{2(10^{m+1})-1}{13}$. 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.