TITLE: A big "smallest" number
QUESTION [5 upvotes]: What is the smallest natural number $N$  such that moving the last digit to the front one gets the number $9N$? In other words, find the least $N$ such that if $N$ has decimal expansion $abc...xyz$, then $9N$ has decimal expansion $zabc...xy$. (Note: the number being asked for is very big!).
HINT: For solving this question I have calculated the powers of $10$ modulo $89$ for which I have used just $89k$ with $k = 1, 2, 3,...,9$ (sorry for the bad English).
Another related question inspired from this one has been asked here.

REPLY [7 votes]: Clearly, the first digit has to be $1$ and the last has to be $9$. Then we have
$$
9N=\frac{N-9}{10}+10^k\cdot 9
$$
where $N$ has $m=k+1$ digits as a total. Solving for $N$, this leads to
$$
N=\frac{9(10^m-1)}{89}
$$
and since $10$ has order $44$ modulo the prime $89$ we get solutions for $m=44s$, ie.
$$
N=\frac{9(10^{44s}-1)}{89}
$$
which for $s=1$ yields
$$
N=\color{blue}{1011235955056179775280898876404494382022471}\color{red}9
$$
which has
$$
9N=\color{red}9\color{blue}{1011235955056179775280898876404494382022471}
$$
as desired.