A big "smallest" number 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.
 A: This in an answer to the original formulation of the question, where the performed operation was swapping the first digit with the last digit, rather than moving the last digit to the first place while rotating the number.
No such $N$ exists. Let $a$ be the first digit of $N$, $b$ the last digit of $N$. We demand that $N + b · 10^n - a · 10^n + a - b = 9N$, where $n + 1$ is the length of $N$. We have $(b - a)(10^n - 1) = 8N$. Since $a, b ∈ \{1, …, 9\}$, we have $b - a = 8$, $b = 9$, $a = 1$ and $N = 10^n - 1$, which is a contradition.
A: 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.
A: No such $N$ exists. The first and last digits of $N$ are $1,9$, respectively.  
$$9\cdot \overline{1a_1a_2\ldots a_n 9}=\overline{9a_1a_2\ldots a_n 1}\iff 8\cdot \overline{a_1a_2\ldots a_n 0}=-80$$
But $\overline{a_1a_2\ldots a_n 0}>0$, impossible.
A: Let $a \neq 0$ be the first digit, $x \ge 0$ be the middle portion (with $L$ digits), and $b$ be the last digit.  Then
$$
N = 10^{L+1}a + 10x + b,
$$
and we need to have
$$
10^{L+1}b + 10x + a = 9N = 9\cdot 10^{L+1}a + 90x + 9b,
$$
or
$$
80x = (10^{L+1} - 9)b - (9\cdot 10^{L+1} - 1)a.
$$
Unless $b=9$ and $a=1$, the right-hand side is negative; so we need
$$
80x = (10^{L+1} - 9)\cdot 9 - (9 \cdot 10^{L+1} - 1) = -80,
$$
which is impossible.
