Striking divisibility

Suppose that the number $N$ is an $m+1$ digit number, and that the first digit is $a$ . We can write $N = 10^ma + b$ for some integer $0 \le b < 10^m$ . The number obtained by moving the first digit of $N$ to the end is $10b + a$ , and so we want $\begin{aligned} 10b + a & = \tfrac32(10^ma + b) \\ 20b + 2a & = 3\times10^ma + 3b \\ 17b & = (3\times10^m - 2)a \end{aligned}$ Since $a$ is a nonzero single digit number, it cannot be divisible by $17$ , and so we deduce that $17$ divides $3\times10^m-2$ , so that $3 \times 10^m \equiv 2 \pmod{17}$ , or $10^m \equiv 12 \pmod{17}$ . The smallest value of $m$ for which this is true is $m=15$ , and we can have $a \; = \; 1 \hspace{2cm} b \; = \; \tfrac{1}{17}(3 \times 10^{15} - 2)$ Choosing $a=1$ with $m=15$ clearly gives us the smallest possible value of $N$ . Thus $N \; = \; 10^{15} + \tfrac{1}{17}(3\times10^{15} - 2) \; = \; \tfrac{1}{17}(20\times10^{15} - 2) \; = \; \frac{2(10^{16} - 1)}{17}$ which makes the answer $\boxed{17}$ .

[This solution is incomplete.]

Let d be the first digit. Then the number is 10^{k}d +r