Shifting the numbers, part 5

Let $x={A_1}$ and $y=\overline{A_2A_3\ldots A_{n-1}A_n}$ . This means that $\displaystyle \frac{\overline{A_1A_2A_3\ldots A_{n-1}A_n}}{\overline{A_2\ldots A_{n-1}A_nA_1}}=\frac{12}{17}$ is equivalent to $\displaystyle \frac{10^{n-1}x+y}{10y+x}=\frac{12}{17}$ which implies that $\displaystyle \frac{y}{x}=\frac{17(10^{n-1})-12}{103}$ .

Since 103 is a prime number, $\displaystyle 17(10^{n-1})-12$ is a multiple of 103. This means that $\displaystyle 17(10^{n-1})\equiv 12 \pmod{103}$ . Now $\displaystyle 17(10^{n})\equiv 120 \equiv 17 \pmod{103}$ . Thus $\displaystyle 10^n \equiv 1 \pmod{103}$ .

Note that $\displaystyle\phi (103)=102= (2)(3)(17)$ and hence $\displaystyle 10^{102} \equiv 1 \pmod{103}$ . Now, $\displaystyle 10^n \equiv 1 \pmod{103}$ if $\displaystyle n \big| 102$ . By checking, $\displaystyle 10^n \not \equiv 1 \pmod{103}$ for $n \in \{2,3,6,17\}$ . This means that $\displaystyle 10^n \not \equiv 1 \pmod{103}$ for $n \le 33$ . Finally, $\displaystyle 10^{34} \equiv 1 \pmod{103}$ . Hence, $\displaystyle 10^{34k} \equiv 1 \pmod{103}$ . The desired value of $n$ is $\color{#3D99F6}{34}$ .