Shifting the numbers, part 4

Let $x=\overline{A_1A_2}$ and $y=\overline{A_3A_4\ldots A_{n-1}A_n}$ . This means that $\displaystyle \frac{\overline{A_1A_2A_3\ldots A_{n-1}A_n}}{\overline{A_3\ldots A_{n-1}A_nA_1A_2}}=\frac{12}{17}$ is equivalent to $\displaystyle \frac{10^{n-2}x+y}{100y+x}=\frac{12}{17}$ which implies that $\displaystyle \frac{y}{x}=\frac{17(10^{n-2})-12}{1183}$ .

Since $x$ is a 2-digit number, $\displaystyle \frac{1183}{\gcd(17(10^{n-2})-12, 1183)}<100$ . Note that $1183 = 7(13^2)$ . Now $x \in \{7,13,91\}$ and hence $\displaystyle \gcd(17(10^{n-2})-12, 1183)=169, 91$ and $13$ respectively. This means that $\displaystyle 17(10^{n-2})\equiv 12 \pmod{169}$ , $\displaystyle 17(10^{n-2})\equiv 12 \pmod{91}$ or $\displaystyle 17(10^{n-2})\equiv 12 \pmod{13}$ .

Suppose $x=13$ . Then $\displaystyle 17(10^{n-2})\equiv 12 \pmod{91}$ . This means that $\displaystyle 17(10^{n})\equiv 12\times 9 \equiv 17 \pmod{91}$ . Thus $\displaystyle 10^n \equiv 1 \pmod{91}$ . By checking, $\displaystyle 10^n \not \equiv 1 \pmod{91}$ for $n\le 5$ and $\displaystyle 10^6 \equiv 1 \pmod{91}$ . Hence, $\displaystyle 10^{6k} \equiv 1 \pmod{91}$ . The desired value of $n$ is $\color{#3D99F6}{12}$ .

Suppose $x=91$ . Similar argument leads to the same answer.

Suppose $x=7$ . Similar argument leads to $\displaystyle 10^n \equiv 1 \pmod{169}$ , which means that the smallest value of $n$ is 78.