The 68th question: Let $\mathrm{N}$ be a positive integer, and its decimal representation contains the consecutive string $11235$. $\mathrm{k}$ is a positive integer, and $10^{\mathrm{k}}>\mathrm{N}$. For all positive integers $\mathrm{N}, \mathrm{k}$ that satisfy the conditions, try to find the minimum possible value of $\frac{10^{\mathrm{k}}-1}{\operatorname{gcd}^{\mathrm{k}}\left(10^{k}-1, N\right)}$.
The 68th question,
Answer: Let $\operatorname{gcd}\left(10^{k}-1, N\right)=d$, and let $10^{k}-1=m d, N=n d$, then $(m, n)=1$, and $\frac{10^{k}-1}{\operatorname{gcd}\left(10^{k}-1, N\right)}=m$.

Notice that $10^{\mathrm{k}}-1>\mathrm{N}$, and $\frac{\mathrm{N}}{10^{\mathrm{k}}-1}=\frac{\mathrm{n}}{\mathrm{m}}$, so $\frac{\mathrm{m}}{\mathrm{m}}$ when expressed in decimal form is a repeating decimal, and its repeating part is $\mathrm{N}$ padded with some consecutive 0s. Therefore, the decimal representation of $\frac{\mathrm{n}}{\mathrm{m}}$ also contains the consecutive characters 11235.

Conversely, we can prove that if $m$ is coprime with 10, and there exists a positive integer $n$ such that $(m, n)=1$, and $\frac{n}{m}$ in decimal form contains the consecutive characters 11235, then $m$ is a possible value of $\frac{10^{k}-1}{\operatorname{gcd}\left(10^{k}-1, N\right)}$.
Indeed, because $(10, m)=1$, there exists a positive integer $k$ such that $\frac{10^{k}-1}{m}=s \in Z^{+}$. Let $N=n s$, then $\frac{\mathrm{N}}{10^{k}-1}=\frac{\mathrm{n}}{\mathrm{m}}$. Since the decimal representation of this fraction contains the string 11235, the decimal representation of $\mathrm{N}$ also contains the string 11245, thus $m$ is a possible value of $\frac{10^{k}-1}{\operatorname{gcd}\left(10^{k}-1, \mathrm{~N}\right)}$.

Next, we look for a fraction $\frac{\mathrm{n}}{\mathrm{m}}$ whose decimal representation contains the string 11235. Since $1, 1, 2, 3, 5$ are the first five terms of the Fibonacci sequence $\left\{F_{n}\right\}$, and $\mathrm{F}_{i}10^{5}\left|10 m^{\prime}-89 n\right| \geq 10^{5} \Rightarrow m^{\prime} > 89$. Therefore, there is no smaller possible value for $m$.
In summary, the smallest value of $\frac{10^{k}-1}{\operatorname{gcd}\left(10^{k}-1, \mathrm{~N}\right)}$ that satisfies the condition is 89.