if $n\in N^{+}$,and such The decimal representation of a number is that$$0.d_{1}d_{2}d_{3}d_{4}\cdots=\dfrac{n}{61} $$(where the digit $d_{i}$ is one of the numbers $0,1,2,\cdots,9)$, and $d_{37}=2,d_{65}=3$,
Find the $n$ and $d_{36}$，
I 've been thinking about this for a long time, and it's hard to find a breakthrough in terms.
We immediately have $0<x<61$. Next, $$\frac{10^{36}x}{61}=\underbrace{d_1\ldots d_{36}}_N\,.2\ldots $$ i.e., $10^{36}x=61N+y$ where $\frac2{10}\le\frac y{61}<\frac3{10}$, i.e., $12.2\le y<18.3$ Likewise, $10^{65}x=61M+z$ where $18.3<z<24.4$. With $10^{34}\equiv 34\equiv 9^{-1}\pmod{61}$, $10^{65}\equiv 10^5\equiv 21=32^{-1}\pmod{61}$, we can find $x$ by looking for common numbers in $9y\bmod 61$, $13\le y\le 18$ and $32z\bmod 61$, $19\le z\le 24$.
Alternatively: Perform a long division to compute $\frac 1{61}$. You will notice that the period length is $60$. Locate $2$'s and $3$'s in the period. find occurrences such that the $3$ is $65-37=28$ places behind the $2$. (There should be only one possibility). The digit to the left of the $2$ is the answer to the second part. Multiply with a suitable power of $10$ (i.e., shift the digits) to move the $2$ to position $37$. Chop off the integer part and multiply by $61$ to find $x$.