Repeating Decimal in Base k

Similar explanation with @Joshua Lowrance 's

Based on how number system work, we have:

$\begin{aligned} 0.\overline{35}_k & = \left(\frac 3k + \frac 5{k^2} + \frac 3{k^3} + \frac 5{k^4} + \cdots \right)_{10} = \left(\frac 5{39}\right)_{10} \end{aligned}$

Therefore in base-10,

$\begin{aligned} \sum_{n=1}^\infty \frac 3{k^n} + \sum_{n=1}^\infty \frac 2{k^{2n}} & = \frac 5{39} \\ \frac 3k \left(\frac 1{1-\frac 1k}\right) + \frac 2{k^2}\left(\frac 1{1-\frac 1{k^2}} \right) & = \frac 5{39} \\ \frac 3{k-1} + \frac 2{k^2-1} & = \frac 5{39} \\ \frac {3(k+1)+2}{k^2-1} & = \frac 5{39} \\ \frac {3k+5}{k^2-1} & = \frac 5{39} \\ 117k + 195 & = 5k^2 - 5 \\ 5k^2 - 117k - 200 & = 0 \\ (5k+8)(k-25) & = 0 \\ \implies k & = \boxed{25} & \small \color{#3D99F6} \text{Since }k \text{ is an integer.} \end{aligned}$

A repeating decimal in base- $k$ has the form $\frac{A}{k}+\frac{B}{k^2}+\frac{C}{k^3}+\cdots$ . This repeating decimal with look like this: $\frac{3}{k}+\frac{5}{k^2}+\frac{3}{k^3}+\frac{5}{k^4}+\frac{3}{k^5}+\cdots$ Separating it out: $\frac{3}{k}+\frac{3}{k^2}+\frac{3}{k^3}+\frac{3}{k^4}+\frac{3}{k^4}+\frac{3}{k^5}+\cdots +\frac{2}{k^2}+\frac{2}{k^4}+\frac{2}{k^6}+\cdots$ Factoring out the numerator: $3(\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k^3}+\frac{1}{k^4}+\cdots)+2(\frac{1}{k^2}+\frac{1}{k^4}+\frac{1}{k^6}+\cdots)$ Changing the last part: $3(\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k^3}+\frac{1}{k^4}+\cdots)+2(\frac{1}{(k^2)}+\frac{1}{(k^2)^2}+\frac{1}{(k^2)^3}+\frac{1}{(k^2)^4}+\cdots)$ Now, we know that $1+r+r^2+r^3+r^4+r^5+\cdots =\frac{1}{1-r}$ , if $0<r<1$ . Because $k$ is a positive integer greater than $1$ (as base $1$ would just be unary, a series of only $1$ 's), $0<\frac{1}{k^2}<\frac{1}{k}<1$ , so this formula applies. However, we are missing the starting $1$ in both equations, but we can just subtract it out. $3(\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k^3}+\frac{1}{k^4}+\cdots)+2(\frac{1}{(k^2)}+\frac{1}{(k^2)^2}+\frac{1}{(k^2)^3}+\frac{1}{(k^2)^4}+\cdots)$ $3(\frac{1}{1-\frac{1}{k}}-1)+2(\frac{1}{1-\frac{1}{k^2}}-1)$ $3(\frac{k}{k-1}-\frac{k-1}{k-1})+2(\frac{k^2}{k^2-1}-\frac{k^2-1}{k^2-1})$ $3(\frac{1}{k-1})+2(\frac{1}{k^2-1})$ We are told that this equals $\frac{5}{39}$ : $3(\frac{1}{k-1})+2(\frac{1}{k^2-1})=\frac{5}{39}$ $\frac{3}{k-1}+\frac{2}{k^2-1}=\frac{5}{39}$ $\frac{3(k+1)}{(k-1)(k+1)}+\frac{2}{k^2-1}=\frac{5}{39}$ $\frac{3k+3}{k^2-1}+\frac{2}{k^2-1}=\frac{5}{39}$ $\frac{3k+5}{(k^2-1)}=\frac{5}{39}$ $117k+195=5k^2-5$ $5k^2-117k-200=0$ $k=\frac{117 \pm \sqrt{117^2+4(200)(5)}}{2(5)}=-1.6,25$ And because $k$ is a positive integer $k=25$