$a$ and $b$ for which $\frac{a}{b}=\overline{a.b}$

Question

Are there natural numbers $a$ and $b$ such that $\frac{a}{b}=\overline{a.b}$?

I know that $\overline{a.b}=a+\sum_{i=1}^{n}b_i10^{-i}$ where $b_i$ is the $i$th digit of $b$ and $b$ has $n$ digits.

I also know that $b=\sum_{i=1}^{n}b_i10^{n-i}$.

Answer 1

So, $10^n\times \overline {a.b}=a\times 10^n+b$, yes? Meaning that you want $10^n\times \frac ab=a\times 10^n+b$. But this implies that $$a\times 10^n\times \left(1-\frac 1b\right)+b=0$$

But this is impossible, given that $b≥1$.