Find the base system, $x$, such that $\frac{1}{5}$ and $\overline{.17}$ are numerals for the same number.

So far I know that $$\frac{1}{5} = 0.2$$ in base 10. However, I'm not really sure how to convert it into another base. Any suggestions would be greatly appreciated.

• do you know how to write a repeating decimal as a sum? – qwr Jul 2 '19 at 4:09

$$0.171717..._x=\dfrac{17_x}{x^2}+\dfrac{17_x}{x^4}+\dfrac{17_x}{x^6}+...$$

$$=\dfrac{17_x/x^2}{1-1/x^2}=\dfrac{17_x}{x^2-1}=\dfrac{x+7}{x^2-1}=\dfrac15$$

• Not OP but do you solve the final quadratic in base $10$? If not, how do you multiply $7$ and $5$ – Anvit Jul 2 '19 at 4:39
• yes, $5(x+7)=x^2-1\iff x^2-5x-36=0 \iff (x-9)(x+4)=0 \iff x=9$ or $x=-4$, and take the positive solution – J. W. Tanner Jul 2 '19 at 4:44

$$(x^2-1)\cdot\dfrac15=(17.17\cdots)_x-(.17\cdots)_x=(17)_x=1\cdot x^1+7$$

\begin{align} (0.1717...)_b =& \left(\frac{1}{b}+\frac{7}{b^2}\right) + \left(\frac{1}{b^3}+\frac{7}{b^4}\right) + \cdots \\ =& \frac{b+7}{b^2}+\frac{b+7}{b^4} +\cdots \\ =& \sum_{n=1}^{\infty} \frac{b+7}{b^{2n}} \\ =& (b+7)\sum_{n=1}^{\infty}(b^{-2})^n \\ =& \frac{b+7}{b^2-1} \end{align} and this is equal to $$1/5$$, so $$b=9$$.

• But why don't you convert 1÷5 in base x as 1÷5=0.2=2×x^-1 – Rajendra Sharma Jul 2 '19 at 15:57
• But, not your request is find $x$ such that $(0.2)_{10}=(0.1717...)_x$? – Azif00 Jul 2 '19 at 16:18
• No 0.2 is also in base x. – Rajendra Sharma Jul 4 '19 at 3:35