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I've come across the following question.

Find $0.\overline{204}_6$ as a base ten fraction.

I understand that is the question asked the repeating decimal in base $10$, I would then say that:

$$x = 0.\overline{204}_{10}$$

$$1000x = 204.\overline{204}_{10}$$

$$999x = 204$$

$$x = \frac{204}{999}$$

Therefore, I can perform a similar task by doing this:

$$x = 0.\overline{204}_{6}$$

$$216x = 204.\overline{204}_{6}$$

$$216x = 204$$

$$x = \frac{204}{216} = \frac{17}{18}$$

Is the following procedure correct? Since initially we had to multiply by $10^3$ in base $10$, I would then assume that you would have to multiply by $6^3$, or $216$ in base $6$.

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  • $\begingroup$ $204_6=76$ so $215x=76$. $\endgroup$ – Jan-Magnus Økland Mar 23 '15 at 13:00
  • $\begingroup$ Note how that's $215x=76$, not the $216x$ shown in the original question. You get $215x$ here for the same reason you got $999x$ in base ten. $\endgroup$ – David K Mar 23 '15 at 13:14
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In your calculations you are confusing $204_6$ with $204_{10}$. If you clear that up, you are on the right track.

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Just a little rectification : $$216x=2\cdot6^2+0+4\cdot6^0+x$$

Or

$$0.\overline{204}_6=\frac26+\frac2{6^4}+\cdots+\dfrac4{6^3}+\dfrac4{6^6}+\cdots$$

$$=\dfrac{2/6}{1-(1/6)^3}+\dfrac{4/6^3}{1-(1/6)^3}$$

$$=\dfrac{2\cdot6^2+4}{215}$$

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