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Which base of numerical system have $\frac{1}{5} = 0.33333\ldots$?

I need assistance in solving this one.

share|cite|improve this question is worth a look. It offers decimal, binary and hex conversions. It might help you see the relationships, ex, enter binary .010101010101 and see what this is in decimal. – JoeTaxpayer Jul 8 '14 at 18:28

If we are working in base $b$ (we must have $b\gt3$), then $0.3333\ldots$ is $$0.3333\ldots = \frac{3}{b} + \frac{3}{b^2} + \frac{3}{b^3}+\cdots$$ Since $$\sum_{n=1}^{\infty}\frac{3}{b^n} = \frac{3}{b}\sum_{n=0}^{\infty}\frac{1}{b^n} = \frac{3}{b}\left(\frac{1}{1-\frac{1}{b}}\right) =\frac{3}{b-1},$$ then...

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Hint: if we multiply $0.33333\ldots$ by $5$ then we get $0.(15)(15)(15)(15)(15)\ldots$. Compare that to what happens when we multiply the same by $3$: $0.99999\ldots$, and its interpretation in decimal.

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A reworking of Arturo's answer: let $x=0.333\dots$, let the base be $b$, then $$bx=3.333\dots$$ so $bx-x=(3.333\dots)-(0.333\dots)$ and you can take it from there.

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That should be $bx-x$, not $bx-b$. – jwodder Oct 30 '11 at 16:34
@jwodder correction made, thanks very much. – Gerry Myerson Oct 30 '11 at 21:44
So irrespective of the Base, the decimal shifts to the right when number is multiplied by the base? – MonK Jul 8 '14 at 19:42
@Sid, yes --- just think about what it means to write a number to a base. – Gerry Myerson Jul 9 '14 at 0:44

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