Given a number with a periodic expansion, identify the the period and use a geometric series trick to identify the number.
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For example, $1/9 = 0.1111111\dots$, means in the number system of base 10, that $$ \frac19 = \frac1{10}+\frac1{100}+\frac1{1000} + \dots = \sum_{n\ge 1} \left(\frac1{10}\right)^n $$ For $1/7=0.142857142857142857\dots$, it can be written as $$\frac17 = 142857\cdot \sum_{n\ge 1}\frac1{{1000000}^n} $$ Note that if $p$ is prime ($p\ne 2,5$), then by Fermat's little theorem, $10^{p-1}\equiv 1 \pmod p$, that is $p|\underbrace{999\dots999}_{(p-1)\text{ pieces}}$. |
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Hint $\rm\ \ r = 0.121212...\, =\, \dfrac{12}{10^2} + \dfrac{12}{10^4} + \dfrac{12}{10^6} + \,\cdots\, =\, 12\,\left(\dfrac{1}{10^2} + \dfrac{1}{10^4} + \dfrac{1}{10^6}+\,\cdots\right)$ Since $\rm\ \dfrac{1}{1-x}\ =\ 1 + x + x^2 + x^3\,+\cdots,\ |x| < 1,\ $ subtracting $1$ yields that $\rm\,\ \ \dfrac{x}{1-x}\ =\ x + x^2 + x^3 +\, \cdots.\ $ Now apply this to the above series. Alternatively $\rm\ r = \overline {0.121212}\:\Rightarrow\:100\, r\, =\, 12 + r,\:$ so $\rm\:99r = 12,\:$ so $\rm\: r = 12/99.$ The same idea works generally, e.g. see here. |
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