Repeated decimal, part 3

Note that $\dfrac{1}{N}=\dfrac{9}{10000}+\dfrac{18}{1000000}+\dfrac{27}{100000000}\cdots$ .

Multiply by 100. $\dfrac{100}{N}=\dfrac{9}{100}+\dfrac{18}{10000}+\dfrac{27}{1000000}\cdots$

$\dfrac{100}{N}-\dfrac{1}{N}=\dfrac{99}{N}=\dfrac{9}{100}+\dfrac{9}{10000}+\dfrac{9}{1000000}\cdots$

$\dfrac{99}{N}=\dfrac{\dfrac{9}{100}}{1-\dfrac{1}{100}}=\dfrac{1}{11}$

$\dfrac{99}{N}=\dfrac{1}{11}\implies N=11\times99=\boxed{1089}$

For a decimal number less than $1$ with repeated digits of an integer $n$ as $0.\dot{n}$ with a period of $p$ is given as below:

$\begin{aligned} 0.\dot{n} & = n \left(\frac{1}{10^p} + \frac{1}{10^{2p}} + \frac{1}{10^{3p}} +... \right) = \frac{n}{10^p}\left(\frac{1}{1-\frac{1}{10^p}} \right) = \frac{n}{10^p-1} \end{aligned}$

For $n = 9182736455463728191$ and $p=22$ , we have:

$\begin{aligned} 0.\ddddot{0009182736455463728191} & = \frac{9182736455463728191}{10^{22}-1} = \frac{1}{N} \\ \Rightarrow N & = \frac{10^{22}-1}{9182736455463728191} \\ & = \frac{9,999,999,999,999,999,999,999}{9,182,736,455,463,728,191} \end{aligned}$

Since $N$ is a 4-digit number, let $N = 1000A + 100B + 10C + D = \overline{ABCD}$ , where $A$ , $B$ , $C$ and $D$ are positive integer less than $10$ . Then, we have:

$\begin{aligned} 9,182,736,455,463,728,191 N & = 9,999,999,999,999,999,999,999 \\ 9,182,736,455,463,728,191 \times \overline{ABCD} & = 9,999,999,999,999,999,999,999 \end{aligned}$

Using long multiplication, we have:

$\begin{array} {rrll} & 9,182,736,455,463,728,191 \\ \times & \overline{ABCD} \\ \hline & 8,2644,628,099,173,553,719 & \color{#3D99F6}{\Leftarrow n \times 9} & \color{#3D99F6}{D = 9 \text{ for the product to end with ......9}} \\ & 734,618,916,437,098,255,280 & \color{#3D99F6}{\Leftarrow n \times 80} & \color{#3D99F6}{C = 8 \text{ for the product to end with .....99}} \\ & 0 & \color{#3D99F6}{\Leftarrow n \times 0} & \color{#3D99F6}{B = 0 \text{ for the product to end with ....999}} \\ & 9,182,736,455,463,728,191,000 & \color{#3D99F6}{\Leftarrow n \times 1000} & \color{#3D99F6}{A = 1 \text{ for the product to end with ...9999}} \\ \hline & 9,999,999,999,999,999,999,999 \end{array}$

$\Rightarrow N = \boxed{1089}$

Just divide 1 by the given number, you will get N (simple maths, if X=1/Y, then Y=1/X)