Reciprocal of 19 from Sum of Powers of 2 Backwards

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Theorem

The decimal expansion of the reciprocal of $19$ can be constructed by summing the powers of $2$, offset progressively backwards by $1$ digit: 

                         1
                        2
                       4
                      8
                    16
                   32
                  64
                128
               256
              512
            1024
           2048
          4096
         8192
       16384
      32768
     65536
   131072
+ 262144
 ......
--------------------------
.......1052631578947368421

while:

$\dfrac 1 {19} = 0 \cdotp \dot 05263 \, 15789 \, 47368 \, 42 \dot 1$


Proof

We confirm that from Reciprocal of $19$:

$\dfrac 1 {19} = 0 \cdotp \dot 05263 \, 15789 \, 47368 \, 42 \dot 1$


The construction above can be expressed as the sum:

$\ds \sum_{k \mathop \ge 0} \paren {2^k \times 10^k} = \sum_{k \mathop \ge 0} 20^k$

Obviously this sum does not converge.

However we can still analyse the last $N$ digits from this construction for every positive integer $N$.

From $k \ge N$ onwards, the last $N$ digits will not be affected by further additions of $20^k$ because:

$\forall k \ge N: 20^k \equiv 20^{k - N} \times 2^N \times 10^N \equiv 0 \pmod {10^N}$

By Number times Recurring Part of Reciprocal gives 9-Repdigit/Generalization, the last $N$ digits of this number coincides with the decimal expansion of the reciprocal of $19$ if and only if:

$\ds \sum_{k \mathop = 0}^{N - 1} 20^k \times 19 \equiv -1 \pmod {10^N}$

We have:

\(\ds \sum_{k \mathop = 0}^{N - 1} 20^k \times 19\) \(=\) \(\ds \frac {20^N - 1} {20 - 1} \times 19\) Sum of Geometric Sequence
\(\ds \) \(=\) \(\ds 20^N - 1\)
\(\ds \) \(\equiv\) \(\ds -1 \pmod {10^N}\)

Hence the result.

$\blacksquare$


Sources

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