Multiples of Reciprocals with Maximum Period form Magic Square

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Theorem

Let $p$ be a prime number whose reciprocal has a decimal expansion which has the maximum period, that is, $p - 1$.

Let the first $p - 1$ digits of $\dfrac 1 p$ be expressed as an integer $n$ (with one or more leading zeroes as it is presented).

Then the first $p - 1$ multiples of $n$, when listed in order of size, arrange themselves as a magic square such that:

the sum of the elements of each row
the sum of the elements in each column

are the same.


Proof


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Examples

Reciprocal of $19$

We have that:

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

Hence:

$\begin{array}{|r|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}

\hline \frac 1 {19} & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 \\ \hline \frac 2 {19} & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 \\ \hline \frac 3 {19} & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 \\ \hline \frac 4 {19} & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 \\ \hline \frac 5 {19} & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 \\ \hline \frac 6 {19} & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 \\ \hline \frac 7 {19} & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 \\ \hline \frac 8 {19} & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 \\ \hline \frac 9 {19} & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 \\ \hline \frac {10} {19} & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 \\ \hline \frac {11} {19} & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 \\ \hline \frac {12} {19} & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 \\ \hline \frac {13} {19} & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 \\ \hline \frac {14} {19} & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 \\ \hline \frac {15} {19} & 7 & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 \\ \hline \frac {16} {19} & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 & 9 & 4 & 7 & 3 & 6 \\ \hline \frac {17} {19} & 8 & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 \\ \hline \frac {18} {19} & 9 & 4 & 7 & 3 & 6 & 8 & 4 & 2 & 1 & 0 & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 8 \\ \hline \end{array}$

As can be seen, each of the rows and columns add to $81$.

As a bonus, so do the numbers along each diagonal.


Sources

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