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Why do these relations hold ? Any underlying theorem/principle/rule/method ?

$ \sum_{k=1}^{n} k \cdot 10^{d(n-k)} \equiv \frac{n(n+1)}{2} \pmod{\text{rep}(1/n)} $

$ \sum_{k=1}^{n-1} k \cdot 10^{d(n-1-k)} \equiv \frac{n(n-1)}{2} \pmod{\text{rep}(1/n)} $

$ \sum_{k=0}^{n-1} k \cdot 10^{kd} + 10^{d(n-1)} \cdot \sum_{k=1}^{n} k \cdot 10^{d(n-k)} \equiv n^{2} \pmod{\text{rep}(1/n)} $

where:

$n$ is not a multiple of $2$ or $5$,

$n>1$,

$\text{rep}(1/n)$ is the repetend of $\frac{1}{n}$,

$\text{rep}(1/n) > 11$,

$d$ is the length or period of $\text{rep}(1/n)$.

Examples:

For $n=3$,

$\text{rep}(1/3)=3$ and $d=1$ because $\frac{1}{3}=0.\overline{3}$

But we need $\text{rep}(1/n) > 11$. So, we can extend it as follows:

$\text{rep}(1/3)=33$ and $d=2$ because $\frac{1}{3}=0.\overline{33}$.

Now, $ \sum_{k=1}^{3} k \cdot 10^{2(3-k)}=10203$

Also, $10203 \equiv \frac{3(3+1)}{2} \pmod{33}$

$\sum_{k=1}^{3} k \cdot 10^{2(3-1-k)}=102$

Also, $102 \equiv \frac{3(3-1)}{2} \pmod{33}$

$\sum_{k=0}^{3-1} k \cdot 10^{2k} + 10^{2(3-1)} \cdot \sum_{k=1}^{3} k \cdot 10^{2(3-k)} = 102030201$

Also, $102030201 \equiv 3^{2} \pmod{33}$

For $n=7$,

$\text{rep}(1/7)=142857$ and $d=6$ because $\frac{1}{7}=0.\overline{142857}$

Now, $ \sum_{k=1}^{7} k \cdot 10^{6(7-k)}=1000002000003000004000005000006000007$

Also, $1000002000003000004000005000006000007 \equiv \frac{7(7+1)}{2} \pmod{142857}$

$\sum_{k=1}^{7} k \cdot 10^{6(7-1-k)}=1000002000003000004000005000006$

Also, $1000002000003000004000005000006 \equiv \frac{7(7-1)}{2} \pmod{142857}$

$\sum_{k=0}^{7-1} k \cdot 10^{6k} + 10^{6(7-1)} \cdot \sum_{k=1}^{7} k \cdot 10^{6(7-k)} = 1000002000003000004000005000006000007000006000005000004000003000002000001$

Also, $1000002000003000004000005000006000007000006000005000004000003000002000001 \equiv 7^{2} \pmod{142857}$

For $n=9$,

$\text{rep}(1/9)=1$ and $d=1$ because $\frac{1}{9}=0.\overline{1}$

But we need $\text{rep}(1/n) > 11$. So, we can extend it as follows:

$\text{rep}(1/9)=111$ and $d=3$ because $\frac{1}{9}=0.\overline{111}$.

Now, $ \sum_{k=1}^{9} k \cdot 10^{3(9-k)}=1002003004005006007008009$

Also, $1002003004005006007008009 \equiv \frac{9(9+1)}{2} \pmod{111}$

$\sum_{k=1}^{9} k \cdot 10^{3(9-1-k)}=1002003004005006007008$

Also, $1002003004005006007008 \equiv \frac{9(9-1)}{2} \pmod{111}$

$\sum_{k=0}^{9-1} k \cdot 10^{3k} + 10^{3(9-1)} \cdot \sum_{k=1}^{9} k \cdot 10^{3(9-k)} = 1002003004005006007008009008007006005004003002001$

Also, $1002003004005006007008009008007006005004003002001 \equiv 9^{2} \pmod{111}$

For $n=11$,

$\text{rep}(1/11)=09$ and $d=2$ because $\frac{1}{11}=0.\overline{09}$

But we need $\text{rep}(1/n) > 11$. So, we can extend it as follows:

$\text{rep}(1/11)=0909$ and $d=4$ because $\frac{1}{11}=0.\overline{0909}$.

Now, $ \sum_{k=1}^{11} k \cdot 10^{4(11-k)}=10002000300040005000600070008000900100011$

Also, $10002000300040005000600070008000900100011 \equiv \frac{11(11+1)}{2} \pmod{0909}$

$\sum_{k=1}^{11} k \cdot 10^{4(11-1-k)}=1000200030004000500060007000800090010$

Also, $1000200030004000500060007000800090010 \equiv \frac{11(11-1)}{2} \pmod{0909}$

$\sum_{k=0}^{11-1} k \cdot 10^{4k} + 10^{4(11-1)} \cdot \sum_{k=1}^{11} k \cdot 10^{4(11-k)} = 100020003000400050006000700080009001000110010000900080007000600050004000300020001 $

Also, $100020003000400050006000700080009001000110010000900080007000600050004000300020001 \equiv 11^{2} \pmod{0909}$

For $n=13$,

$\text{rep}(1/13)=076923$ and $d=6$ because $\frac{1}{13}=0.\overline{076923}$

Now, $ \sum_{k=1}^{13} k \cdot 10^{6(13-k)}=1000002000003000004000005000006000007000008000009000010000011000012000013$

Also, $1000002000003000004000005000006000007000008000009000010000011000012000013 \equiv \frac{13(13+1)}{2} \pmod{076923}$

$ \sum_{k=1}^{13} k \cdot 10^{6(13-1-k)}=1000002000003000004000005000006000007000008000009000010000011000012$

Also, $1000002000003000004000005000006000007000008000009000010000011000012\equiv \frac{13(13-1)}{2} \pmod{076923}$

$\sum_{k=0}^{13-1} k \cdot 10^{6k} + 10^{6(13-1)} \cdot \sum_{k=1}^{13} k \cdot 10^{6(13-k)} = 1000002000003000004000005000006000007000008000009000010000011000012000013000012000011000010000009000008000007000006000005000004000003000002000001 $

Also, $1000002000003000004000005000006000007000008000009000010000011000012000013000012000011000010000009000008000007000006000005000004000003000002000001 \equiv 13^{2} \pmod{076923}$

Mr. Dev
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1 Answers1

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Hint e.g. for $\,n=7\,$ the modulus $\,m = 142857 = (\color{#0a0}{10^6-1})/7,\,$ so $\bmod m\,$ we have $\,\color{#0a0}{10^6\equiv 1},\,$ so $\,\color{#c00}{10^{6j}}\equiv (\color{#0a0}{10^6})^j\equiv \color{#0a0}1^j\equiv \color{#c00}1,\,$ so $\,\sum_{k=1}^7 k\cdot \color{#c00}{10^{6j}}\equiv \sum_{k=1}^7 k\cdot\color{#c00}1\equiv \large \frac{7(7+1)}2.\,\ $ $[\:\!\color{#c00}j = 7\!-\!k\,$ in OP]

Generally, by this answer $\,m = {\rm rep}(1/n) = (\color{#0a0}{10^d-1})/n,\,$ so, as above, $\,\color{#0a0}{10^d\equiv 1},\,$ so $\,\color{#c00}{10^{\:\!d\:\!j}\equiv 1},\,$ yielding the claimed result as above.

Similarly your newly added third equation evaluates to $\,\frac{n(n-1)}2+\frac{n(n+1)}2 = n^2\,$ after replacing all powers of $\,\color{#c00}{10^d\,}$ by $\:\!\color{#c00}1$.

Mr. Dev
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Bill Dubuque
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