Take any number \(m\) (let's say 3).

Write all the numbers starting from \(1\) in one line.

Eliminate numbers at the index \(k\) (\(k \equiv 0 \pmod{m}\))

Now write for each index, partial sum of the numbers till that index.

Eliminate numbers at the index \(k\) (\(k \equiv m-1 \pmod{m}\)).

Now write for each index, partial sum of the numbers till that index.

\(\vdots\)

\(\quad\)2m-1. Eliminate numbers at the index \(k\) (\(k \equiv 2 \pmod{m}\)).

\(\quad\)2m. Now write for each index, partial sum of the numbers till that index.

For example, for \(m=3\),

\[
\begin{array}{ c c c c c c c c c c c c c}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\
1 & 2 & & 4 & 5 & & 7 & 8 & & 10 & 11 & & 13\\

1 & 3 & & 7 & 12 & & 19 & 27 & & 37 & 48 & & 61\\

1 & & & 7 & & & 19 & & & 37 & & & 61\\
1 & & & 8 & & & 27 & & & 64 & & & 125\\
\end{array}
\]

\(\displaystyle \text{We get a sequence of} \ n^m \text{!}\)

Prove that this happens for any number of numbers and for any natural number \(m\).

PS - This is not a challenge. This is seeking help.

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## Comments

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TopNewestThis is seriously magic. Another math-e-magical thing with which you can fascinate a primary school student and can "irritate" a mathematician.

(Another such thing is the Goldbach's Theorem).

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(1) I have fixed your LaTeX. The correct code here is "\begin{array}".

(2) This result is known as Moessner's Theorem.

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Oh I see. Thanks! I will look into the references. BTW, can you give a simplified proof?

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I'm afraid I don't know the proof. I think it's a hard result.

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@Jon Haussmann

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@Mark Hennings @Pi Han Goh @Ishan Singh Please also fix the latex!

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