Square of Repunit times Sum of Digits

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Theorem

The following pattern emerges:

\(\ds 121 \times \paren {1 + 2 + 1}\) \(=\) \(\ds 22^2\)
\(\ds 12 \, 321 \times \paren {1 + 2 + 3 + 2 + 1}\) \(=\) \(\ds 333^2\)
\(\ds 1 \, 234 \, 321 \times \paren {1 + 2 + 3 + 4 + 3 + 2 + 1}\) \(=\) \(\ds 4444^2\)

and so on, up until $999 \, 999 \, 999^2$ after which the pattern breaks down.


Proof

From Square of Repunit:

\(\ds 121\) \(=\) \(\ds 11^2\)
\(\ds 12 \, 321\) \(=\) \(\ds 111^2\)
\(\ds 1 \, 234 \, 321\) \(=\) \(\ds 1111^2\)

and so on.


Then from 1+2+...+n+(n-1)+...+1 = n^2:

\(\ds 1 + 2 + 1\) \(=\) \(\ds 2^2\)
\(\ds 1 + 2 + 3 + 2 + 1\) \(=\) \(\ds 3^2\)
\(\ds 1 + 2 + 3 + 4 + 3 + 2 + 1\) \(=\) \(\ds 4^2\)

and so on.


Then:

\(\ds 11^2 \times 2^2\) \(=\) \(\ds 22^2\)
\(\ds 111^2 \times 3^2\) \(=\) \(\ds 333^2\)
\(\ds 1111^2 \times 4^2\) \(=\) \(\ds 4444^2\)


The pattern breaks down after $9$:

$1 \, 111 \, 111 \, 111^2 = 1 \, 234 \, 567 \, 900 \, 987 \, 654 \, 321$

$\blacksquare$


Sources

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