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
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $1,234,321$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1,234,321$