Smallest Pandigital Square
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Theorem
The smallest pandigital square is $1 \, 026 \, 753 \, 849$:
- $1 \, 026 \, 753 \, 849 = 32 \, 043^2$
Proof
We check all the squares of numbers from $\ceiling {\sqrt {1 \, 023 \, 456 \, 789} } = 31 \, 992$ up to $32 \, 042$, with the following constraints:
Since all these squares has $10$ as its two leftmost digits, the number cannot end with $0$, $1$ or $9$.
A pandigital number is divisible by $9$, so our number must be divisible by $3$.
These constraints leaves us with the following $12$ candidates:
\(\ds 31 \, 992^2\) | \(=\) | \(\ds 1 \, 023 \, 488 \, 064\) | ||||||||||||
\(\ds 31 \, 995^2\) | \(=\) | \(\ds 1 \, 023 \, 680 \, 025\) | ||||||||||||
\(\ds 31 \, 998^2\) | \(=\) | \(\ds 1 \, 023 \, 872 \, 004\) | ||||||||||||
\(\ds 32 \, 004^2\) | \(=\) | \(\ds 1 \, 024 \, 256 \, 016\) | ||||||||||||
\(\ds 32 \, 007^2\) | \(=\) | \(\ds 1 \, 024 \, 448 \, 049\) | ||||||||||||
\(\ds 32 \, 013^2\) | \(=\) | \(\ds 1 \, 024 \, 832 \, 169\) | ||||||||||||
\(\ds 32 \, 016^2\) | \(=\) | \(\ds 1 \, 025 \, 024 \, 256\) | ||||||||||||
\(\ds 32 \, 022^2\) | \(=\) | \(\ds 1 \, 025 \, 408 \, 484\) | ||||||||||||
\(\ds 32 \, 025^2\) | \(=\) | \(\ds 1 \, 025 \, 600 \, 625\) | ||||||||||||
\(\ds 32 \, 028^2\) | \(=\) | \(\ds 1 \, 025 \, 792 \, 784\) | ||||||||||||
\(\ds 32 \, 034^2\) | \(=\) | \(\ds 1 \, 026 \, 177 \, 156\) | ||||||||||||
\(\ds 32 \, 037^2\) | \(=\) | \(\ds 1 \, 026 \, 369 \, 369\) |
By inspection, none of these numbers are pandigital.
$\blacksquare$
Historical Note
In his Curious and Interesting Numbers, 2nd ed. of $1997$, David Wells attributes this result to Michal Stajsczak, but gives no context.
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1,026,753,849$