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What is the smallest $9$-digit number that has all the digits from $1$ to $9$ exactly once and is also a perfect square?

Please give me a method that doesn't involve programming.

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  • $\begingroup$ Have you tried anything? $\endgroup$ Sep 17, 2015 at 15:30
  • $\begingroup$ The last digit of the number, say $n$, should be in the set $\{1,4,5,6,9\}$, that can work as a starting point. $\endgroup$ Sep 17, 2015 at 15:32
  • $\begingroup$ Sure, it would be helpful if you could tell me how to proceed from there. $\endgroup$ Sep 17, 2015 at 15:48
  • $\begingroup$ Hmm, that seems to be difficult, it seems not to pan out well. $\endgroup$ Sep 17, 2015 at 15:49
  • $\begingroup$ It's hard to imagine a 'mathematical method' since the property of being pandigital is base ten dependent. Using a handheld calculator, I used brute force to find the answer in about 15 or 20 minutes. $\endgroup$
    – paw88789
    Sep 17, 2015 at 15:58

1 Answer 1

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Just look up A036744, the penholodigital squares.

The possible solutions are
139854276, 152843769, 157326849, 215384976, 245893761, 254817369,
326597184, 361874529, 375468129, 382945761, 385297641, 412739856,
523814769, 529874361, 537219684, 549386721, 587432169, 589324176,
597362481, 615387249, 627953481, 653927184, 672935481, 697435281,
714653289, 735982641, 743816529, 842973156, 847159236, 923187456

With square roots
11826, 12363, 12543, 14676, 15681, 15963,
18072, 19023, 19377, 19569, 19629, 20316,
22887, 23019, 23178, 23439, 24237, 24276,
24441, 24807, 25059, 25572, 25941, 26409,
26733, 27129, 27273, 29034, 29106, 30384

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