Arrange numbers to prove question

Question

Prove that 100 0's, 100 1's, 100 2's, 100 3's, 100 4's, 100 5's, 100 6's, 100 7's, 100 8's and 100 9's cannot be used in any form to make a perfect square.

I have no idea how to do this question. I was asked this question in a competition. A simple solution would be better.

May we take it that $(1 + 0 \cdot \mbox{all the other digits})$ is an unacceptable form for some reason? :-) — David K, Nov 15 '14 at 04:16
Actually this was the question I was asked in the competition. Any form probably means all arrangement works. — Arpit Saxena, Nov 15 '14 at 04:38

score 2 · Answer 1 · answered Nov 15 '14 at 05:43

2

1000 numbers (10 distinct digits * 100 each) can't be arranged in a square.

answered Nov 15 '14 at 05:43

Simpson17866

522

This seems the simple solution that was requested. +1 – Ross Millikan Nov 15 '14 at 05:45
I don't understand. Can you explain a little further? – Arpit Saxena Nov 17 '14 at 07:59
If you have 31 copies of each of 31 numbers, then you could make a 961-item square. If you had 32 copies of each of 32 numbers, then you could make a 1024-item square. You can't do this with 100 copies of each of 10 numbers. – Simpson17866 Nov 17 '14 at 14:53
But 121 is a square but the digits cannot be arranged in a square. – Arpit Saxena Nov 28 '14 at 09:29

Arrange numbers to prove question

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