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 Sums of squares and stuff. (Posted on 2005-03-26)

Arrange two of each digit (0-9) into a 20 digit long number, which does not begin with a zero, then score your number as follows:

For every 2 digit sequence that forms a perfect square score 2 points, for every 3 digit sequence that forms a perfect square score 3 points. A four digit square would score 4 points, and so on.

For example, if your number was 58738219024971503664, then you would score 2 points each for 49, 36, and 64 and 6 points for 219024 giving a total of 12 points. You may not count 036 as a three digit square.

What is the maximum number of points you can score?

 See The Solution Submitted by Erik O. No Rating

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 re(3): And some more | Comment 10 of 16 |
(In reply to re(2): And some more by Erik O.)

You only have 20!/2^10 which is more like 2.3759 x 10^15 permutations (divide by 2 for each pair)

This is still over a million billion.  I hope you have a fast computer and/or a long weekend.   This would take over a month at a billion numbers per second.

Still, I am curious.  You should do it anyways.

 Posted by Jer on 2005-03-29 19:02:48

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