Adrian asked some friends to each make up a list of four perfect squares greater than zero, each having no more than four digits, that together use all ten digits exactly once.

Alice's and Brian's lists contained exactly two squares in common. Carol's list had no square in common with either Alice's or Brian's. David found a different set.

What was Carol's list and what was David's list?

There are only 58 squares between 1 and 99^2 with no repeated digits. Out of these, the only subsets of size 4 which exactly partition the digits 0-9 are:

1, 36, 784, 9025
9, 16, 784, 3025
9, 81, 324, 7056
9, 81, 576, 2304

Now, Alice's and Brian's lists must be different, so at least one of them contains a 9. Therefore Carol must have list #1. This limits Alice and Brian to lists #3 and #4. (So the fact that A and B have 2 squares in common is unnecessary, as long as they're distinct sets.) Finally, David must have list #2.

Mathematica code:
<<DiscreteMath`Combinatorica`
S=Select[IntegerDigits/@(Range@99^2),Length@Union@#==Length@#&];
Select[KSubsets[S,4],Sort[Join@@#]==Range[0,9]&]
Edited on March 17, 2008, 1:55 am

Posted by Eigenray on 2008-03-17 01:28:10