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Another Square (Posted on 2010-07-06) Difficulty: 4 of 5

No Solution Yet Submitted by brianjn    
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Some Thoughts re(4): Solution (as interpreted) | Comment 7 of 9 |
(In reply to re(3): Solution (as interpreted) by brianjn)

I had arrived at the same solution(s) as ed and Charlie. But, due to the large number of possible solutions, I tried to see if there could be an interpretation that might limit the bulk to a smaller, if not unique, set.

With the assumption that a solution might be sought such that each digit appeared only once in the set of bases, I arrived at the following:

  • 1614, 3636, 4920, 6457, 10098
  • 1614, 3636, 4920, 6458, 10097
  • 1616, 3630, 4924, 6457, 10098
  • 1616, 3630, 4924, 6458, 10097
  • 1616, 3634, 4920, 6457, 10098
  • 1616, 3634, 4920, 6458, 10097
The sum of the bases add to the perfect square 225 (152).
Of these there are two sets with the minimum total of their base 10 values. This total is 76. (The other four sets total 80 each):
  • 1616=10, 3630=16, 4924=21, 6457=17, 10098=12
  • 1616=10, 3630=16, 4924=21, 6458=16, 10097=13

      Posted by Dej Mar on 2010-07-07 08:16:52
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