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Tricky Triangle (Posted on 2003-12-03) Difficulty: 4 of 5
Replace the letters in the diagram with a different number from 0 to 9, such that the sum of the four numbers on all edges are the same, and the sum of the three numbers on all three corners are the same.
        B   C
      D   E   F
    G   H   I   J

In other words, A+C+F+J = A+B+D+G = G+H+I+J and
A+B+C = D+G+H = F+I+J.

No Solution Yet Submitted by Kelsey    
Rating: 3.2500 (4 votes)

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Solution re(2): Multiple solutions.--down to two | Comment 7 of 15 |
(In reply to re: Multiple solutions. by wonshot)

A bug in my original program (failure to set taken(g)), resulted in those duplicate digits. Correcting that, we're left with 2 solutions:

8 9
2 4 0 4
6 3 5 2 6 5
1 7 9 0 8 3 7 1

Note that the 3's across the center do add up to a constant in each solution: 14 and 13 respectively, the same as the corner triangles.

These solutions, when rotated and reflected, produce the same twelve as found by Federico Kereki.
Edited on December 3, 2003, 2:18 pm
  Posted by Charlie on 2003-12-03 14:12:03
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