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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.
          A
        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)

Comments: ( Back to comment list | You must be logged in to post comments.)
wording change | Comment 4 of 15 |
I seem to remember that the three number sums were originally different when this problem was waiting in the queue - possibly when it was re-formatted the intent was altered? I could easily be wrong of course, but the three number equivalent sums were originally;

D+E+F = B+E+I = C+E+H

I also don't know if this would reduce the redundancies, as I never worked through this one.
  Posted by Cory Taylor on 2003-12-03 11:31:00
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