Take one 1, two 2s, three 3s, four 4s, five 5s, six 6s, seven 7s, and eight 8s and place them in a 6-by-6 grid, one digit per square, such that each row, column, and major diagonal sums to 34.

In my solution, only one major diagonal adds to 34.  The other adds to 33.  Notice, however, that in the problem, he only says one major diagonal has to sum to 34 if you interpret the 'each' as applying to 'row' and 'column' only.

3 8 6 3 8 6

4 7 2 8 8 5

8 5 6 3 6 6

4 7 7 7 1 8

8 2 7 5 7 5

7 5 6 8 4 4

Edited on October 30, 2004, 2:56 am

Posted by Dustin on 2004-10-30 02:31:16