Source: Reader Shivani Goyal.
Solution from Joseph DeVincentis: Use the 4-4, 4-3, and all dominoes with sum 9 or more. Sum 19 is the largest possible since no 8 dominoes add to 80. 6-6 4-3 4-4 5-6 4 6 4 5 | | | | 5 3 6 5
Claudio Baiocchi's solution: Let us try to minimize the common sum (thus working with smaller numbers); then replacing each entry X with 6-X we will get the maximum solution. The smaller pieces in a double-six dominoes are: - the 0-sum piece (0,0); - the 1-sum piece (0,1); - the 2-sum pieces (0,2), (1,1); - the 3-sum pieces (0,3), (1,2); - the 4-sum pieces (0,4), (1,3), (2,2); - the 5-sum pieces (0,5), (1,4), (2,3). In particular a magic sum 4, requiring 8 pieces with total sum 16, is impossible; while many choices of 8 pieces give raise to a total sum 20, thus a priori suitable for a magic sum 5. In fact, almost any 8-tuple of pieces with total sum 20 gives raise to (many) magic-squares with magic sum 5; the unique exception being that the piece (0,5) must be avoided. As an example of magic-sum 5 solution and its 6-complement with magic-sum 19 we have the "fully horizontal" solution: (0,0),(2,3) (6,6),(4,3) (1,2),(1,1) (5,4),(5,5) (3,0),(2,0) (3,6),(4,6) (1,3),(0,1) (5,3),(6,5)