so that the sum of the numbers along each side of the triangle is the same AND so that the product of the numbers at the vertices (pink cells) equals the product of the numbers in the middle (green cells) ? Hint: The sum of the numbers in the center and the sum of the numbers at the vertices have to be equivalent modulo 3.

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Here is my solution.

The numbers in the interior of the triangle are: 3,8,10

The numbers on the vertices are: 4,5,12

3+8+10 = 4+5+12 = 21

3*8*10 = 4*5*12 = 240

The sides of the triangle are (4,7,11,13,5), (5,2,6,15,12), & (12,9,14,1,4) which all sum up to 40.

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Very nice!

I put 15, 14, and 3 in the middle and 7, 9, and 10 at the corners since 15*14*3 = 7*9*10 = 630, and since 15 + 14 + 3 = 32 and 7 + 9 + 10 = 26 have the same remainder mod 3. Then I determined that the sides had to add up to 38 since

(1+2+...+14+15) - (15+14+3) + (7+9+10) = 114

and 114/3 = 38. The side numbers are {7, 4, 6, 12, 9}, {9, 1, 5, 13, 10}, and {10, 2, 8, 11, 7}.

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