A J B   I C \ / H\ /D GFE 

Source: Reader Chuck Johnson, citing his grandson.
Let the common sum of each side be SUM . Then, (A+B+C)+(C+D+E)+(E+F+G)+(G+H+I)+(I+J+A) = 5 * SUM => A+C+E+G+I+55 = 5 * SUM ( as the sum of the first 10 integers = 55 ) => A+C+E+G+I = 5 * ( SUM  11 ) Now, 1+2+3+4+5 < = A+C+E+G+I < = 6+7+8+9+10 => 14 < = SUM < = 19 The only possible solutions are : SUM = 14 1 10 9 3 4 6 8 5 7 2 SUM = 16 1 1 10 8 8 5 5 7 7 10 2 6 6 2 9 4 3 3 9 4 SUM = 17 2 1 9 7 9 6 6 8 7 10 1 5 2 3 10 3 4 8 5 4 SUM = 19 6 5 4 8 9 1 3 10 2 7 The last 3 solutions can be obtained from the first 3 on replacing each number by ( 11  NUMBER ) .