[IBM Ponder This] July 2018 Solution 

Thank you for showing your solution. I did something very similar to yours as first attempt and got into the same memory problems. It did not scale. So I did another approach. Notice a solution with 7 obscure triplets (if it exists!) will have sum n,n+3,n+6,n+9,n+12,n+15,n+18. So if you instead make a loop for (a,b,c) that will have a given sum, you will only need 7 separate data structures to store the information. When you done with 'n' etc, you can garbage collect the first structure and create a new 'n+21'. Since you step 3 numbers every time, you will need to run it 3 times for n=0,1,2 (mod 3). Another nice benefit here is if your problem stops, you can always continue from where you got to, since this only needs to know the starting sum n. I checked all triplets (a,b,c) having sum < 32000 and only found those listed on solution page. All of them was found within 1 hour calculation, the largest with sum=14963. Almost two days later had checked up to 32000 and not found another solution with 6 obscure triplets in a row, and none with 7 of course. The 32000 limit was due to storing shorts requires a little more memory. Also I do not need to have 7 different data structures, since 6 will do. If I found another solution with 6 triplets I could just test if was 7 solution relative fast. At sum 32000 it took around 30 GB memory(I had too run it it 3 times for values 0,1,2 (mod 3).

Notice that for a candidate triplet the product needs t least 5 factors, otherwise the sum cannot be the same.

Consider it as a LP problem with a nonlineair component. Solve it with AMPL an KNITRO; here is the model:: param Top = 70; var A integer >= 1 = 2 = 3 = 1 = 2 = 3 = 1 = 2 = 3 = 1 = 2 = 3 = 1 = 2 = 3 = 1; s.t. Q02: C - B >= 1; s.t. Q03: F - E >= 1; s.t. Q04: G - F >= 1; s.t. Q05: abs(A - E) >= 1; s.t. Q06: A + B + C = E + F + G; s.t. Q07: A * B * C = E * F * G; s.t. Q12: F1 - E1 >= 1; s.t. Q13: G1 - F1 >= 1; s.t. Q14: abs(A + 1- E1) >= 1; s.t. Q15: A + B + C + 3 = E1 + F1 + G1; s.t. Q16: A * B * C + B * C + A * C + C + A * B + B + A + 1 = E1 * F1 * G1; s.t. Q22: F2 - E2 >= 1; s.t. Q23: G2 - F2 >= 1; s.t. Q24: abs(A + 2 - E2) >= 1; s.t. Q25: A + B + C + 6 = E2 + F2 + G2; s.t. Q26: A * B * C + 2 * B * C + 2 * A * C + 4 * C + 2 * A * B + 4 * B + 4 * A + 8 = E2 * F2 * G2; s.t. Q32: F3 - E3 >= 1; s.t. Q33: G3 - F3 >= 1; s.t. Q34: abs(A + 3 - E3) >= 1; s.t. Q35: A + B + C + 9 = E3 + F3 + G3; s.t. Q36: A * B * C + 3 * B * C + 3 * A * C + 9 * C + 3 * A * B + 9 * B + 9 * A + 27 = E3 * F3 * G3; And the commands to solve it: model c:\temp\ibm1807.mod; option solver knitroampl; solve; display A,B,C,E,F,G >IBM1807.out; display E1,F1,G1 >IBM1807.out; display E2,F2,G2 >IBM1807.out; display E3,F3,G3 >IBM1807.out; close IBM1807.out; This program gives also the alternative triplets.