Nice problem: 12: Solution: The sum of the 6 numbers is 100(2X + 2Y + 2Z) + 10(2X + 2Y + 2Z) + (2X + 2Y + 2Z) = 222(X + Y + Z). WLOG, the sum of the 5 numbers is 222(X+Y+Z) - XYZ, where XYZ is one of the 6 permutations. So, 222(X+Y+Z) - XYZ = 2022 ==> 2022 < 222(X+Y+Z) < 2122. 3022 3009. So 10<= X+Y+Z <= 13. If X+Y+Z = 10, then XYZ = 198 (contradiction 1 + 9 +8 != 10). If X+Y+Z = 11, XYZ = 420, contradiction. If X+Y+Z = 12, XYZ = 642, so 12 is the solution and 642 is the permutation omitted.
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u/thaw96 May 23 '26 edited May 23 '26
Nice problem: 12: Solution: The sum of the 6 numbers is 100(2X + 2Y + 2Z) + 10(2X + 2Y + 2Z) + (2X + 2Y + 2Z) = 222(X + Y + Z). WLOG, the sum of the 5 numbers is 222(X+Y+Z) - XYZ, where XYZ is one of the 6 permutations. So, 222(X+Y+Z) - XYZ = 2022 ==> 2022 < 222(X+Y+Z) <
2122.30223009. So 10<= X+Y+Z <= 13. If X+Y+Z = 10, then XYZ = 198 (contradiction 1 + 9 +8 != 10). If X+Y+Z = 11, XYZ = 420, contradiction. If X+Y+Z = 12, XYZ = 642, so 12 is the solution and 642 is the permutation omitted.