(related to Problem: The Barrel of Beer)
Here the digital roots of the six numbers are $6, 4, 1, 2, 7, 9,$ which together sum to $29,$ whose digital root is $2.$ As the contents of the barrels sold must be a number divisible by $3,$ if one buyer purchased twice as much as the other, we must find a barrel with root $2, 5,$ or $8$ to set on one side. There is only one barrel, that containing $20$ gallons, that fulfills these conditions. So the man must have kept these $20$ gallons of beer for his own use and sold one man $33$ gallons (the $18$-gallon and $15$-gallon barrels) and sold the other man $66$ gallons (the $16,$ $19,$ and $31$ gallon barrels).
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