Re: Gold Bar Riddle

As promised, click here the solution.

At the end of day two, the pharaoh is able to make one more cut and must pay the mason 2/7ths of the gold bar. The only way to do this is to use the larger (6/7th) piece and cut out a 2/7th piece. The pharaoh takes back the 1/7th piece and gives the 2/7th piece. Now the pharaoh has all made all of his alloted cuts and there are 3 pieces of gold bar: a 1/7th piece, a 2/7th piece, and the remaining 4/7th piece.

With these three pieces, the pharaoh is able to pay the mason every day the fraction amount of the entire gold bar that corresponds to the number of days work. I think the trickiest part was realizing that the mason can give back pieces as change.

Day 1: Pay with the 1/7th piece
Day 2: Pay with the 2/7th piece
Day 3: Pay with the 1/7th and 2/7th pieces (1/7 + 2/7 = 3/7)
Day 4: Pay with the 4/7th piece
Day 5: Pay with the 1/7th and 4/7th piece (1/7 + 4/7 = 5/7)
Day 6: Pay with the 2/7th and 4/7th pieces (2/7 + 4/7 = 6/7)
Day 7: Pay with all the pieces (1/7 + 2/7 + 4/7 = 7/7)

Of course, this is only possible if the mason doesn’t go off and spend his pay before the end of the week 🙂

Re: Gold Bar Riddle

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