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A father on his death bed gave his three sons 17 silver coins. However he had some conditions:

1. The first son will get half of the coins

2. The second son will get one third of the coins

3. Third son will get one ninth of the coins

Knowing that 17 cannot be divided by 2, 6 or 9, three sons needs your help before the father dies. :P

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One of the sons should say, "Here, I'll donate a coin to the stash. If there are any coins left over after we divvy them up, I'll take my donated coin back." Then there will be 18 coins. The first son gets 9, the second son gets 6, and the third son gets 2. That adds up to 17, so, there's one left which the donor son then gets back!

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:(:(
the three brothers got the idea. However unfortunately none of them had a spare silver coin. It was too late, so the father died taking the 17 coins away with him.
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Isn't that cheating?

It seems like the father meant for 1/18 of his money to be set aside for something else.

1 - 1/2 - 1/3 - 1/9 = (1/18)*(18-9-6-2) = 1/18 of sum (less than one coin)

The sons are just helping themselves to this bit of money.

There is 17/18 of a coin left,

Son 1 steals 1/2 of a coin, leaving 8/18 left

Son 2 steals 1/3 of a coin, leaving 2/18 left

Son 3 steals 1/9 of a coin, leaving nothing left

If I hadn't been told this was the answer, I would have assumed that this scheme is invalid.

Edited by mmiguel1
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