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The father's coins

#1 bonanova

bonanova

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Posted 09 November 2007 - 10:35 AM

A man divided his collection of silver coins among his sons and their wives.

To the eldest son he said, select for yourself the choicest of the coins, and your wife may take one ninth of the remaining coins.
The next oldest son was given as many coins as the first son, plus one more since the eldest son had first pick, and his wife also received one ninth of the remaining coins.
The third son received one coin more the the second son, and his wife received one ninth of the rest.
Each of the remaining sons also received as many coins as the previous son, plus one coin, and each wife took one ninth of the rest.

Until the last son took his coins. At that point, there were none left for his wife.

So the father said, here are seven gold coins, each worth twice the value of a silver coin.
Divide these coins among yourselves so that each family will own coins of equal value.

Every coin remained intact. No coin was cut.

How many silver coins did the man originally have?
How many sons does the man have?

Spoiler for hint

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#2 belegoth

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Posted 09 November 2007 - 10:59 PM

I'm not sure, because my solution is a bit confusing (to me, that is <!-- s:) --><!-- s:) --> ), so here it goes...:

Spoiler for solution

Edit: yay, correct!
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