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# The father's coins

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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?

56 silver coins; 7 sons.

The eldest son selected two coins.

Each family received one gold coin.

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

No coins were cut, therefore each time one of the wives took 1/9 of the coins, the current number of coins must have been a multiple of 9. This results in the fact that the coins of Wife1+Son2 are an exact multiple of 9 (lets say 9*k). This is, therefore, also true for Wife2+Son3, Wife3+Son4... Wife(n-1)+Son(n) = 9*k. (where n is the number of sons=number of wives). If x is the total number of silver coins, then:

x=Son1+(n-1)*9*k;

Since each son takes 1 coin more than the one before him, then Son2=Son1+1, Son3=Son2+2 ... Son(n)=Son(n-1)+1.

But, as Wife(i)+Son(i+1)=9*k=Wife(i+1)+Son(i+2) (for any integer 1<=i<=n); that means that

a.Wife(i)=Wife(i-1)-1;

b.Wife(i)+Son(i)=9*k-1; which means that all the pairs took equal amounts of silver coins. They were additionally given 7 gold coins, which were divided equally without cutting, but 7 is a prime number, thus n=7 and every couple took 1 gold coin.

Its easy from here on:

Wife7=0; therefore Wife6=1;Wife5=2 ... Wife1=6;

6=1/9 * (x-Son1)

x-Son1 = 54;

but since

x=(n-1)*9*k+Son1 -----> (n-1)*9*k=54 -----> 6*9*k=54 -----> k=1

9*k=Wife1+Son2 -----> 9=6+Son2 -----> Son2=3 -----> Son1=Son2-1=2

x=54+2=56

Answer: x=56 silver coins, n=7 sons.

Edit: yay, correct! :D

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