# pirates treasure divide problem

A group of 12 pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{th}$ pirate receive?

• What have you tried? This is not a 'do my homework for free' service! You could by the least indicate a restriction on the initial number of coins in the chest. Dec 18 '16 at 20:47

Suppose that the chest has $A$ coins, after the $k$'th pirate passes the chest has $\frac{(12-k)A}{12}$ coins.
So if the chest initially has $A$ coins, after the first $11$ pirates pass it will have $\frac{11!A}{12^{12}}$, this number must be an integer. Therefore $12^{12}$ must divide $A\times(12^{12},11!)=A\times 2^83^4$, so $A$ is at least $\frac{12^{12}}{2^84^3}$, and clearly it works.
So the $12^{th}$ will take $\frac{12!}{12^{12}}$,which is $\frac{1925}{35831808}$. So the smallest number should be 1925.