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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?

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    $\begingroup$ 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. $\endgroup$ Commented Dec 18, 2016 at 20:47

2 Answers 2

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We can get below table for each pirate if assume total number as 1:

\begin{array} {|l|l|l|} \hline N & Taken & Remaining \\ \hline 1 & \frac{1}{12} & \frac{11}{12} \\ \hline 2 & \frac{11}{12}\cdot \frac{2}{12} & \frac{11}{12}\cdot \frac{10}{12} \\ \hline 3 & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{3}{12} & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12} \\ \hline 4 & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{4}{12} & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12} \\ \hline 5 & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{5}{12} & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12} \\ \hline 6 & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12} & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12} \\ \hline 7 & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{7}{12} & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12} \\ \hline 8 & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{8}{12} & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{4}{12} \\ \hline 9 & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{4}{12}\cdot \frac{9}{12} & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{4}{12}\cdot \frac{3}{12} \\ \hline 10 & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{4}{12}\cdot \frac{3}{12}\cdot \frac{10}{12} & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{4}{12}\cdot \frac{3}{12}\cdot \frac{2}{12} \\ \hline 11 & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{4}{12}\cdot \frac{3}{12}\cdot \frac{2}{12}\cdot \frac{11}{12} & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{4}{12}\cdot \frac{3}{12}\cdot \frac{2}{12}\cdot \frac{1}{12} \\ \hline 12 & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{4}{12}\cdot \frac{3}{12}\cdot \frac{2}{12}\cdot \frac{1}{12}\cdot \frac{12}{12} & \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{4}{12}\cdot \frac{3}{12}\cdot \frac{2}{12}\cdot \frac{1}{12}\cdot \frac{0}{12} \\ \hline \end{array}

So the $12^{th}$ will take $\frac{12!}{12^{12}}$,which is $\frac{1925}{35831808}$. So the smallest number should be 1925.

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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.

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