2013 AMC 10A Problem 21

Work backward. If $n$ coins remain for the $12$th pirate, then before pirate $k$ took a share, the chest had $\frac{12}{12-k}$ times as many coins as it had afterward. Therefore the initial number of coins is $n\cdot\frac{12^{11}}{11!}$. Since $\frac{12^{11}}{11!}=\frac{2^{14}3^7}{5\cdot7\cdot11}$, the smallest $n$ that makes the initial number an integer is $5^2\cdot7\cdot11=1925$. This value is attainable. The number of coins present just before pirate $k$ takes a share is $$2^{14}3^7\cdot \frac{11!}{(12-k)!\,12^{k-1}}.$$ For $k=1,2,\ldots,12$, all these amounts are integers. Each pirate’s share is the difference between two consecutive remaining amounts, so every share is an integer as well. Thus, the $12$th pirate receives $1925$ coins, and D is the correct answer.