2013 AMC 12A Problem 17

For $1 \le k \le 11,$ the number of coins before the $k$th pirate takes a share is $\dfrac{12}{12 - k}$ times the number afterward. So if $n$ coins are left for the $12$th pirate, the initial count is $$\dfrac{12^{11}\, n}{11!} = \dfrac{2^{14}\cdot 3^{7}\, n}{5^2\cdot 7\cdot 11}.$$ The smallest $n$ making this a positive integer is $5^2\cdot7\cdot11=1925.$ Before pirate $k,$ the remaining count is the initial count multiplied by $\frac{11!}{(12-k)!\,12^{k-1}};$ substituting this $n$ shows it is an integer for every $k.$ Hence all shares are integral, and the $12$th pirate receives $1925$ coins. Thus, the correct answer is D.