imc

Let $x$ be the number of coins. After the $k^{\text{th}}$ pirate takes his share, $\frac{12-k}{12}$ of the original amount is left. Thus, we know that $x\cdot \frac{11}{12}\cdot \frac{10}{12}\cdot \frac{9}{12}\cdot \frac{8}{12}\cdot \frac{7}{12}\cdot \frac{6}{12}\cdot \frac{5}{12}\cdot \frac{4}{12}\cdot \frac{3}{12}\cdot \frac{2}{12}\cdot \frac{1}{12}$ must be an integer. Simplifying, we get $x\cdot \frac{11}{12}\cdot \frac{5}{6}\cdot \frac{1}{2}\cdot \frac{7}{12}\cdot \frac{1}{2}\cdot \frac{5}{12}\cdot \frac{1}{3}\cdot \frac{1}{4}\cdot \frac{1}{6}\cdot \frac{1}{12}$. Now, the minimal $x$ is the denominator of this fraction multiplied out, obviously. We mentioned before that this product must be an integer. Specifically, it is an integer and it is the amount that the $12^{\text{th}}$ pirate receives, as he receives $\frac{12}{12}=1=$ all of what is remaining. Thus, we know the denominator is canceled out, so the number of gold coins received is going to be the product of the numerators, $11\cdot 5\cdot 7\cdot 5=\boxed{1925}$.