Problem:
A group of $ 12 $ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $ k^\text{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^{\text{th}} $ pirate receive?
$ \textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850 $
Solution:
We work backwards.
Let there be $c$ coins before the 12th pirate takes his share. Then, since the 11th pirate takes $\frac{11}{12}^{\text{th}}$ of whatever he is given, he leaves $\frac{1}{12}^{\text{th}}$ of it behind. So, $c$ must be one twelfth of the amount before the 11th pirate. Hence, there were $\frac{12}{1}c$ coins before the 11th pirate. Similarly, the 10th pirate takes $\frac{10}{12}^{\text{th}}$ of what he receives and leaves behind $\frac{2}{12}^{\text{th}}$ of it, so the number of coins before the 10th pirate would be $\frac{12}{2}$ times the number of coins before the 11th pirate, which is $\left(\frac{12}{1}\right)\left(\frac{12}{2}\right)c$. Continuing this pattern, we get that in terms of $c$, the original number of coins must have been$$\left(\frac{12}{1}\right)\left(\frac{12}{2}\right)\left(\frac{12}{3}\right)\cdots\left(\frac{12}{11}\right)c.$$Now we need to pick the smallest $c$ such that this value is a positive integer. We can write it as$$\frac{12^{10}}{11!}c=\frac{2^{20} \cdot 3^{10}}{2^8 \cdot 3^4 \cdot 5 \cdot 7 \cdot 5 \cdot 11}c=\frac{2^{12} \cdot 3^6}{5^2 \cdot 7 \cdot 11}c.$$Then $c$ must cancel the $5^2 \cdot 7 \cdot 11$ such that this value is an integer. The smallest positive integer $c$ that can do this is $c=5^2 \cdot 7 \cdot 11$ itself, which is equal to $1925 \longrightarrow \boxed{\textbf{(D)}}$.
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