2013 AMC 10A #21

 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?

$\mathbf{\text{(A)}}720\mathbf{\text{(B)}}1296\mathbf{\text{(C)}}1728\mathbf{\text{(D)}}1925\mathbf{\text{(E)}}3850$

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 $1925⟶\boxed{{\displaystyle \mathbf{\text{(D)}}}}$.