Question

A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k$ th pirate to take a share takes $\dfrac{k}{12}$ 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 $12$ th pirate receive?

$720$

$1296$

$1728$

$1925$

$3850$

Solution:

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 $5^2\cdot 7\cdot 11 = 1925,$ and one checks each earlier pirate then receives a whole number of coins. The $12$ th pirate receives $1925$ coins.

Thus, the correct answer is D.

2013 AMC 12A Problem 17

Solution:

Problem 17 in Other Years