Question

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 $\dfrac{k}{12}$ 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 $12^{\text{th}}$ pirate receive?

$720$

$1296$

$1728$

$1925$

$3850$

Solution:

Work backward. If $n$ coins remain for the $12$ th pirate, then before pirate $k$ took a share, the chest had $\frac{12}{12-k}$ times as many coins as it had afterward.

Therefore the initial number of coins is $n\cdot\frac{12^{11}}{11!}$ .

Since $\frac{12^{11}}{11!}=\frac{2^{14}3^7}{5\cdot7\cdot11}$ , the smallest $n$ that makes the initial number an integer is $5^2\cdot7\cdot11=1925$ .

Thus, the $12$ th pirate receives $1925$ coins, and D is the correct answer.

2013 AMC 10A Problem 21

Solution:

Problem 21 in Other Years