2013 AMC 12A Problems/Problem 17

Problem 17

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

The first pirate takes $\frac{1}{12}$ of the $x$ coins, leaving $\frac{11}{12} x$ .

The second pirate takes $\frac{2}{12}$ of the remaining coins, leaving $\frac{10}{12}*\frac{11}{12}*x$ .

Continuing this pattern, the eleventh pirate must take $\frac{11}{12}$ of the remaining coins after the first ten pirates have taken their share, which leaves $\frac{11!}{12^{12}}*x$ . The twelfth pirate takes all of this.

Note that

$12^{12} = (2^2 * 3)^{12} = 2^{24} * 3^{12}$

$11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2$

All the 2s and 3s cancel out of $11!$ , leaving

$11 * 5 * 7 * 5 = 1925$

in the numerator.

We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, $x$ is the denominator, leaving $1925$ coins for the twelfth pirate.

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2013 AMC 12A Problems/Problem 17

Problem 17

Solution

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