2013 AMC 12A Problem 17

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

Answer: D

Concepts:divisibility prime factorization factorial

Difficulty rating: 2050

Small hint:

Before the

k

th pirate takes a share, the chest holds

\dfrac{12}{12 - k}

times what remains after

Big hint:

If

n

coins remain for the last pirate, the initial count is

\dfrac{12^{11}n}{11!};

make it an integer with smallest

n

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\cdot7\cdot11=1925.

Before pirate

k,

the remaining count is the initial count multiplied by

\frac{11!}{(12-k)!\,12^{k-1}};

substituting this

n

shows it is an integer for every

k.

Hence all shares are integral, and the

12

th pirate receives

1925

coins.

Thus, the correct answer is D.