2013 AMC 10A Problem 21

Let $N$ be the initial number of coins. When pirate $k$ takes $\frac{k}{12}$ of the pile, the pile shrinks to $\frac{12-k}{12}$ of its size. After pirates $1$ through $11$ have gone, the chest holds
$$N \cdot \frac{11}{12} \cdot \frac{10}{12} \cdots \frac{1}{12} = \frac{N \cdot 11!}{12^{11}},$$
and pirate $12$ takes $\frac{12}{12}$ of this, i.e. all of it.

Every pirate's share is a whole number exactly when the pile is a whole number after each step; the tightest requirement is the final one, that $\frac{N \cdot 11!}{12^{11}}$ be an integer (the earlier piles are this quantity multiplied back by fractions with smaller denominators, and one checks they stay integral for the $N$ found below).

Factor: $12^{11} = 2^{22} \cdot 3^{11}$ and $11! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7 \cdot 11$. So $N$ must supply $2^{22-8} \cdot 3^{11-4} = 2^{14} \cdot 3^{7}$, and the smallest choice is $N = 2^{14} \cdot 3^7$.

Then the twelfth pirate receives
$$\frac{2^{14} \cdot 3^7 \cdot 11!}{2^{22} \cdot 3^{11}} = \frac{11!}{2^8 \cdot 3^4} = 5^2 \cdot 7 \cdot 11 = 1925.$$

The answer is $\boxed{\textbf{(D)}\ 1925}$.

Tracking what remains rather than what is taken turns twelve steps into one product. The whole-number condition is then a question about which prime powers of $12^{11}$ survive after cancelling with $11!$; the minimal $N$ is precisely the leftover, and the final share is the part of $11!$ made of primes other than $2$ and $3$.

Answer-choice check: the twelfth pirate's share is $\frac{N \cdot 11!}{12^{11}}$ where $N$ is chosen as small as possible, so the share is $11!$ with all factors of $2$ and $3$ stripped away, and must be a multiple of $5^2 \cdot 7 \cdot 11 = 1925$ with no factor $2$ or $3$. Only (D) is odd and not divisible by $3$.

• Setting N = 12^11 (or checking only that each fraction has denominator 12) instead of cancelling the factors of 2 and 3 that 11! already supplies.
• Miscounting the powers of $2$ in $11!$ (it is $8$: from $2, 4, 6, 8, 10$ contribute $1 + 2 + 1 + 3 + 1$).