2013 AMC 10A Problem 21

Going forward, the pile keeps shrinking by awkward fractions. Going backwards from the last pirate is easier.

1. What each pirate leaves. Pirate $k$ takes $\tfrac{k}{12}$ of the pile, so he leaves $\tfrac{12 - k}{12}$ of it. Pirate $1$ leaves $\tfrac{11}{12}$, pirate $2$ leaves $\tfrac{10}{12}$, and so on, down to pirate $11$, who leaves $\tfrac{1}{12}$. Pirate $12$ takes $\tfrac{12}{12}$, which is everything left.
2. Call the last share $F$. Pirate $12$ gets the whole final pile, $F$ coins.
3. Step backwards. Pirate $11$ left $\tfrac{1}{12}$ of his pile, so his pile was $12$ times as big: $F \times \tfrac{12}{1}$. Pirate $10$ left $\tfrac{2}{12}$, so his pile was that times $\tfrac{12}{2}$. Keep going:

4. What must be true. Every pile must be a whole number. (Then each pirate's share, the pile before him minus the pile after, is whole too.) The starting pile grows when $F$ grows, so the smallest start means the smallest $F$.
5. The 2s and 3s take care of themselves. Each $12 = 2 \times 2 \times 3$ brings two 2s and one 3. The numbers we divide by bring fewer. For example, $1 \times 2 \times \cdots \times 11$ contains only eight 2s and four 3s, while eleven 12s bring twenty-two 2s and eleven 3s. The same holds at every row of the table.
6. The other primes do not. We also divide by $5$, $7$, $10 = 2 \times 5$ and $11$. Twelves never contain $5$, $7$ or $11$, so $F$ itself must supply them: two 5s (one for $5$, one for $10$), one $7$ and one $11$. So $F$ is a multiple of
$$5 \times 5 \times 7 \times 11 = 1925.$$
7. 1925 works. With $F = 1925$ every factor in the table cancels, so every pile is a whole number.

The twelfth pirate receives $\boxed{\textbf{(D)}\ 1925}$ coins.

When the rule is "take a fraction of what is left", work with what is left, not with what is taken. Walking backwards turns each step into "multiply by 12, divide by a small number." Then the whole question is which prime factors the 12s cannot supply: here $5, 5, 7$ and $11$.

Use the answer choices. The last share is what is left of $1 \times 2 \times \cdots \times 11$ after all its 2s and 3s are cancelled by twelves, so it cannot be even or a multiple of $3$. Choices (A) $720$, (B) $1296$, (C) $1728$ and (E) $3850$ are all even. Only (D) $1925$ is left.

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 $\boxed{\textbf{(D)}\ 1925}$.