AMC12 2013 A

AMC12 2013 A · Q17

AMC12 2013 A · Q17. It mainly tests Fractions, Basic counting (rules of product/sum).

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 $\frac{k}{12}$ 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 12th pirate receive?

一共有 12 个海盗，他们同意按照以下方式分配一箱金币。第 $k$ 个海盗取份时，取走箱中剩余金币的 $\frac{k}{12}$。箱子最初的金币数是最小使得这种分配方式每个海盗都能得到正整数枚金币的数。第 12 个海盗得到多少枚金币？

(A) 720 720

(B) 1296 1296

(D) 1925 1925

(E) 3850 3850

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): For $1\le k\le 11$, the number of coins remaining in the chest before the $k$th pirate takes a share is $\frac{12}{12-k}$ times the number remaining afterward. Thus if there are $n$ coins left for the $12$th pirate to take, the number of coins originally in the chest is $$ \frac{12^{11}\cdot n}{11!} =\frac{2^{22}\cdot 3^{11}\cdot n}{2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7\cdot 11} =\frac{2^{14}\cdot 3^{7}\cdot n}{5^{2}\cdot 7\cdot 11}. $$ The smallest value of $n$ for which this is a positive integer is $5^{2}\cdot 7\cdot 11=1925$. In this case there are $$ 2^{14}\cdot 3^{7}\cdot \frac{11!}{(12-k)!\cdot 12^{k-1}} $$ coins left for the $k$th pirate to take, and note that this amount is an integer for each $k$. Hence the $12$th pirate receives 1925 coins.

答案（D）：对于 $1\le k\le 11$，第 $k$ 个海盗分配前箱中剩余的金币数是分配后剩余金币数的 $\frac{12}{12-k}$ 倍。因此，若留给第 $12$ 个海盗的金币数为 $n$，则箱中最初的金币总数为 $$ \frac{12^{11}\cdot n}{11!} =\frac{2^{22}\cdot 3^{11}\cdot n}{2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7\cdot 11} =\frac{2^{14}\cdot 3^{7}\cdot n}{5^{2}\cdot 7\cdot 11}. $$ 使其成为正整数的最小 $n$ 为 $5^{2}\cdot 7\cdot 11=1925$。在这种情况下，留给第 $k$ 个海盗领取的金币数为 $$ 2^{14}\cdot 3^{7}\cdot \frac{11!}{(12-k)!\cdot 12^{k-1}}, $$ 并且注意对每个 $k$ 该数都是整数。因此，第 $12$ 个海盗得到 1925 枚金币。

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