AMC10 2013 A

AMC10 2013 A · Q21

AMC10 2013 A · Q21. It mainly tests Rational expressions, Fractions.

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名海盗，他们同意按照如下方式分配一箱金币。第$k$名海盗拿走剩余金币的$\frac{k}{12}$。箱子中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金币的数。第12名海盗得到多少金币？

(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 $k^{\text{th}}$ pirate takes a share is $\frac{12}{12-k}$ times the number remaining afterward. Thus if there are $n$ coins left for the $12^{\text{th}}$ 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 $\frac{2^{14}\cdot 3^7\cdot 11!}{(12-k)!\cdot 12^{k-1}}$ coins left for the $k^{\text{th}}$ pirate to take, and note that this amount is an integer for each $k$. Hence the $12^{\text{th}}$ pirate receives 1925 coins.

答案（D）：对 $1 \le k \le 11$，在第 $k^{\text{th}}$ 个海盗分得一份之前，箱子里剩下的硬币数是他分完之后剩余硬币数的 $\frac{12}{12-k}$ 倍。因此，如果留给第 $12^{\text{th}}$ 个海盗的硬币数为 $n$，那么箱子里最初的硬币数为 $\frac{12^{11}\cdot n}{11!}=\frac{2^{22}\cdot 3^{11}\cdot n}{2^8\cdot 3^4\cdot 5^2\cdot 7\cdot 11}=\frac{2^{14}\cdot 3^7\cdot n}{5^2\cdot 7\cdot 11}.$ 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 $k^{\text{th}}$ 个海盗取走的硬币数为 $\frac{2^{14}\cdot 3^7\cdot 11!}{(12-k)!\cdot 12^{k-1}}$ 并且注意对每个 $k$ 该数都是整数。因此，第 $12^{\text{th}}$ 个海盗得到 1925 枚硬币。

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