Question

一组 $12$ 名海盗同意按如下方式分一箱金币。第 $k$ 个取份额的海盗拿走箱中剩余金币的 $\dfrac{k}{12}$ 。箱中最初的金币数是能使每名海盗都得到正整数枚金币的最小数。第 $12$ 名海盗得到多少枚金币？

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?

$720$

$1296$

$1728$

$1925$

$3850$

答案：D

知识点：整除性质因数分解阶乘

难度评级：2050

解答：

对 $1 \le k \le 11$ ，第 $k$ 名海盗取走份额之前的金币数，是取走之后金币数的 $\dfrac{12}{12 - k}$ 倍。所以若留给第 $12$ 名海盗的是 $n$ 枚金币，初始金币数为 $\dfrac{12^{11}\, n}{11!} = \dfrac{2^{14}\cdot 3^{7}\, n}{5^2\cdot 7\cdot 11}.$

使其成为正整数的最小 $n$ 为 $5^2\cdot7\cdot11=1925$ ，并且可以检查每个更早的海盗也都得到整数枚金币。第 $12$ 名海盗得到 $1925$ 枚金币。 $k$ ， $11!/((12-k)!\,12^{k-1})$ ； $n$ $k$ 。

因此，正确答案是 D。

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 $11!/((12-k)!\,12^{k-1});$ 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.

2013 AMC 12A 第 17 题

答案：D

难度评级：2050

解答：

其他年份的第 17 题