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2013 AMC 12A 第 17 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

17.

一组 1212 名海盗同意按如下方式分一箱金币。第 kk 个取份额的海盗拿走箱中剩余金币的 k12\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 1212 名海盗得到多少枚金币？

A group of 1212 pirates agree to divide a treasure chest of gold coins among themselves as follows. The kkth pirate to take a share takes k12\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 1212th pirate receive?

720720

12961296

17281728

19251925

38503850

答案：D
知识点：整除性质因数分解阶乘
难度评级：2050
小提示：

在第 kk 名海盗取走份额之前，箱中金币数是取走之后剩余金币数的 1212−k\dfrac{12}{12 - k} 倍

Before the kkth pirate takes a share, the chest holds 1212−k\dfrac{12}{12 - k} times what remains after

大提示：

若最后一名海盗拿到 nn 枚金币，则初始金币数为 1211n11!\dfrac{12^{11}n}{11!}；令它成为整数并使 nn 最小

If nn coins remain for the last pirate, the initial count is 1211n11!;\dfrac{12^{11}n}{11!}; make it an integer with smallest nn

解答：

对 1≤k≤111 \le k \le 11，第 kk 名海盗取走份额之前的金币数，是取走之后金币数的 1212−k\dfrac{12}{12 - k} 倍。所以若留给第 1212 名海盗的是 nn 枚金币，初始金币数为 1211 n11!=214⋅37 n52⋅7⋅11。 \dfrac{12^{11}\, n}{11!} = \dfrac{2^{14}\cdot 3^{7}\, n}{5^2\cdot 7\cdot 11}\text{。}

使其成为正整数的最小 nn 为 52⋅7⋅11=19255^2\cdot7\cdot11=1925。在第 kk 名海盗取金币之前，剩余的金币数等于初始数目乘以 11!(12−k)! 12k−1\frac{11!}{(12-k)!\,12^{k-1}}；代入这个 nn 便可看出它对每个 kk 都是整数。因此每个人分到的都是整数枚金币，而第 1212 名海盗得到 19251925 枚金币。

因此，正确答案是 D。

For 1≤k≤11,1 \le k \le 11, the number of coins before the kkth pirate takes a share is 1212−k\dfrac{12}{12 - k} times the number afterward. So if nn coins are left for the 1212th pirate, the initial count is 1211 n11!=214⋅37 n52⋅7⋅11. \dfrac{12^{11}\, n}{11!} = \dfrac{2^{14}\cdot 3^{7}\, n}{5^2\cdot 7\cdot 11}.

The smallest nn making this a positive integer is 52⋅7⋅11=1925.5^2\cdot7\cdot11=1925. Before pirate k,k, the remaining count is the initial count multiplied by 11!(12−k)! 12k−1;\frac{11!}{(12-k)!\,12^{k-1}}; substituting this nn shows it is an integer for every k.k. Hence all shares are integral, and the 1212th pirate receives 19251925 coins.

Thus, the correct answer is D.

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