Question

$12$ 名海盗同意按如下方式分一箱金币。第 $k^{\text{th}}$ 个拿份额的海盗取走箱中剩余金币的 $\dfrac{k}{12}$ 。箱中最初的金币数是使得每名海盗都能得到正整数枚金币的最小数。第 $12^{\text{th}}$ 个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 $k^{\text{th}}$ 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 $12^{\text{th}}$ pirate receive?

$720$

$1296$

$1728$

$1925$

$3850$

答案：D

知识点：整除性质因数分解逆推法

难度评级：2300

解答：

反向计算。若第 $12$ 个海盗分到时还剩 $n$ 枚金币，那么在第 $k$ 个海盗拿走份额之前，箱中的金币数是他拿完之后的 $\frac{12}{12-k}$ 倍。

因此最初的金币数为 $n\cdot\frac{12^{11}}{11!}$ 。

由于 $\frac{12^{11}}{11!}=\frac{2^{14}3^7}{5\cdot7\cdot11}$ ，使最初金币数为整数的最小 $n$ 是 $5^2\cdot7\cdot11=1925$ 。

这个值确实可以达到。在第 $k$ 个海盗拿走份额之前，金币数为 $2^{14}3^7\cdot \frac{11!}{(12-k)!\,12^{k-1}}.$ 对 $k=1,2,\ldots,12$ ，这些数全是整数。每个海盗拿到的份额是相邻两个剩余量之差，所以每份也都是整数。

因此第 $12$ 个海盗得到 $1925$ 枚金币，正确答案是 D。

Work backward. If $n$ coins remain for the $12$ th pirate, then before pirate $k$ took a share, the chest had $\frac{12}{12-k}$ times as many coins as it had afterward.

Therefore the initial number of coins is $n\cdot\frac{12^{11}}{11!}$ .

Since $\frac{12^{11}}{11!}=\frac{2^{14}3^7}{5\cdot7\cdot11}$ , the smallest $n$ that makes the initial number an integer is $5^2\cdot7\cdot11=1925$ .

This value is attainable. The number of coins present just before pirate $k$ takes a share is $2^{14}3^7\cdot \frac{11!}{(12-k)!\,12^{k-1}}.$ For $k=1,2,\ldots,12$ , all these amounts are integers. Each pirate's share is the difference between two consecutive remaining amounts, so every share is an integer as well.

Thus, the $12$ th pirate receives $1925$ coins, and D is the correct answer.

2013 AMC 10A 第 21 题

答案：D

难度评级：2300

解答：

其他年份的第 21 题