2013 AMC 10A 第 21 题

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

小提示：

从第十二个海盗得到的金币数反向推。

Work backward from the number left for the last pirate

大提示：

反推第

k

个海盗之前的数量时，要乘以

\frac{12}{12-k}

。

The remaining amount is multiplied by

\frac{12}{12-k}

when reversing pirate

k

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}}\text{。}

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

小提示：

大提示：

解答：

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