渡轮停靠：每站下 3/4 再上 7 人

Ferry Stops

专题: Brainteaser / 脑筋急转弯
难度: L2
来源: QuantQuestion

题目详情

渡轮上初始有未知人数。每次停靠后有 $3/4$ 的人下船，并有 7 人上船。如此再发生 2 次（共 3 次停靠），到达终点。

问：终点时船上人数的最小可能值是多少？

There are an unknown amount of people on a ferry. After the first stop, $\dfrac{3}{4}$ of them get off and $7$ people get on. This happens again at $2$ more stops until you reach the last stop. What is the smallest possible amount of people that could be aboard the ferry after this process is completed (last stop)?

解析

设初始人数为 $x$ 。

三次停靠后人数为

\frac{1}{4}\left(\frac{1}{4}\left(\frac{1}{4}x+7\right)+7\right)+7 =\frac{x+588}{64}.

需为整数且最小。令 $x+588$ 为 64 的最小倍数且大于等于 588，即 640。

则 $x=52$ ，终点人数为 $640/64=10$ 。

Original Explanation

Denote $x$ as the amount of people we started with on the ferry. After the first round, there are $\dfrac{1}{4}x + 7$ people on the ferry. After the second stop, there are $\dfrac{1}{4}\left(\dfrac{1}{4}x + 7\right) + 7 = \dfrac{1}{16}x + \dfrac{35}{4}$ . After the last stop, there are $\dfrac{1}{4}\left(\dfrac{1}{16}x + \dfrac{35}{4}\right) + 7 = \dfrac{1}{64}x + \dfrac{147}{16} = \dfrac{x + 588}{64}$ people on the ferry. Now, we must find the smallest integer $x$ such that $x + 588$ is divisible by $64$ . Note that: $10 \cdot 64 = 640$ which is the smallest integer larger than $588$ that is divisible by $64$ . Thus, there are $640 - 588 = 52$ people on the ferry at the start, meaning there are $10$ people on at the end.