Question

六棵苹果树上结的苹果数构成一个等差数列，其中这六棵树中最多的苹果数是最少的苹果数的两倍。六棵树上的苹果总数为 $990$ 。求任意一棵树上最多的苹果数。

The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples growing on any of the six trees.

答案

答案：220

知识点：等差数列一次方程

难度评级：1890

解答：

设六棵树上的苹果数为 $a, a+d, \ldots, a+5d$ ，公差为 $d \ge 0$ 。最多的苹果数是最少的两倍，所以 $a + 5d = 2a$ ，从而 $a = 5d$ 。总数为 $\begin{aligned} 6a + 15d &= 30d + 15d \\ &= 45d = 990, \end{aligned}$ 因此 $d = 22$ 。

最多的苹果数是 $a + 5d = 10d = 220$ 。

Let the six counts be $a, a+d, \ldots, a+5d$ with common difference $d \ge 0.$ The greatest count is double the least, so $a + 5d = 2a,$ which gives $a = 5d.$ The total is $\begin{aligned} 6a + 15d &= 30d + 15d \\ &= 45d = 990, \end{aligned}$ so $d = 22.$

The greatest number of apples is $a + 5d = 10d = 220.$

2023 AIME II 第 1 题

答案：220

难度评级：1890

解答：

其他年份的第 1 题