Question

求所有整数等差数列的第 $10$ 项之和，这些数列的首项都等于 $4$ ，并且都包含 $24$ 和 $34$ 作为其中的项。

Find the sum of the $10$ th terms of all arithmetic sequences of integers that have first term equal to $4$ and include both $24$ and $34$ as terms.

答案

答案：178

知识点：等差数列整除性最大公约数

难度评级：1840

解答：

设公差为 $d$ 。因为首项是 $4$ ，且 $24$ 和 $34$ 都出现，所以 $d \mid 20$ 且 $d \mid 30$ ，也就是 $d$ 整除 $24 - 4 = 20$ 和 $34 - 4 = 30$ ，因此 $d$ 整除 $\gcd(20, 30) = 10$ 。为了从 $4$ 到达 $24$ 和 $34$ ，公差必须为正，所以 $d \in \{1, 2, 5, 10\}$ ；这些值也都可行，因为它们都同时整除二十和三十。

第 $10$ 项为 $4 + 9d$ ，所以所求和为 $\begin{aligned} &\sum_{d \in \{1,2,5,10\}} (4 + 9d) \\ &= 4 \cdot 4 + 9(1 + 2 + 5 + 10) \\ &= 16 + 162 = 178. \end{aligned}$

Let the common difference be $d.$ Since the first term is $4$ and both $24$ and $34$ appear, $d$ divides $24 - 4 = 20$ and $34 - 4 = 30,$ so $d$ divides $\gcd(20, 30) = 10.$ The difference must be positive to reach $24$ and $34$ from $4,$ so $d \in \{1, 2, 5, 10\}$ (and each of these works, since $d \mid 20$ and $d \mid 30$ put both targets in the sequence).

The $10$ th term is $4 + 9d,$ so the requested sum is $\begin{aligned} &\sum_{d \in \{1,2,5,10\}} (4 + 9d) \\ &= 4 \cdot 4 + 9(1 + 2 + 5 + 10) \\ &= 16 + 162 = 178. \end{aligned}$

2026 AIME II 第 1 题

答案：178

难度评级：1840

解答：

其他年份的第 1 题