Question

和为 $45$ 的连续整数最多可以有多少个？

What is the greatest number of consecutive integers whose sum is $45?$

$9$

$25$

$45$

$90$

$120$

答案：D

知识点：等差数列极端原理

难度评级：1070

解答：

设连续整数共有 $k$ 项，首项为 $a$ 。它们的和为 $\dfrac{k(2a+k-1)}{2}=45$ ，所以 $k$ 必须整除 $90$ 。因此项数不可能超过 $90$ 。

这个上界确实可以达到：共有 $90$ 项，其和为 $45$ 。 $-44, -43, \cdots, 44, 45$

所以连续整数的最大项数是 $90$ 。

所以正确答案是 D。

Suppose there are $k$ consecutive integers with first term $a.$ Their sum is $\dfrac{k(2a+k-1)}{2}=45,$ so $k$ must divide $90.$ Therefore the number of terms cannot exceed $90.$

This bound is attained by $-44, -43, \cdots, 44, 45$ which has $90$ terms and sum $45.$

Thus the greatest possible number of consecutive integers is $90.$

Thus, D is the correct answer.

2019 AMC 10A 第 5 题

答案：D

难度评级：1070

解答：

其他年份的第 5 题