Question

$5$ 个连续偶数的和比前 $8$ 个连续正奇数的和少 $4$ 。这些偶数中最小的是多少？

The sum of $5$ consecutive even integers is $4$ less than the sum of the first $8$ consecutive odd counting numbers. What is the smallest of the even integers?

$6$

$8$

$10$

$12$

$14$

答案：B

知识点：求和一次方程

难度评级：880

解答：

前 $8$ 个正奇数的和是 $1+3+\cdots+15=64$ 。

设最小偶数为 $n$ ，则因此 $n=8$ 。 $\begin{gathered} n+(n+2)+(n+4) \\ {}+(n+6)+(n+8) \\ = 5n+20 = 60, \end{gathered}$

所以正确答案是 B。

The first $8$ odd counting numbers sum to $1+3+\cdots+15=64.$

Letting $n$ be the smallest even integer, $\begin{gathered} n+(n+2)+(n+4) \\ {}+(n+6)+(n+8) \\ = 5n+20 = 60, \end{gathered}$ so $n=8.$

Thus, the correct answer is B.

2003 AMC 10B 第 3 题

答案：B

难度评级：880

解答：

其他年份的第 3 题