2003 AMC 10B 第 3 题

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

前

8

个正奇数的和为

8^2

。

The first

8

odd counting numbers sum to

8^2

将这些偶数写成

n

、

n+2

、

n+4

、

n+6

、

n+8

，并令它们的和等于

64-4

。

Write the even integers as

n,

n+2,

n+4,

n+6,

n+8

and set their sum equal to

64-4

前

8

个正奇数的和是

1+3+\cdots+15=64

。

设最小偶数为

n

，则

\begin{gathered} n+(n+2)+(n+4) \\ {}+(n+6)+(n+8) \\ = 5n+20 = 60 \end{gathered}\text{，}

因此

n=8

。

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.

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