1996 AMC 12 第 14 题

Question

令

E(n)

表示

n

的偶数数字之和。例如，

E(5681)=6+8=14

。求

\begin{gathered} E(1)+E(2)+E(3)+\cdots\\ {}+E(100) \end{gathered}\text{。}

Let

E(n)

denote the sum of the even digits of

n.

For example,

E(5681)=6+8=14.

Find

\begin{gathered} E(1)+E(2)+E(3)+\cdots\\ {}+E(100). \end{gathered}

200

360

400

900

2250

在数字前补零，并考虑从

00

到

99

的所有整数

Include leading zeros and consider the integers from

00

through

99

在每一个数位上，每个数字出现的次数都相同

In each digit position, every digit occurs equally often

从

00

到

99

，每个数字在两个数位上都各出现

10

次。正偶数数字的和为

2+4+6+8=20

，所以总贡献为

2\cdot10\cdot20=400

。数

100

不贡献任何正偶数数字，所以正确答案是 C。

From

00

through

99,

each digit occurs

10

times in each of the two positions. The positive even digits sum to

2+4+6+8=20,

so the total contribution is

2\cdot10\cdot20=400.

The number

100

contributes no even positive digit, so the correct answer is C.

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