2005 AMC 12B 第 15 题

Question

四个两位数的和是

221

。这八个数字中没有

0

，且互不相同。下列哪一个数字没有出现在这八个数字中？

The sum of four two-digit numbers is

221.

None of the eight digits is

0

and no two of them are the same. Which of the following is not included among the eight digits?

1

2

3

4

5

答案：D

知识点：数字位值分类讨论

难度评级：1660

小提示：

使用的八个不同非零数字的总和在

36

到

44

之间。

The eight distinct nonzero digits used have a total between

36

and

44

大提示：

若个位数字和为

U

，十位数字和为

T

，则

10T + U = 221

，所以

U

的个位是

1

。

If the units digits sum to

U

and the tens digits to

T,

then

10T + U = 221,

so

U

ends in

1

解答：

八个数字来自

1

到

9

，全部非零数字的和为

45

，所以这八个数字的总和在

45 - 9 = 36

与

45 - 1 = 44

之间。

设个位数字和为

U

，十位数字和为

T

，则

10T + U = 221

，所以

U

的个位为

1

。又

1+2+3+4 = 10

\le U \le

6+7+8+9 = 30

，故

U = 11

或

U = 21

。

若

U = 11

则

10T = 210

，所以

T = 21

，八个数字总和为

32

，小于

36

，不可能。因此

U = 21

，

T = 20

，总和为

41

。

缺失数字为

45 - 41 = 4

。例如

13 + 25 + 86 + 97 = 221

。

所以正确答案是 D。

The eight digits are distinct and chosen from

1

through

9,

whose total is

45.

So the eight used digits sum to between

45 - 9 = 36

and

45 - 1 = 44.

Let the four units digits sum to

U

and the four tens digits sum to

T.

Then

10T + U = 221,

so

U

ends in

1.

Since

1+2+3+4 = 10

\le U \le

6+7+8+9 = 30,

we have

U = 11

or

U = 21.

If

U = 11,

then

10T = 210,

so

T = 21

and the eight digits sum to

32,

which is below

36.

So

U = 21,

giving

T = 20

and total

41.

The missing digit is

45 - 41 = 4.

For example,

13 + 25 + 86 + 97 = 221.

Thus, the correct answer is D.

2005 AMC 12B 第 15 题

答案：D

难度评级：1660

小提示：

大提示：

解答：

其他年份的第 15 题