2007 AMC 12A 第 16 题

Question

有多少个三位数由三个互不相同的数字组成，并且其中一个数字是另外两个数字的平均数？

How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

96

104

112

120

256

答案：C

知识点：等差数列数字分类讨论

难度评级：1630

小提示：

这三个数字必须按某种顺序构成一个等差数列。

The three digits must form an arithmetic progression in some order

大提示：

按公差数出等差数列组，再排列每组数字；含

0

的组排列数较少。

Count the progressions by common difference, then arrange each set, giving fewer arrangements for sets containing

0

解答：

三个互不相同的数字构成一个递增等差数列。按公差计数：公差为

1

有

8

组，公差为

2

有

6

组，公差为

3

有

4

组，公差为

4

有

2

组，共

20

组。

其中

4

组含有

0

（即

\{0,1,2\}

，

\{0,2,4\}

，

\{0,3,6\}

，

\{0,4,8\}

）；每组产生

2\cdot 2!=4

个有效三位数，因为

0

不能放在首位。

另外

16

组各产生

3!=6

个数。总数为

4\cdot 4+16\cdot 6=112

。

因此，正确答案是 C。

The three distinct digits form an increasing arithmetic progression. Counting by common difference:

8

with difference

1,

6

with difference

2,

4

with difference

3,

and

2

with difference

4,

for

20

sets.

Of these,

4

sets contain

0

(namely

\{0,1,2\},

\{0,2,4\},

\{0,3,6\},

\{0,4,8\}

); each yields

2\cdot 2!=4

valid numbers since

0

cannot lead.

The other

16

sets each yield

3!=6

numbers. The total is

4\cdot 4+16\cdot 6=112.

Thus, the correct answer is C.

2007 AMC 12A 第 16 题

答案：C

难度评级：1630

小提示：

大提示：

解答：

其他年份的第 16 题