1991 AMC 8 第 20 题

Question

在所示加法算式中，每个数字都被一个字母代替。若不同字母表示不同数字，那么

C =

\begin{array}{cccc} & A & B & C \\ & & A & B \\ + & & & A \\ \hline & 3 & 0 & 0 \end{array}

In the addition problem shown, each digit has been replaced by a letter. If different letters represent different digits, then

C =

\begin{array}{cccc} & A & B & C \\ & & A & B \\ + & & & A \\ \hline & 3 & 0 & 0 \end{array}

1

3

5

7

9

用

A, B, C

写出这个加法和

Write the sum in terms of

A, B, C

把三个数相加，得到

100A + 10B + C

+ 10A + B + A

= 111A + 11B + C

= 300

这个等式

100A + 10B + C

+ 10A + B + A

= 111A + 11B + C

= 300

三个数相加为

111A + 11B + C = 300

。若

A = 1

，数值太小；若

A \ge 3

，数值太大，所以

A = 2

。

此时

11B + C = 78

，只能有

B = 7

且

C = 1

。所以

C = 1

。

所以正确答案是 A。

The three numbers add to

111A + 11B + C = 300.

Since

A = 1

is too small and

A \ge 3

is too large,

A = 2.

Then

11B + C = 78,

which forces

B = 7

and

C = 1.

So

C = 1.

Thus, the correct answer is A .

其他年份的第 20 题