1978 AMC 12 第 4 题

Question

若

a=1

、

b=10

、

c=100

且

d=1000

，则

\begin{aligned} &(a+b+c-d)+(a+b-c+d)\\ &\quad+(a-b+c+d)\\ &\quad+(-a+b+c+d) \end{aligned}

等于

If

a=1,

b=10,

c=100

and

d=1000,

then

\begin{aligned} &(a+b+c-d)+(a+b-c+d)\\ &\quad+(a-b+c+d)\\ &\quad+(-a+b+c+d) \end{aligned}

is equal to

1111

2222

3333

1212

4242

数一数每个变量分别以正号和负号出现了多少次

Count how many times each variable appears with each sign

整个和式可化简为

a+b+c+d

的两倍

The entire sum simplifies to twice

a+b+c+d

每个变量以正号出现三次、以负号出现一次，所以该式为

\begin{gathered} 2(a+b+c+d)\\ =2a+2b+2c+2d\\ =2+20+200+2000\\ =2222\text{。} \end{gathered}

因此，正确答案是 B。

Each variable appears positively three times and negatively once, so the expression is

\begin{gathered} 2(a+b+c+d)\\ =2a+2b+2c+2d\\ =2+20+200+2000\\ =2222. \end{gathered}

Therefore, the correct answer is B.

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