If a=1, b=10, c=100 and d=1000, then &(a+b+c-d)+(a+b-c+d) & +(a-b+c+d) & +(-a+b+c+d) is equal to

1978 AMC 12 Problem 4

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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

Count how many times each variable appears with each sign

The entire sum simplifies to twice

a+b+c+d

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.