1.13 If $a=1, b=10, c=100$ and $d=1000$, then
$$
\begin{array}{l}
(a+b+c-d)+(a+b-\dot{c}+d)+(a-b+c+d)+(-a+ \\
b+c+d)
\end{array}
$$

equals
(A) 1111.
(B) 2222 .
(C) 3333 .
(D) 1212 .
(E) 4242 .
(29th American High School Mathematics Examination, 1978)
$$
\text { [Solution] } \begin{aligned}
& (a+b+c-d)+(a+b-c+d)+(a-b+c+d) \\
& +(-a+b+c+d) \\
= & 2(a+b+c+d)=2222 .
\end{aligned}
$$

Therefore, the answer is $(B)$.