$29 \cdot 24$ Let $b=1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-\cdots-1988^{2}+1989^{2}$. The remainder when $b$ is divided by 1991 is
(A) 0.
(B) 1.
(C) 2.
(D) 4.
(Changsha, Jiangxi Province, Junior High School Mathematics Competition, 1991)
[Solution 1] Notice the following transformation:
$$
\begin{aligned}
& 1^{2}-2^{2}+3^{2}-4^{2}+\cdots+1987^{2}-1988^{2}+1989^{2} \\
= & 1+\left(1989^{2}-2^{2}\right)-\left(1988^{2}-3^{2}\right)+\cdots-\left(996^{2}-995^{2}\right) \\
= & 1+1991 \cdot 1987-1991 \cdot 1985+\cdots-1991 \cdot 1 \\
= & 1+1991 \cdot(1987-1985+\cdots+3-1) .
\end{aligned}
$$

Then 1991 divides $b$ with a remainder of 1.
Therefore, the answer is $(B)$.
[Solution 2] $b=\left(1^{2}-2^{2}\right)+\left(3^{2}-4^{2}\right)+\cdots+\left(1987^{2}-1988^{2}\right)$
$$
\begin{aligned}
& +1989^{2} \\
= & -3-7-11-\cdots-3975+1989^{2} \\
= & -\frac{(3+3975) \cdot 994}{2}+1989^{2} \\
= & 1989 \cdot 995=(1991-2) \cdot 995 \\
= & 1991 \cdot 995-1990=1991 \cdot 995-1991+1 .
\end{aligned}
$$

Then 1991 divides $b$ with a remainder of 1.
Therefore, the answer is $(B)$.