$29 \cdot 169^{1990}$ when divided by 11 leaves a remainder of
(A) 1 .
(B) 3 .
(C) 5 .
(D) 7 .
(China Shanxi Province Taiyuan City High School Mathematics Competition, 1990)
[Solution 1] $9^{1990}-1=9^{5 \cdot 398}-1=\left(9^{5}-1\right)\left[\left(9^{5}\right)^{397}+\cdots+1\right]$,
and
$$
\begin{aligned}
9^{5}-1 & =9^{2} \cdot 9^{2} \cdot 9-1 \\
& =[(7 \cdot 11+4)(7 \cdot 11+4) \cdot 9]-1 \\
& =[11 M+16 \cdot 9]-1 \\
& =\left[11 M^{\prime}+5 \cdot 9\right]-1 \\
& =11 M^{\prime}+44,
\end{aligned}
$$

which means
$$
11 \text { divides }\left(9^{5}-1\right) \text {, so } 11 \text { divides }\left(9^{1990}-1\right) \text {. }
$$

Therefore, $9^{1990}$ leaves a remainder of 1 when divided by 11.
Thus, the answer is $(A)$.
[Solution 2]
$$
\begin{aligned}
9^{1990} & =(11-2)^{1990}=11 \cdot M+2^{1990}=11 M+32^{398} \\
& =11 M+(33-1)^{398}=11 \cdot N+1^{398} \\
& =11 N+1,
\end{aligned}
$$

Therefore, $9^{1990}$ leaves a remainder of 1 when divided by 11.
Thus, the answer is $(A)$.