$29 \cdot 182^{1000}$ divided by 13, the remainder is 
(A) 1.
(B) 2.
(C) 3.
(D) 7.
(E) 11.
(23rd American High School Mathematics Examination, 1972)
［Solution］Notice the following fact:
If $N_{1}=Q_{1} D+R_{1}, N_{2}=Q_{2} D+R_{2}$, then
$$
N_{1} N_{2}=\left(Q_{1} Q_{2} D+Q_{1} R_{2}+Q_{2} R_{1}\right) D+R_{1} R_{2} \text {. }
$$

That is, if $N_{1}, N_{2}$ are divided by $D$ with remainders $R_{1}, R_{2}$, then $N_{1} N_{2}$ divided by $D$ has a remainder of $R_{1} R_{2}$.
Since $2^{6}=64$ divided by 13 has a remainder of -1, then $2^{12}$ divided by 13 has a remainder of 1. Also, $2^{1000}=2^{996} \cdot 2^{4}=\left(2^{12}\right)^{83} \cdot 16$, when divided by 13, leaves a remainder of $1^{83} \cdot 3=3$.
Therefore, the answer is $(C)$.