$29 \cdot 7$ Given that $m$ is the smallest natural number which leaves a remainder of 1 when divided by 3, a remainder of 5 when divided by 7, and a remainder of 4 when divided by 11. Then the remainder when $m$ is divided by 4 is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] Let $m=3a+1=7b+5=11c+4$, where $a, b, c$ are all non-negative integers. Thus,
$$
\begin{array}{l}
a=\frac{2c}{3}+3c+1 \\
b=\frac{4c-1}{7}+c
\end{array}
$$

Now find the smallest $c$ such that $a, b$ are both non-negative integers:
From (1), (2), we know $3 \mid c$, and $7 \mid (4c-1)$, so $c=9$, thus
$$
\begin{array}{c}
m=11 \cdot 9+4=103 \\
103 \div 4=25 \cdots \cdots \cdot 3
\end{array}
$$

Therefore, the answer is $(D)$.