$29 \cdot 54$ If $a, b$ are natural numbers, and $a$ divided by 7 leaves a remainder of $2, b$ divided by 7 leaves a remainder of 5. When $a^{2}>3 b$, the remainder of $a^{2}-3 b$ divided by 7 is
(A) 1 .
(B) 3 .
(C) 4 .
(D) 6 .
(China Jiangxi Province Nanchang Junior High School Mathematics Competition, 1989)
[Solution] Let $a=7 m+2, b=7 n+5 \quad(m, n \in \mathbb{N})$, then
$$
\begin{aligned}
a^{2}-3 b= & (7 m+2)^{2}-3(7 n+5) \\
& =7\left(7 m^{2}+4 m-3 n\right)-11 \\
& =7\left(7 m^{2}+4 m-3 n\right)-14+3,
\end{aligned}
$$

we know that $a^{2}-3 b$ leaves a remainder of 3 when divided by 7.
Therefore, the answer is $(B)$.