$10 \cdot 9$ Given two sequences of numbers:
$$
\begin{array}{c}
2,5,8,11,14,17, \cdots, 2+(200-1) \cdot 3 \\
5,9,13,17,21,25, \cdots, 5+(200-1) \cdot 4
\end{array}
$$

Both sequences have 200 terms. The number of identical terms in these two sequences is
(A) 49 terms.
(B) 50 terms.
(C) 51 terms.
(D) 149 terms.
(China Zhejiang Province Junior High School Mathematics Competition, 1991)
［Solution 1］If there are identical terms in the two sequences, then we have
$$
2+3(n-1)=5+4(m-1), \quad m, n=1,2, \cdots, 199,200 \text {. }
$$

which simplifies to
$$
4 m-3 n=-2 \text {. }
$$

Let $m=3 k+x, x=0,1,2$; and $n=4 k+y, y=0,1,2,3$, substituting into the above equation gives $4 x-3 y=-2$, leading to $x=1, y=2$.
Thus, $m=3 k+1, n=4 k+2$. (where $k$ is an integer)
Also, $1 \leqslant m, n \leqslant 200$,

then
$$
1 \leqslant 3 k+1 \leqslant 200, \quad \text { and } 1 \leqslant 4 k+2 \leqslant 200,
$$

which gives $0 \leqslant k \leqslant 49$. Therefore, there are 50 pairs of $m, n$ that satisfy the equation $*$.
Hence, the answer is $(B)$.
[Solution 2] According to the problem, the two sequences can be represented as
$$
2+(n-1) \cdot 3 \text { and } 5+(m-1) \cdot 4 \text {. }
$$

To make $2+(n-1) \cdot 3=5+(m-1) \cdot 4$, then $n=\frac{4 m+2}{3}$.
Thus, $1 \leqslant \frac{4 m+2}{3} \leqslant 200$,
so $1 \leqslant m \leqslant 149$. Also, 3 divides $(m-1)$.
Therefore, $m=1,4,7, \cdots, 145,148$, a total of 50 terms.
Hence, the answer is $(B)$.