$10 \cdot 8$ Given two sequences of numbers
$$
\begin{array}{l}
1,3,5,7, \cdots, 1991 \\
1,6,11,16, \cdots, 1991
\end{array}
$$

The number of numbers that appear in both sequences is
(A) 201 .
(B) 200 .
(C) 199 .
(D) 198 .
(3rd "Five Sheep Cup" Junior High School Mathematics Competition, 1991)
[Solution] The two sequences can be represented as $2 m+1$ and $5 n+1$ (where $m, n$ are non-negative integers). The least common multiple of the primes 5 and 2 is 10.
Thus, the numbers that appear in both sequences are
$$
1, 11, 21, 31, \cdots, 1991 \text{, }
$$

a total of 200 numbers.
Therefore, the answer is $(B)$.