$36 \cdot 3$ If the sum of 11 consecutive odd numbers is 1991, and these numbers are arranged in ascending order, the sixth number is
(A) 179.
(B) 181.
(C) 183.
(D) 185.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] Let the 11 consecutive odd numbers be $2 n+1, 2 n+3, \cdots, 2 n+21$, then 
$$(2 n+1)+(2 n+3)+\cdots+(2 n+21)=1991,$$

which means
$$
11(2 n+11)=1991 \text {. }
$$

Solving for $n$ gives
$$
n=85 \text {. }
$$

Then the sixth number is $2 \cdot 85+11=181$.
Therefore, the answer is $(B)$.