21. The number of three-digit positive integers that leave a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11 is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
21. E.

Notice that, $11-7=9-5=6-2=4$.
Then the number that meets the requirement is 4 less than the common multiple of 11, 9, and 6. The least common multiple of these numbers is $11 \times 3^{2} \times 2=198$, so the number that meets the requirement is $198k-4\left(k \in \mathbf{Z}_{+}\right)$.

Since this number is a three-digit number, we know that $k$ can take $1,2, \cdots, 5$, i.e., 194, 392, 590, 788, and 986, a total of 5 numbers.