$32 \cdot 32$ A two-digit natural number has a one-digit number (including 0) inserted in the middle, turning it into a three-digit number. Some two-digit numbers, when a one-digit number is inserted in the middle, become a three-digit number that is exactly 9 times the original two-digit number. The number of such two-digit numbers is
(A) 1 .
(B) 4 .
(C) 10 .
(D) more than 10 .
("Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1993)
[Solution] If a "three-digit number" is 9 times a "two-digit number", then the sum of such a "three-digit number" and a "two-digit number" should be 10 times the "two-digit number".
Therefore, the unit digit of such a "two-digit number" can only be 0 and 5.
Also, because the inserted digit should be less than the unit digit of the "two-digit number", the unit digit of the "two-digit number" can only be 5, and the inserted unit digit can only be $0,1,2,3,4$.
And a "three-digit number" is a multiple of 9 $\Longleftrightarrow$ the sum of its digits can be divided by 9.
Thus, the inserted digit can only be $0,1,2,3$, and the corresponding "two-digit number" is $45,35,25$ and 15, four in total.
Hence, the answer is $(B)$.