2. (10 points) In the following four equations: $\overline{\mathrm{AB}} \div \overline{\mathrm{CD}}=2, E \times F=0, G-H=1, I+J=4, A \sim J$ represent different digits from $0 \sim$ 9, then the two-digit number $\overline{\mathrm{AB}}$ cannot be ( )
A. 54
B. 58
C. 92
D. 96
【Analysis】Since $E \times F=0$, at least one of $E F$ must be 0, and the other can be any number; also, because $I+J=4$, one of $I$ and $J$ is 3 and the other is 1; since $A \sim J$ represent different digits from $0 \sim 9$, and $G-H=1$, we can discuss the following cases: (1) $G H$ is 9, 8; (2) $G H$ is 8, 7; (3) $G H$ is 7, 6; (4) $G H$ is 6, 5; (5) $G H$ is 5, 4. Based on this, we can solve the problem.

【Solution】From the conditions, we know: $E$ and $F$ must have at least one as 0, assume $E$ is 0; the other can be any number; one of $I$ and $J$ is 3, and the other is 1; then the remaining digits from $0 \sim 9$ are $2, 4, 5, 6, 7, 8, 9$;
Since: $G-H=1$
(1) When $G H$ is 9, 8

then $54 \div 27=2$
at this time $F=6$
(2) When $G H$ is 8, 7

then $92 \div 46=2$
at this time $F=5$
(3) When $G H$ is 7, 6

then $58 \div 29=2$
at this time $F=4$
(4) When $G, H$ is 6, 5

this does not satisfy the condition
(5) When $G, H$ is 5, 4,

this does not satisfy the condition
Therefore, the two-digit number $\overline{\mathrm{AB}}$ could be $54, 58, 92$; it cannot be 96
So the answer is: D.