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Return your final response within \boxed{}. A set consists of five different odd positive integers, each greater than 2. When these five integers are multiplied together, their product is a five-digit integer of the form $AB0AB$, where $A$ and $B$ are digits with $A \neq 0$ and $A \neq B$. (The hundreds digit of the product is zero.) In total, how many different sets of five different odd positive integers have these properties?
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<|begin_of_thought|>

Deep Thought is thinking for 7.5 million years.

<|end_of_thought|>

<|begin_of_solution|>


1. **Observe the Structure of \( N \)**:
    Let \( N = AB0AB \) and let \( t \) be the two-digit integer \( AB \).

    We recognize that \( N = 1001 \cdot t \), where \( 1001 = 11 \cdot 91 = 11 \cdot 7 \cdot 13 \).

    Thus, 
    \[
    N = t \cdot 7 \cdot 11 \cdot 13 
    \]

2. **Formulate the Problem**:
    We need to write \( N \) as the product of 5 distinct odd integers, each greater than 2, and count the possible sets \( S \) of such integers.

3. **Case Analysis**:
   
    - **Case 1: \( S = \{7, 11, 13, m, n\} \)**:
        - Here, 
        \[
        N = 7 \cdot 11 \cdot 13 \cdot m \cdot n 
        \]
        This implies \( t = m \cdot n \). Since \( t \) is a two-digit number,
        \[
        m \cdot n < 100
        \]
        Analyzing possible values of \( m \) and corresponding \( n \):
        - If \( m = 3 \), then \( n \) can be \( 5, 9, 15, 17, 19, 21, 23, 25, 27, 29, 31 \) yielding \( mn \) values: \( 15, 27, 45, 51, 57, 63, 69, 75, 81, 87, 93 \).
        - If \( m = 5 \), then \( n \) can be \( 9, 15, 17, 19 \) yielding \( mn \) values: \( 45, 75, 85, 95 \).
        - Higher \( m \) values result in \( mn \geq 135 \), which is invalid.
        
        There are 15 sets in this case.

    - **Case 2: \( S = \{7q, 11, 13, m, n\} \) where \( q \) is an odd integer \( > 1 \)**:
        - Here, 
        \[
        N = 7q \cdot 11 \cdot 13 \cdot m \cdot n 
        \]
        So, \( t = mnq \). This constraint is:
        \[
        mnq \leq 99
        \]
        - If \( q = 3 \), then \( mn \leq 33 \). Possible \( mn \) pairs:
            - If \( m = 3 \), \( n = 5, 9 \), giving 2 potential sets.
        - If \( q = 5 \), then \( mn \leq \frac{99}{5} = 19 \), 
            - Only \( m = 3 \) and \( n = 5 \) satisfy this condition.
        - \( q \geq 7 \) is invalid.
        
        There are 3 sets in this case.

    - **Case 3: \( S = \{7, 11q, 13, m, n\} \)**, similar constraints apply:
        - If \( q = 3 \), \( mn \leq 33 \) with possibilities:
            - If \( m = 3 \), \( n = 5, 9 \), 2 sets are valid.
        - If \( q = 5 \), only valid option is \( m = 3 \) and \( n = 5 \).
        - \( q \geq 7 \) is invalid.

        There are 3 sets in this case.

    - **Case 4: \( S = \{7, 11, 13 q, m, n\} \)**:
        - If \( q = 3 \), valid sets: \( S = \{3,5,7,11,13\} \) with 2 possible sets.
        - If \( q = 5 \), possibility: \( S = \{3,5,7,11,13\} \), only 1 set.
        - \( q \geq 7 \) is invalid.

        There are 3 sets in this case.

    - **Cases 5 and beyond: Combinations involving \( q, r (>1) \equiv \text{odd}\)**:
        - Trying combinations like \( \{7q, 11r, 13, m, n\} \) converge to the realization that the product \( mnr \leq 99 \) can’t hold up with distinct odd integers under errors faced previously.
        
    - **Case 6: \( S = \{77, 13, m, n, l\} \)**:
        - Recognize that \( 77 = 7 \cdot 11 \) and set \( mn \leq 99 \) but \( mnr \geq 105 \). No favorable solutions detected.

Consolidate final conclusion:
There are a total of \( 15 + 3 + 3 + 3 = 24 \) possible sets with distinct odd positive integers greater than \(2\) each.

\[
\boxed{24}
\]

<|end_of_solution|>