Solution 1: Let $N=AB0AB$ and let $t$ be the two-digit integer $AB$. We note that $N=1001t$, and that $1001=11 \cdot 91=11 \cdot 7 \cdot 13$. Therefore, $N=t \cdot 7 \cdot 11 \cdot 13$. We want to write $N$ as the product of 5 distinct odd integers, each greater than 2, and to count the number of sets $S$ of such odd integers whose product is $N$. There are several situations to consider. First, we look at possible sets $S$ that include the integers 7, 11 and 13 (which we know are divisors). Second, we look at possible sets $S$ that include two of these integers and an odd multiple of the third. Third, we rule out possible sets $S$ that include one of these integers and odd multiples of the second and third. Fourth, we rule out possible sets $S$ that include the product of two or three of these integers and additional integers. Case 1: $S=\{7,11,13, m, n\}$ where $m<n$ and $m, n \neq 7,11,13$ Here, $N=7 \cdot 11 \cdot 13 \cdot m \cdot n=mn \cdot 1001$ and so $t=mn$. This tells us that $mn$ is less than 100. If $m=3$, then the possible values for $n$ are $5,9,15,17,19,21,23,25,27,29,31$. These give the following corresponding values of $mn$: $15,27,45,51,57,63,69,75,81,87,93$. Note that $n \neq 33$, since $m=3$ and $n=33$ gives $mn=99$ which has two equal digits and so is not possible. If $m=5$, then the possible values for $n$ are $9,15,17,19$. If $m \geq 9$, then $n \geq 15$ since the integers in $n$ are odd and distinct, and so $mn \geq 135$, which is not possible. Therefore, in this case, there are 15 possible sets. Case 2: $S=\{7q, 11,13, m, n\}$ where $m<n$ and $q>1$ is odd and $m, n \neq 7q, 11,13$ Here, we have $N=7q \cdot 11 \cdot 13 \cdot m \cdot n$ and so $N=1001 \cdot mnq$ which gives $t=mnq$. Note that $mnq \leq 99$. Suppose that $q=3$. This means that $mn \leq 33$. If $m=3$, then the possible values of $n$ are 5 and 9 since $m$ and $n$ are odd, greater than 2, and distinct. ($n=7$ is not possible, since this would give the set $\{21,11,13,3,7\}$ which is already counted in Case 1 above.) If $m \geq 5$, then $n \geq 7$ which gives $mn \geq 35$, which is not possible. Suppose that $q=5$. This means that $mn \leq \frac{99}{5}=19 \frac{4}{5}$. If $m=3$, then $n=5$. There are no further possibilities when $q=5$. Since $mn \geq 3 \cdot 5=15$ and $mnq \leq 99$, then we cannot have $q \geq 7$. Therefore, in this case, there are 3 possible sets. Case 3: $S=\{7,11q, 13, m, n\}$ where $m<n$ and $q>1$ is odd and $m, n \neq 7,11q, 13$ Suppose that $q=3$. This means that $mn \leq 33$. If $m=3$, then the possible values of $n$ are 5 and 9. (Note that $n \neq 7$.) We cannot have $n=11$ as this would give $mnq=99$ and a product of 99099 which has equal digits $A$ and $B$. We cannot have $m \geq 5$ since this gives $mn \geq 45$. Suppose that $q=5$. This means that $mn \leq \frac{99}{5}$. If $m=3$, then $n=5$. As in Case 2, we cannot have $q \geq 7$. Therefore, in this case, there are 3 possible sets. Case 4: $S=\{7,11,13q, m, n\}$ where $m<n$ and $q>1$ is odd and $m, n \neq 7,11,13q$ Suppose that $q=3$. This means that $mn \leq 33$. If $m=3$, the possible values of $n$ are 5 and 9. (Again, $n \neq 11$ in this case.) We cannot have $m \geq 5$ when $q=3$ otherwise $mn \geq 45$. If $q=5$, we can have $m=3$ and $n=5$ but there are no other possibilities. As in Cases 2 and 3, we cannot have $q \geq 7$. Therefore, in this case, there are 3 possible sets. Case 5: $S=\{7q, 11r, 13, m, n\}$ where $m<n$ and $q, r>1$ are odd and $m, n \neq 7q, 11r, 13$ Here, $mnqr \leq 99$. Since $q, r>1$ are odd, then $qr \geq 9$ which means that $mn \leq 11$. Since there do not exist two distinct odd integers greater than 1 with a product less than 15, there are no possible sets in this case. A similar argument rules out the products $N=7q \cdot 11 \cdot 13r \cdot m \cdot n$, $N=7 \cdot 11q \cdot 13r \cdot m \cdot n$, $N=7q \cdot 11r \cdot 13s \cdot m \cdot n$ where $q, r, s$ are odd integers greater than 1. Case 6: $S=\{77,13, m, n, \ell\}$ where $m<n<\ell$ and $m, n, \ell \neq 77,13$ Note that $77=7 \cdot 11$ since we know that $N$ has divisors of 7 and 11. Here, $mn\ell \leq 99$. Since $mn\ell \geq 3 \cdot 5 \cdot 7=105$, there are no possible sets in this case, nor using $7 \cdot 143$ or $11 \cdot 91$ in the product or 1001 by itself or multiples of 77,91 or 143. Having considered all cases, there are $15+3+3+3=24$ possible sets. Solution 2: We note first that $AB0AB=AB \cdot 1001$, and that $1001=11 \cdot 91=11 \cdot 7 \cdot 13$. Therefore, $AB0AB=AB \cdot 7 \cdot 11 \cdot 13$. Since $AB0AB$ is odd, then $B$ is odd. Since $A \neq 0$ and $A \neq B$ and $B$ is odd, then we have the following possibilities for the two-digit integer $AB$: $13,15,17,19,21,23,25,27,29,31,35,37,39,41,43,45,47,49,51,53,57,59,61,63,65,67,69,71,73,75,79,81,83,85,87,89,91,93,95,97$. If the integer $AB$ is a prime number, then $AB0AB$ cannot be written as the product of five different positive integers each greater than 2, since it would have at most four prime factors. Using this information, we can eliminate many possibilities for $AB$ from our list to obtain the shorter list: $15,21,25,27,35,39,45,49,51,57,63,65,69,75,81,85,87,91,93,95$. Several of the integers in this shorter list are the product of two distinct prime numbers neither of which is equal to 7,11 or 13. These integers are $15=3 \cdot 5$ and $51=3 \cdot 17$ and $57=3 \cdot 19$ and $69=3 \cdot 23$ and $85=5 \cdot 17$ and $87=3 \cdot 29$ and $93=3 \cdot 31$ and $95=5 \cdot 19$. Thinking about each of these as $p \cdot q$ for some distinct prime numbers $p$ and $q$, we have $AB0AB=p \cdot q \cdot 7 \cdot 11 \cdot 13$. To write $AB0AB$ as the product of five different positive odd integers greater each greater than 2, these five integers must be the five prime factors. For each of these 8 integers $(15,51,57,69,85,87,93,95)$, there is 1 set of five distinct odd integers, since the order of the integers does not matter. This is 8 sets so far. This leaves the integers $21,25,27,35,39,45,49,63,65,75,81,91$. Seven of these remaining integers are equal to the product of two prime numbers, which are either equal primes or at least one of which is equal to 7,11 or 13. These products are $21=3 \cdot 7$ and $25=5 \cdot 5$ and $35=5 \cdot 7$ and $39=3 \cdot 13$ and $49=7 \cdot 7$ and $65=5 \cdot 13$ and $91=7 \cdot 13$. In each case, $AB0AB$ can then be written as a product of 5 prime numbers, at least 2 of which are the same. These 5 prime numbers cannot be grouped to obtain five different odd integers, each larger than 1, since the 5 prime numbers include duplicates and if two of the primes are combined, we must include 1 in the set. Consider, for example, $21=3 \cdot 7$. Here, $21021=3 \cdot 7 \cdot 7 \cdot 11 \cdot 13$. There is no way to group these prime factors to obtain five different odd integers, each larger than 1. Similarly, $25025=5 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ and $91091=7 \cdot 13 \cdot 7 \cdot 11 \cdot 13$. The remaining three possibilities (35, 49 and 65) give similar situations. This leaves the integers $27,45,63,75,81$ to consider. Consider $27027=3^{3} \cdot 7 \cdot 11 \cdot 13$. There are 6 prime factors to distribute among the five odd integers that form the product. Since there cannot be two 3's in the set, the only way to do this so that they are all different is $\{3,9,7,11,13\}$. Consider $81081=3^{4} \cdot 7 \cdot 11 \cdot 13$. There are 7 prime factors to distribute among the five odd integers that form the product. Since there cannot be two 3 s or two 9 s in the set and there must be two powers of 3 in the set, there are four possibilities for the set $S$: $S=\{3,27,7,11,13\},\{3,9,21,11,13\},\{3,9,7,33,13\},\{3,9,7,11,39\}$. Consider $45045=3^{2} \cdot 5 \cdot 7 \cdot 11 \cdot 13$. There are 6 prime factors to distribute among the five odd integers that form the product. Since two of these prime factors are 3, they cannot each be an individual element of the set and so one of the 3 s must always be combined with another prime giving the following possibilities: $S=\{9,5,7,11,13\},\{3,15,7,11,13\},\{3,5,21,11,13\},\{3,5,7,33,13\},\{3,5,7,11,39\}$. Consider $75075=3 \cdot 5^{2} \cdot 7 \cdot 11 \cdot 13$. Using a similar argument to that in the case of 45045, we obtain $S=\{15,5,7,11,13\},\{3,25,7,11,13\},\{3,5,35,11,13\},\{3,5,7,55,13\},\{3,5,7,11,65\}$. Finally, consider $63063=3^{2} \cdot 7^{2} \cdot 11 \cdot 13$. There are 6 prime factors to distribute among the five odd integers that form the product. Since we cannot have two 3 s or two 7 s in the product, the second 3 and the second 7 must be combined, and so there is only one set in this case, namely $S=\{3,7,21,11,13\}$. We have determined that the total number of sets is thus $8+1+4+5+5+1=24$.