Solution. $\overline{a b a b a b}=10101 \cdot \overline{a b}=3 \cdot 7 \cdot 13 \cdot 37 \cdot \overline{a b}$. According to the conditions, $\overline{a b}$ is a two-digit number that is the product of two different prime numbers, none of which are $3,7,13,37$. Therefore, the possible products for $\overline{a b}$ are: $2 \cdot 5, 2 \cdot 11$, $2 \cdot 17, 2 \cdot 19, 2 \cdot 23, 2 \cdot 29, 2 \cdot 31, 2 \cdot 41, 2 \cdot 43, 2 \cdot 47, 5 \cdot 11, 5 \cdot 17$. Any further products would result in numbers with at least three digits, so there are 12 valid numbers.
How many six-digit numbers of the form ababab are there, which are the product of six different prime numbers?