12. List all positive integers that are coprime with 105 in ascending order, and find the 100th term of this sequence.
12. First, it can be proven that if $a$ is a number coprime with 105, then $a+105k\left(k \in \mathbf{N}_{+}\right)$ is also coprime with 105. In the set $S_{1}=\{1,2, \cdots, 105\}$, we can see that there are 48 numbers $a_{1}=1, a_{2}=2, a_{3}=4, a_{4}=8 \cdots, a_{48}=104$ that are coprime with 105. Thus, $S_{n}=\{1+105(n-1), 2+105(n-1), \cdots, 105n\}$ also has 48 numbers that are coprime with 105. Therefore, the 100th number that is coprime with 105 must be in $S_{3}$, and it is the fourth number, which is $a_{4}=4+105 \cdot 2=a_{100}=218$.