12. (10 points) The sum of three-digit numbers $n$ for which $\frac{3 \mathrm{n}+2}{5 \mathrm{n}+1}$ is not in its simplest form is equal to what,


【Analysis】 $\frac{3 n+2}{5 n+1}$ is not in simplest form, indicating that $(5 n+1,3 n+2)=a \neq 1$. According to the principle of the Euclidean algorithm, we have $1 \neq a \mid (5 n+1) \times 3 - (3 n+2) \times 5$, which simplifies to $1 \neq a \mid 7$. Therefore, $a$ can only be 7. We can try to determine the answer using $5 n+1$. One attempt shows that $5 n+1=1$ or 6 or 11 or 16 or 21. Since $21=3 \times 7$, when $5 n+1=21$, $7 \mid 5 n+1$ holds true, and at this point, $n$ is the smallest value, which is 4. Other values can be found by incrementing 4 by 7. The problem asks for the sum of all three-digit numbers that satisfy the condition, which are at least 102 and at most 998. We can then use the sum of an arithmetic series to find the answer.【Solution】Solution: $\frac{3 n+2}{5 n+1}$ is not in simplest form, indicating that $(5 n+1,3 n+2)=a \neq 1$. According to the principle of the Euclidean algorithm, we have $1 \neq a \mid (5 n+1) \times 3 - (3 n+2) \times 5$, which simplifies to $1 \neq a \mid 7$. Therefore, $a$ can only be 7. One attempt shows that $5 n+1=1$ or 6 or 11 or 16 or 21. Since $21=3 \times 7$, when $5 n+1=21$, $7 \mid 5 n+1$ holds true, and at this point, $n$ is the smallest value, which is 4. Incrementing 4 by 7,

The three-digit numbers that satisfy the condition are at least 102 and at most 998,
$$
\begin{array}{l}
102+109+116+\cdots+998 \\
=(102+998) \times 129 \div 2 \\
=70950
\end{array}
$$

Answer: The sum of all three-digit numbers $n$ that make $\frac{3 n+2}{5 n+1}$ not in simplest form is 70950.