12. 10 consecutive natural numbers are arranged in ascending order. If the sum of the last 6 numbers is 15 more than twice the sum of the first 4 numbers, then the smallest number among these 10 numbers is . $\qquad$
【Analysis】This problem mainly examines arithmetic sequences.
【Solution】Solution: Let the smallest number be $x$, then the remaining natural numbers are $x+1, x+2, \cdots, x+9$. According to the problem, we have $2(4 x+1+2+3)+15=6 x+4+5+6+7+8+9$,

Simplifying gives $8 x+27=6 x+39$
$$
\therefore x=6 \text {, }
$$

【Comment】This problem can be solved by setting up equations, letting the smallest number be $x$, and expressing other consecutive natural numbers using $x$. The solution can be found based on the equal relationship.