12. If to a certain three-digit number, first the digit 7 is appended on the left, and then on the right, the first of the resulting four-digit numbers will be 3555 more than the second. Find this three-digit number.
12. Let the desired three-digit number be $x$. If we prepend the digit 7, we get the number $7000+x$; if we append the digit 7, we get the number $10x+7$. According to the problem,

$$
(7000+x)-(10x+7)=3555
$$

from which $9x=3438$ and $x=382$. The problem can also be solved by trial and error.