8. (10 points) Today is December 17, 2011, and in this date there are 4 ones, 2 twos, 1 zero, and 1 seven. Using these 8 digits to form several composite numbers and then find their sum (each digit must be used exactly once, the leading digit cannot be 0, for example, the sum of 21110 and 217 is 21327), the minimum value of the sum of these composite numbers is $\qquad$
【Solution】Solution: (1) A single-digit number, since $0,1,2,7$ are not composite numbers, there are no single-digit composite numbers formed by these, which is impossible.
(2) A two-digit number, is also unlikely, because there are 4 ones. If it is 11, it is not a composite number, so the 4 ones must be placed in four different two-digit composite numbers. Among them, there must be a 1 and 7 in the same composite number, but $17, 71$ are both prime numbers, leading to a contradiction.
(3) Two three-digit numbers and one two-digit number, let these numbers be $\overline{\mathrm{ABC}}, \overline{\mathrm{DEF}}$ and $\overline{\mathrm{GH}}$ (three numbers in total). Organizing, we get: as long as we try to make $A$ and $D$ as small as possible, then ensure $B, E$ are as small as possible, and then make adjustments to meet the result. It is easy to get, $\overline{\mathrm{ABC}}+\overline{\mathrm{DEF}}+\overline{\mathrm{GH}}=100(A+D)+10(B+E+G)+C+F+H \geqslant 100(1+1)+10(0+1+1)$ $+2+2+7=231$

At this point, $\overline{\mathrm{ABC}}=102, \overline{\mathrm{DEF}}=117, \overline{\mathrm{GH}}=12$, satisfying the problem's requirements and being the minimum.
Therefore, the answer is: 231.