8. Three positive integers greater than 1000 satisfy: the unit digit of the sum of any two numbers is equal to the unit digit of the third number. Then, the last three digits of the product of these 3 numbers have $\qquad$ possible values.
【Answer】4
【Solution】
Let the units of the three numbers be $a, b, c$
(1) If $a, b, c$ are all equal, then they can only all be 0;
(2) If two of $a, b, c$ are equal,
(1) $a, a, c$ and $a \neq c$, then $c+a=a, a+a=10+c$, so $a=5, c=0$;
(3) If $a, b, c$ are all different, let $a<b<c$, then $c+b=10+a, c+a=10+b$, so $c=10$, which contradicts $c$ being a digit; In summary, the units of the three numbers are $0,0,0$ or $0,5,5$;
(1) If they are all 0, then the last 3 digits of the product are 000;
(2) If they are $0,5,5$
(1) If the number with the unit 0 has the last 3 digits all 0, then the last 3 digits of the product are 000;
(2) If the number with the unit 0 has the last 2 digits all 0, then the last 3 digits of the product are 500 or 000;
(3) If the number with the unit 0 has the last 1 digit 0, let the last two digits be $\overline{c 0}$, and let the last two digits of the other two be $\overline{a 5}, \overline{b 5}$, then $\overline{a 5} \times \overline{b 5}=100 a b+50(a+b)+25$, if $(a+b)$ is odd, then the last 3 digits of the product are 75; if $(a+b)$ is even, then the product is 25, when multiplied by $\overline{c 0}$, regardless of what $c$ is, the last three digits can only be $000,250,500,750$. In summary, the last 3 digits of the product have 4 possibilities: $000,500,250,750$.