6. If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are distinct positive odd numbers, satisfying
$$
\begin{array}{l}
\left(2005-x_{1}\right)\left(2005-x_{2}\right)\left(2005-x_{3}\right) . \\
\left(2005-x_{4}\right)\left(2005-x_{5}\right)=24^{2},
\end{array}
$$

then, the last digit of $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}$ is ( ).
(A) 1
(B) 3
(C) 5
(D) 7
6. A.

It is evident that the five factors on the left side of the equation are different even numbers. Decomposing \(24^2\) into the product of five distinct even numbers, there is only one unique form:
$$
24^{2}=2 \times(-2) \times 4 \times 6 \times(-6) .
$$

Therefore, the five factors \(2005-x_{i} (i=1,2,3,4,5)\) should be equal to \(2, -2, 4, 6, -6\) respectively. Thus,
$$
\begin{array}{l}
\left(2000-x_{1}\right)^{2}+\left(2000-x_{2}\right)^{2}+\cdots+\left(2005-x_{5}\right)^{2} \\
=2^{2}+(-2)^{2}+4^{2}+6^{2}+(-6)^{2}=96,
\end{array}
$$

i.e., \(\square\)
$$
\begin{array}{l}
5 \times 2005^{2}-4010\left(x_{1}+x_{2}+\cdots+x_{5}\right)+ \\
\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{5}^{2}\right)=96 . \\
\text { Hence } x_{1}^{2}+x_{2}^{2}+\cdots+x_{5}^{2} \equiv 1(\bmod 10) .
\end{array}
$$