5. Let $a, b, c, d, e$ be distinct positive odd numbers. If the equation
$$
(x-a)(x-b)(x-c)(x-d)(x-e)=2009
$$

has an integer root $x$, then the last digit of $a+b+c+d+e$ is ( ).
(A) 1
(B) 3
(C) 7
(D) 9
5.I).

For any integer $x, x-a, x-b, x-c, x-$ $d, x-e$ are $I$ distinct integers. And expressing 2009 as the product of 7 $I$ distinct integers has only one unique form:
$$
2009=1 \times(-1) \times 7 \times(-7) \times 41.
$$

From this, $x$ is even.
Let $x=2 m$. Then
$$
\begin{array}{l}
\{2 m-a, 2 m-b, 2 m-c, 2 m-d, 2 m-e\} \\
=\{-1,1,-7,7,41\}.
\end{array}
$$

Thus, from the sum of the elements
$$
10 m-(a+b+c+d+e)=41,
$$

we get $a+b+c+d+e=10 m-41$, whose last digit is 9.