5. Reverse the order of the three digits of a three-digit number, and add the resulting number to the original number. If none of the digits in the sum is even, then the number is called a "Odd Sum Number". Then, among all three-digit numbers, the number of Odd Sum Numbers is ( ).
(A) 100
(B) 120
(C) 160
(D) 200
5.A.

Let the three-digit number be $\overline{a_{1} a_{2} a_{3}}$. Then
$$
\begin{array}{l}
\overline{a_{1} a_{2} a_{3}}+\overline{a_{3} a_{2} a_{1}} \\
=100\left(a_{1}+a_{3}\right)+10\left(a_{2}+a_{2}\right)+\left(a_{1}+a_{3}\right) .
\end{array}
$$

If $a_{1}+a_{3}$ does not carry over, then the tens digit of the sum must be even, which does not meet the requirement. Therefore,
$$
a_{1}+a_{3}=11,13,15,17 \text {. }
$$

Since $11=9+2=8+3=7+4=6+5$, there are $4 \mathrm{~A}_{2}^{2}$ possible values for $a_{1} 、 a_{3}$;

Since $13=9+4=8+5=7+6$, there are $3 \mathrm{~A}_{2}^{2}$ possible values for $a_{1} 、 a_{3}$;

Since $15=9+6=8+7$, there are $2 \mathrm{~A}_{2}^{2}$ possible values for $a_{1} 、 a_{3}$;

Since $17=9+8$, there are $\mathrm{A}_{2}^{2}$ possible values for $a_{1} 、 a_{3}$.

Also, $a_{2}+a_{2}$ cannot carry over, so $a_{2}$ can only be 0, $1$, $2$, $3$, or $4$.
Therefore, the number of numbers that satisfy the condition is
$$
5\left(4 \mathrm{~A}_{2}^{2}+3 \mathrm{~A}_{2}^{2}+2 \mathrm{~A}_{2}^{2}+\mathrm{A}_{2}^{2}\right)=100 \text { (numbers). }
$$