5. It is known that the sum of two three-digit numbers, in the notation of which there are no identical digits, is a three-digit number, where the number of hundreds does not exceed 7 and is greater than 5, the number of tens is odd, and the number of units is 1. In one of the addends, the number of tens is 9, and in the other, the number of units is 0. Find these numbers.
5. According to the condition of the problem, the first addend is of the form $A 9 B$, and the second is $C K 0$. The number of units in the sum is 1, so this number has the form: $M X 1$, i.e., we get the example

$$
A 9 B+C K 0=M X 1 .
$$

From this, we can conclude that $B=1$

$$
A 91+C K 0=M X 1 .
$$

The number of tens in the sum must be odd, and this is possible only when $K=2,4,6,8$:

$$
\begin{aligned}
& A 91+C 20=M 11, A 91+C 40=M 31 \\
& A 91+C 60=M 51, A 91+C 80=M 71
\end{aligned}
$$

Moreover, it is known that the number of hundreds in the sum is no more than 7 and exceeds 5. Therefore, the number of hundreds in the sum is 6 or 7.

Consider the first case - $A 91+C 20=611$. Since

$$
9+2=11 \text {, then } A+C=6-1, A+C=5
$$

This is possible only if $A$ and $C$ are 1 and 4 or 2 and 3. None of these pairs satisfy the condition that the digits in the addends do not repeat. The last case $-A 91+C 80=671$ also does not satisfy this condition.

Similarly, considering the cases $-A 91+C 40=631, A 91+$ $+C 60=651$, we get $240+391=631,340+291=631,260+391=651,360+291=651$.

Now consider the case when the number of hundreds in the sum is 7:

$$
\begin{aligned}
& A 91+C 20=711, A 91+C 40=731 \\
& A 91+C 60=751, A 91+C 80=771
\end{aligned}
$$

Sequentially analyzing each of these cases, we get:

$$
280+491=771,480+291=771,260+491=751460+291=751
$$

Answer: $240+391=631,340+291=631,260+391=651$,

$$
\begin{aligned}
& 360+291=651,280+491=771,480+291=771 \\
& 260+491=751,460+291=751
\end{aligned}
$$