7. Find the sum of all possible different two-digit numbers, all digits of which are odd.
7. To find out how many such numbers there are, let's try to determine which digits can stand in the tens place. There are five such digits: $1,3,5,7,9$.

The second digit must also be odd, so it can also be chosen in five ways.
![](https://cdn.mathpix.com/cropped/2024_05_21_e53d4ff4f2278a1c0d12g-028.jpg?height=272&width=1268&top_left_y=1880&top_left_x=384)
$\begin{array}{llll}13 & 5 & 7 & 9\end{array}$
$\begin{array}{llllllllll}1 & 3 & 5 & 7 & 9 & 13 & 5 & 9\end{array}$
$\begin{array}{lllll}135 & 5 & 9\end{array}$
$\begin{array}{llllll}1 & 3 & 5 & 7 & 9\end{array}$

In total, there will be $5 \times 5=25$ such numbers. Let's list these numbers and find their sum.

$11+13+15+17+\ldots+93+95+97+99=(11+99) \times 12+55=1375$.

Answer: 1375.

# FIFTH