What is the sum of those three-digit numbers, all of whose digits are odd?
I. solution. The three-digit numbers that contain only odd digits can be paired as follows:

| 1. | 2. |
| :---: | :---: |
| 111 | 999 |
| 113 | 997 |
| 115 | 995 |
| $\vdots$ | $\vdots$ |
| 549 | 561 |
| 551 | 559 |
| 553 | 557 |
| 555 | - |

The sum of the numbers in the 1st and 2nd columns is 1110 for each pair, except for 555, which has no pair. The number of numbers that meet the criteria is $5^{3}=125$, so $\frac{5^{3}-1}{2}=\frac{124}{2}=62$ pairs of numbers give the sum of 1110. Therefore, the sum of the three-digit numbers consisting of odd digits is: $62 \cdot 1110+555=69375$.

II. solution. There are five odd digits: $1,3,5,7,9$. We form the three-digit numbers from these. An odd digit will appear in all three positions of the numbers, and there are five different odd digits in each position.

A digit appears 25 times in a given position because the other two positions can be filled with the five digits in $5 \cdot 5=25$ ways.

If every digit appears 25 times in every position, then the sum is:

$$
\begin{aligned}
& 25 \cdot(100+10+1)+25 \cdot(300+30+3)+25 \cdot(500+50+5)+ \\
+ & 25 \cdot(700+70+7)+25 \cdot(900+90+9)= \\
= & 25 \cdot(111+333+555+777+999)=69375
\end{aligned}
$$