B3. A digital clock displays times from 00:00:00 to 23:59:59 throughout a day. Every second of the day, you can add up the digits, resulting in an integer. At 13:07:14, for example, the sum is $1+3+0+7+1+4=16$. If you write down the sum for every possible time on the clock and then take the average of all these numbers, what is the result?
B3. Let the total number of possible clock positions be $S$ (for this problem, it is not necessary to explicitly calculate $S$). If we write down only the last digit of each possible position, we would write each of the digits $\frac{1}{10} S$ times. The total sum of these last digits is thus $\frac{1}{10} S(0+1+2+\ldots+9)=\frac{45}{10} S=\frac{9}{2} S$.

If we do the same for the second-to-last digit, we would only write the digits 0 through 5, each exactly $\frac{1}{6} S$ times. The total sum of these digits is then $\frac{1}{6} S(0+1+\ldots+5)=\frac{15}{6} S=\frac{5}{2} S$.

In the same way, the sum of the digits in the minute positions is equal to $\frac{9}{2} S+\frac{5}{2} S$.

For the first two digits, which indicate the hours, we need to be more careful since not every digit occurs equally often. However, it is true that each of the digit pairs $00,01, \ldots, 23$ occurs equally often: $\frac{1}{24} S$ times. The total sum of the first two digits is then equal to

$$
\begin{aligned}
& \frac{1}{24} S((0+0)+(0+1)+(0+2)+\ldots+(2+3)) \\
= & \frac{1}{24} S((0+1+\ldots+9)+10 \cdot 1+(0+1+\ldots+9)+4 \cdot 2+(0+1+2+3)) \\
= & \frac{114}{24} S=\frac{19}{4} S .
\end{aligned}
$$

The total sum of all digits of all possible positions is now $2 \cdot \frac{9}{2} S+2 \cdot \frac{5}{2} S+\frac{19}{4} S=$ $\frac{75}{4} S$. The average is the total divided by the number, $S$, and that is thus equal to $\frac{75}{4}\left(=18 \frac{3}{4}\right)$.