103. a) Prove that the sum of all n-digit numbers $(n>2)$ is $494 \underbrace{99 \ldots 9}_{(n-3) \text { times }} 5 \underbrace{00 \ldots 0}_{(n-2) \text { times }}$ (thus, the sum of all three-digit numbers is 494550, and the sum of all six-digit numbers is 494999550000).

b) Find the sum of all four-digit even numbers that can be written using the digits $0,1,2,3,4,5$ (the same digit can be repeated in a number).
103. a) The $2n$-digit number given in the condition of the problem can be expressed as follows:

$$
\begin{gathered}
4 \cdot 10^{2n-1} + 9 \cdot 10^{2n-2} + 4 \cdot 10^{2n-3} + 9(10^{2n-4} + 10^{2n-5} + \cdots \\
\ldots + 10^n) + 5 \cdot 10^{n-1} + 5 \cdot 10^{n-2} = 4 \cdot 10^{2n-1} + 9 \cdot 10^{2n-2} + \\
+ 4 \cdot 10^{2n-3} + 9 \cdot 10^n \frac{10^{n-3} - 1}{9} + 5 \cdot 10^{n-1} + 5 \cdot 10^{n-2} = \\
-4 \cdot 10^{-2n-1} + 9 \cdot 10^{2n-2} + 5 \cdot 10^{2n-3} - 10^n + 5 \cdot 10^{n-1} \div \\
+ 5 \cdot 10^{n-2} = \frac{1}{2} \left(8 \cdot 10^{2n-1} + 18 \cdot 10^{2n-2} + 10 \cdot 10^{2n-3} - \right. \\
\left. - 2 \cdot 10^n + 10^n + 10^{n-1} \right) = \frac{1}{2} \left(9 \cdot 10^{2n-1} + 9 \cdot 10^{2n-2} - 9 \cdot 10^{n-1} \right) = \\
= \frac{\left( (10^n - 1) + 10^{n-1} \right) \cdot 9 \cdot 10^{n-1}}{2}
\end{gathered}
$$

But this is also equal to the sum of an arithmetic progression with a common difference of 1, the first term being $10^{n-1}$ and the last term being $10^n - 1$ (the number of terms in the progression is $10^n - 10^{n-1} = 8 \cdot 10^{n-1}$), which is the sum of all $n$-digit numbers.

b) The number of the considered numbers that start with a given digit $a$ (where $a$ can be 1, 2, 3, 4, or 5) is $6 \cdot 6 \cdot 3 = 108$ (because on the second and third places, any of the six digits $0, 1, 2, 3, 4, 5$ can stand, and on the third place, any of the three digits $0, 2, 4$, since only even numbers are considered). Therefore, the total sum of all thousands contained in all our numbers is $(1+2+3+4+5) \cdot 108 \cdot 1000 = 1620000$.

Similarly, the number of numbers in which a given digit stands on the second place is $5 \cdot 6 \cdot 3 = 90$, because on the first place, one of the five digits $1, 2, 3, 4,$ or $5$ can stand. Therefore, the total sum of all hundreds (after subtracting all thousands) contained in all our numbers is $(1+2+3+4+5) \cdot 90 \cdot 100 = 135000$.

In the same way, we find that the total sum of all tens is $(1+2+3+4+5) \cdot 90 \cdot 10 = 13500$ and, finally, the sum of units is $(2+4) \cdot 5 \cdot 6 \cdot 6 \cdot 1 = 1080$.

The desired sum is

$$
1620000 + 135000 + 13500 + 1080 = 1769580
$$