1. Determine whether among all eight-digit multiples of 4, there are more of those that contain the digit 1 in their decimal representation, or those that do not contain the digit 1.
1. First, we determine the number $u$ of all eight-digit numbers divisible by four. Each such number has a non-zero digit in its first position from the left. Thus, we have 9 possibilities. For the next five positions, it can have any digit of the decimal system, i.e., 10 possibilities for each position, and it ends with a two-digit number divisible by four, i.e., 00, 04, 08, 12, 16, 20, 24, ..., 96, a total of 25 possibilities. Therefore,

$$
u=9 \cdot 10^{5} \cdot 25=22500000
$$

A similar consideration can be applied to finding the number $v$ of all eight-digit numbers divisible by four, which in their decimal notation do not contain the digit 1. For the first position from the left, we now have 8 possibilities, and for each of the next five positions, we have 9 possibilities. The last two digits on the right must be a two-digit number divisible by four, but it must not contain the digit 1. These are all the two-digit numbers from the previous paragraph except for 12 and 16, i.e., 23 possibilities. Therefore,

$$
v=8 \cdot 9^{5} \cdot 23=10865016.
$$

Conclusion. Since $u>2 v$, there are more eight-digit multiples of 4 that contain the digit 1 in their (decimal) notation than those that do not.

Note. The number $u$ of all eight-digit multiples can also be determined by a simple consideration: the smallest multiple is $A=10000000$, the largest is $B=99999996$, so the desired number is $\frac{1}{4}(B-A)+1=\frac{1}{4}(B+4-A)=22500000$.

To prove the inequality $u>2 v$, it is not necessary to calculate $v$, because the ratio

$$
\frac{u}{v}=\frac{9 \cdot 10^{5} \cdot 25}{8 \cdot 9^{5} \cdot 23}=\frac{9}{8} \cdot\left(\frac{10}{9}\right)^{5} \cdot \frac{25}{23}
$$

can be well estimated using the binomial theorem

$$
\left(\frac{10}{9}\right)^{5}=\left(1+\frac{1}{9}\right)^{5}>1+5 \cdot \frac{1}{9}+10 \cdot \frac{1}{9^{2}}=\frac{136}{81}=\frac{8 \cdot 17}{9^{2}}
$$

thus

$$
\frac{u}{v}=\frac{9}{8} \cdot\left(\frac{10}{9}\right)^{5} \cdot \frac{25}{23}>\frac{9}{8} \cdot \frac{8 \cdot 17}{9^{2}} \cdot \frac{25}{23}=\frac{17 \cdot 25}{9 \cdot 23}=\frac{425}{207}>2.
$$

For a complete solution, award 6 points, of which 2 points for the correct expression of $u$, 3 points for the correct expression of $v$, and 1 point for the proof of the inequality $u>2 v$ or an equivalent inequality. Deduct no more than 1 point for numerical errors in the calculation.