From the given digits, Iveta created 4 one-digit numbers. All these numbers occupy 4 places in Iveta's long number.

$Z$ from the given digits, Iveta created $4^{2}=16$ two-digit numbers. All these numbers occupy $2 \cdot 16=32$ places in Iveta's long number. The last digit of the last two-digit number is at the 36th place $(4+32=36)$.

From the given digits, Iveta created $4^{3}=64$ three-digit numbers. All these numbers occupy $3 \cdot 64=192$ places in Iveta's long number. The last digit of the last three-digit number is at the 228th place $(36+192=228)$.

From the given digits, Iveta created $4^{4}=256$ four-digit numbers. All these numbers occupy $4 \cdot 256=1024$ places in Iveta's long number. The last digit of the last four-digit number is at the 1252nd place $(228+1024=1252)$.

To reach the 1286th place, 34 places are missing, and these are occupied by five-digit numbers. Since $34: 5$ is 6 with a remainder of 4, the sought digit is the 4th digit in the 7th five-digit number that Iveta created. These numbers, arranged in ascending order, are:

$$
11111,11113,11115,11117,11131,11133,111 \underline{3} 5 .
$$

In Iveta's long number, the 1286th place is occupied by the digit 3.

Evaluation. 2 points for the counts of one-, two-, three-, and four-digit numbers; 2 points for the counts of places these numbers occupy in Iveta's long number; 2 points for determining the sought digit and the quality of the commentary.
Iveta gradually listed natural numbers composed of the digits 1, 3, 5, 7. She used no other digits, proceeded in ascending order from the smallest number, and did not omit any number. She wrote the numbers immediately after each other, thus forming one extraordinarily long number:

$$
1357111315173133 \ldots
$$

Which digit is in the 1286th place of this number?

(E. Novotná)