11.1. Given two five-digit numbers without the digits 0 and 1 in their notation. The absolute value of their difference is a four-digit number \( S \). It is known that if each digit of one of the original numbers is decreased by 1, then the absolute value of the difference becomes 10002. What values can the number \( S \) take?
Answer: 1109.

Solution: Let $A$ and $B$ be the two given numbers, and $C$ be the number obtained from $B$ by decreasing each of its digits by 1, that is, $C = B - 11111$. If $A10000 > B - A$ (this number is four-digit), then $C > B$. This is a contradiction. Therefore, $A > C$. Also, the case $A > B$ is impossible (then $A - C = A - (B - 11111) = (A - B) + 11111 > 10002 = A - C)$. Thus, the only possible case is: $C < A < B$. And then $A - C = A - (B - 11111) = 10002$, that is, $A - B = -1109$. Hence, $S = |A - B| = B - A = 1109$.

Comment: The correct answer without justification - 0 points.

The correct answer obtained by considering an example -1 point.

The order of numbers $A, B, C$ (in the notation of the solution) is established - 3 points.

An extraneous solution is obtained - no more than 4 points for the problem.

Note: It is not necessary to provide examples of suitable numbers $A, B$, as it has been proven that only one answer is possible, and it follows from the condition that a suitable pair exists. Points are not deducted for the lack of such an example.