To solve the problem involving the large numbers consisting of many repetitions of the digits '5' and '9', we will find the remainder when [tex]\( n \)[/tex] is divided by 11 using properties of congruences and the properties of individual digits' congruences. Here's a step-by-step solution:
1. Represent the original problem clearly:
[tex]\[
n = (\underbrace{555 \ldots 555}_{30185 \text{ times}})^2 - (\underbrace{999 \ldots 999}_{50069 \text{ times}})^2
\][/tex]
2. Simplify and reduce the numbers modulo 11:
- Step into the congruence world:
- Since we only need the remainder modulo 11, we convert both large numbers into their congruent forms under modulo 11.
3. Understand the patterns with single digit numbers:
- Any number consisting of repeated '5's can be reduced because:
[tex]\[
555 \ldots 555 \equiv 5 \mod 11
\][/tex]
- Any number consisting of repeated '9's can be reduced similarly:
[tex]\[
999 \ldots 999 \equiv 9 \mod 11
\][/tex]
This is due to the fact that any number that has consistent digits can be "broken down" to just considering that repetitive digit.
4. Calculate Squares Modulo 11:
- Find out what these digits square to, modulo 11.
[tex]\[
5^2 = 25 \implies 25 \mod 11 = 3
\][/tex]
[tex]\[
9^2 = 81 \implies 81 \mod 11 = 4
\][/tex]
5. Using the properties:
[tex]\[
(\underbrace{555 \ldots 555}_{30185 \text{ times}})^2 \equiv 5^2 \mod 11 \equiv 3 \mod 11
\][/tex]
[tex]\[
(\underbrace{999 \ldots 999}_{50069 \text{ times}})^2 \equiv 9^2 \mod 11 \equiv 4 \mod 11
\][/tex]
6. Formulate [tex]\( n \mod 11 \)[/tex]:
- According to the problem:
[tex]\[
n = (\underbrace{555 \ldots 555}_{30185 \text{ times}})^2 - (\underbrace{999 \ldots 999}_{50069 \text{ times}})^2
\][/tex]
- Substituting the squares' modulo results:
[tex]\[
n \equiv 3 - 4 \mod 11
\][/tex]
- Simplifying that:
[tex]\[
n \equiv -1 \mod 11
\][/tex]
- Adjust the negative result by adding 11:
[tex]\[
-1 + 11 = 10
\][/tex]
Therefore, the remainder when [tex]\( n \)[/tex] is divided by 11 is:
[tex]\[
\boxed{10}
\][/tex]
