We are tasked with finding the remainder when $265^{4081} + 9$ is divided by 266.

This can be tackled using modular arithmetic and simplification techniques. Let's break it down:

Step 1: Simplify the base in mod 266

We know that $265 \equiv -1 \ (\text{mod} \ 266)$. Therefore, we can rewrite the expression:

$265^{4081} + 9 \equiv (-1)^{4081} + 9 \ (\text{mod} \ 266)$

Step 2: Simplify the power of -1

Since $4081$ is an odd number, we know that:

$(-1)^{4081} = -1$

Thus, the expression becomes:

$265^{4081} + 9 \equiv -1 + 9 \ (\text{mod} \ 266)$

$\equiv 8 \ (\text{mod} \ 266)$

Conclusion:

The remainder when $265^{4081} + 9$ is divided by 266 is 8.

---

Would you like more details on any part of this solution, or have any questions? Here are some related problems to expand on this concept:

1. What is the remainder when $265^{4082} + 9$ is divided by 266?
2. How can we calculate $265^{4081} \ \text{mod} \ 266$ for different moduli?
3. What happens if the base was 266 instead of 265 in the problem above?
4. How can modular arithmetic simplify other large exponentiation problems?
5. How does Fermat's Little Theorem help in similar modular arithmetic problems?

Tip: When working with large powers in modular arithmetic, reducing the base first can significantly simplify the calculations!