The expression shown in the image is $(2025^{2025}) \mod 24$.

To solve this efficiently without directly calculating $2025^{2025}$, we can simplify the expression by using properties of modular arithmetic.

Step-by-Step Solution:

1. Simplify 2025 mod 24: First, find $2025 \mod 24$: $2025 \div 24 = 84 \text{ remainder } 9$ So, $2025 \equiv 9 \mod 24$.

2. Rewrite the original expression: Now, the original expression $(2025^{2025}) \mod 24$ becomes: $(9^{2025}) \mod 24$

3. Use modular exponentiation: We now need to compute $9^{2025} \mod 24$. However, we can apply patterns in powers of 9 modulo 24. Let's check the first few powers of 9 mod 24 to see if any cycle appears:

   • $9^1 \mod 24 = 9$
   • $9^2 \mod 24 = 81 \mod 24 = 9$
   • $9^3 \mod 24 = 9 \times 9 \mod 24 = 81 \mod 24 = 9$

   We see that $9^k \mod 24 = 9$ for any $k \geq 1$.

4. Conclusion: Since $9^{2025} \mod 24 = 9$, the original expression simplifies to: $(2025^{2025}) \mod 24 = 9$

Thus, the value of $(2025^{2025}) \mod 24$ is 9.

Would you like more details or have any questions?

Here are five related questions to consider:

1. What is modular arithmetic, and why is it useful in number theory?
2. How can the Chinese Remainder Theorem help in solving larger modulus problems?
3. What are cycles in modular exponentiation, and how do they simplify computations?
4. What are some practical applications of modular arithmetic in cryptography?
5. How does Fermat's Little Theorem apply to problems involving prime modulus?

Tip: When dealing with large powers in modular arithmetic, look for repeating cycles in the powers to simplify the calculations!