Introduction

This article aims to explore the method of finding the last three digits of 799^{800} using modular arithmetic. This technique is particularly useful in number theory and helps in solving complex problems efficiently. We will use both brute-force and theoretical methods to find the solution.

Brute-Force Approach

The brute-force approach involves direct computation of 799^{800} and then taking the modulo 1000. However, this approach can be computationally expensive and time-consuming. The provided steps in the original content demonstrate a more efficient method to find the last three digits of the given number.

Step-by-Step Computation

Start with the basic calculation:

799^2 800 - 1^2 800 - 1 799

799^2 equiv -1 (mod 1000)

Compute higher powers of 799:

799^4 (799^2)^2 equiv (-1)^2 1 (mod 1000)

799^8 (799^4)^2 equiv 1^2 1 (mod 1000)

799^{16} (799^8)^2 equiv 1^2 1 (mod 1000)

799^{32} (799^{16})^2 equiv 1^2 1 (mod 1000)

799^{64} (799^{32})^2 equiv 1^2 1 (mod 1000)

799^{128} (799^{64})^2 equiv 1^2 1 (mod 1000)

799^{256} (799^{128})^2 equiv 1^2 1 (mod 1000) 799^{512} (799^{256})^2 equiv 1^2 1 (mod 1000) 799^{768} 799^{512} cdot 799^{256} equiv 1 cdot 1 1 (mod 1000)

799^{800} 799^{768} cdot 799^{32} equiv 1 cdot 1 1 (mod 1000)

Theoretical Approach using Euler's Theorem

To further simplify the process, we can use Euler's theorem which states that if a and n are coprime, then a^{phi(n)} equiv 1 (mod n). Here, phi(1000) 400 (since 1000 2^3 cdot 5^3 and phi(p^k) p^k - p^{k-1}). Therefore, we can use:

799^{400} equiv 1 (mod 1000)

799^{800} (799^{400})^2 equiv 1^2 1 (mod 1000)

Conclusion

In conclusion, the last three digits of 799^{800} are 001. The key takeaway from this exercise is the effective use of modular arithmetic and number theory concepts such as Euler's theorem to simplify complex computations. This approach not only provides the answer but also an efficient method to tackle similar problems.