Introduction to Euler's Theorem in Modular Arithmetic

Euler's Theorem is a powerful tool in number theory that helps us determine remainders of large powers when divided by a given number. In this article, we will apply Euler's Theorem to find the last two digits of 72016.

Understanding the Problem: Finding Last Two Digits

The problem is to find the last two digits of 72016, which is equivalent to determining 72016 mod 100. This translates to finding the value of 72016 when taken modulo 100.

Euler's Theorem: A Brief Overview

Euler's Theorem states that if (a) and (n) are coprime, then (a^{phi(n)} equiv 1 pmod{n}), where (phi(n)) is the Euler's totient function, which counts the number of integers up to (n) that are coprime with (n).

Application of Euler's Theorem

To apply Euler's Theorem, we first need to determine the value of (phi(100)). Since (100 2^2 times 5^2), we can calculate (phi(100)) using the formula for the Euler's totient function:

(phi(100) 100 times left(1 - frac{1}{2}right) times left(1 - frac{1}{5}right) 100 times 1/2 times 4/5 40)

Since (7) and (100) are coprime, we can use Euler's Theorem to say that (7^{40} equiv 1 pmod{100}).

Reducing the Exponent

We need to reduce the exponent (2016) modulo (40):

(2016 div 40 50) with a remainder of (16)

Hence, (2016 equiv 16 pmod{40}), which means (7^{2016} equiv 7^{16} pmod{100}).

Calculating 7^16 Modulo 100

To find (7^{16} mod 100), we can use repeated squaring:

(7^1 equiv 7 pmod{100}) (7^2 equiv 49 pmod{100}) (7^4 equiv 49^2 2401 equiv 1 pmod{100})

Since (7^4 equiv 1 pmod{100}), we can simplify (7^{16}) further:

(7^{16} (7^4)^4 equiv 1^4 1 pmod{100})

Conclusion

Therefore, the last two digits of (7^{2016}) are 01.

Verification Through Alternative Methods

For verification, we can also observe the periodicity in the last two digits of powers of 7. Starting from 71, the last two digits cycle every 4 numbers: 7, 49, 43, 01, 07, 49,.... Since (2016 mod 4 0), the last two digits of 72016 would be the same as the last two digits of 74, which is 01.

Therefore, the last two digits of 72016 are 01.

Thanks for reading! If you have any questions, feel free to leave a comment.