The remainder when $(2021{)}^{2023}$ is divided by 7 is :

Complete Solution:

To solve for the remainder when 2021 ²⁰²³ is divided by 7, we can use modular arithmetic and Fermat's Little Theorem.

Step 1: Simplify the Base Using Modulo

We need to find 2021²⁰²³ mod 7.

First, simplify 2021 mod 7:

2021 ÷ 7 = 288

with a remainder of 5, so 2021 ≡ 5 mod 7.

Therefore, 2021²⁰²³ mod 7 is equivalent to 5²⁰²³ mod 7.

Step 2: Use Fermat's Little Theorem

According to Fermat's Little Theorem, if p is a prime number and a is an integer not divisible by p, then:

a^p-1 ≡ 1 mod p

Since 7 is a prime number, we can apply this theorem:

5⁶ ≡ 1 mod 7

Step 3: Reduce the Exponent Modulo 6

Now we reduce the exponent 2023 modulo 6, because 5⁶ ≡ 1 mod 7:

2023 ÷ 6 = 337

with a remainder of 1, so 2023 ≡ 1 mod 6.

This means that 5²⁰²³ ≡ 5¹ mod 7.

Step 4: Calculate the Result

5¹ = 5

Thus, the remainder when 2021²⁰²³ is divided by 7 is 5.