Key Concepts and Formulas:

This problem is a classic application of Modular Arithmetic, which deals with remainders after division. The goal is to find the remainder of a large number raised to a large power when divided by a smaller number.

The key principles we will use are:

1. Congruence Relation: We say $a \equiv b \pmod{n}$ if $a$ and $b$ have the same remainder when divided by $n$. This also implies that $a-b$ is a multiple of $n$.
2. Properties of Congruence:
   • If $a \equiv b \pmod{n}$, then $a^k \equiv b^k \pmod{n}$ for any positive integer $k$.
   • If $a \equiv b \pmod{n}$ and $c \equiv d \pmod{n}$, then $ac \equiv bd \pmod{n}$.
3. Cyclicity of Remainders: Powers of an integer modulo $n$ exhibit a repeating (cyclic) pattern of remainders.
4. Fermat's Little Theorem: If $p$ is a prime number, then for any integer $a$ not divisible by $p$, we have $a^{p-1} \equiv 1 \pmod{p}$. This theorem is very useful for finding the cycle length of powers modulo a prime number. In this problem, $p=7$.

Step-by-Step Working with Explanations:

1. Simplify the Base of the Expression using Modular Arithmetic

The problem asks for the remainder when $(2021)^{2023}$ is divided by $7$. The first step is to simplify the base, $2021$, by finding its remainder when divided by $7$.

• What: Calculate $2021 \pmod{7}$.

• Why: According to the congruence property, if we can reduce the base $2021$ to a smaller equivalent number (its remainder) modulo $7$, then the entire expression $(2021)^{2023}$ will have the same remainder as (simplified base)$^{2023} \pmod{7}$. This drastically simplifies the calculation.

  Let's perform the division: $2021 \div 7$ We find that $2021 = 7 \times 288 + 5$. This means that $2021$ leaves a remainder of $5$ when divided by $7$. In modular arithmetic notation: $2021 \equiv 5 \pmod{7}$

  Now, applying the property $a^k \equiv b^k \pmod{n}$ if $a \equiv b \pmod{n}$: $(2021)^{2023} \equiv (5)^{2023} \pmod{7}$ So, our problem is now reduced to finding the remainder of $5^{2023}$ when divided by $7$.

2. Simplify the Exponent using Cyclicity and Fermat's Little Theorem

Our next task is to find the remainder of $5^{2023}$ when divided by $7$. We will use the concept of cyclic patterns of remainders and Fermat's Little Theorem.

• What: Determine the effective exponent for $5^{2023} \pmod{7}$.

• Why: When taking powers modulo a number, the remainders repeat in a cycle. Identifying this cycle allows us to reduce the large exponent to a much smaller, equivalent exponent, making the calculation manageable. Since $7$ is a prime number and $5$ is not divisible by $7$, Fermat's Little Theorem provides an efficient way to find a key point in this cycle.

  According to Fermat's Little Theorem, for a prime $p=7$ and an integer $a=5$ (since $7$ does not divide $5$): $5^{7-1} \equiv 5^6 \equiv 1 \pmod{7}$ This is a crucial result: it tells us that every $6$ powers of $5$ will produce a remainder of $1$ when divided by $7$. This means the cycle length for powers of $5$ modulo $7$ is $6$.

  Now, we need to find out where $2023$ falls within this cycle of $6$. We do this by finding the remainder of $2023$ when divided by $6$: $2023 \div 6$ We perform the division: $2023 = 6 \times 337 + 1$. This means that $2023$ is equivalent to $1$ modulo $6$: $2023 \equiv 1 \pmod{6}$

  We can now rewrite $5^{2023}$ using this information: $5^{2023} = 5^{(6 \times 337) + 1} = (5^6)^{337} \times 5^1$ Substituting this back into our modular congruence: $5^{2023} \equiv (5^6)^{337} \times 5^1 \pmod{7}$ From Fermat's Little Theorem, we know $5^6 \equiv 1 \pmod{7}$: $5^{2023} \equiv (1)^{337} \times 5 \pmod{7}$ Since any positive integer power of $1$ is $1$: $5^{2023} \equiv 1 \times 5 \pmod{7}$ $5^{2023} \equiv 5 \pmod{7}$

3. Final Remainder

Combining the results from Step 1 and Step 2: From Step 1: $(2021)^{2023} \equiv 5^{2023} \pmod{7}$ From Step 2: $5^{2023} \equiv 5 \pmod{7}$

Therefore, by transitivity: $(2021)^{2023} \equiv 5 \pmod{7}$ The remainder when $(2021)^{2023}$ is divided by $7$ is $5$.

Tips and Common Mistakes to Avoid:

• Always Simplify the Base First: The most common mistake is attempting to work with large numbers like $2021$ before reducing them modulo the divisor. Always reduce the base modulo the divisor first to simplify the problem.
• Understanding Exponent Modulo Cycle Length: When $a^k \pmod n$ is to be calculated, you need to calculate $k \pmod{\text{cycle length}}$. Do NOT calculate $k \pmod n$. For prime moduli, Fermat's Little Theorem provides the cycle length ($p-1$).
• Using Negative Remainders: Sometimes it's more convenient to use negative remainders. For example, $5 \equiv -2 \pmod{7}$. Then $5^3 \equiv (-2)^3 \equiv -8 \equiv -1 \pmod{7}$. This can sometimes make calculations quicker, as $(-1)^{\text{even}} = 1$ and $(-1)^{\text{odd}} = -1$.
• Binomial Theorem vs. Modular Arithmetic: While the binomial theorem can be applied (as in the original solution), using direct modular arithmetic properties and theorems like Fermat's Little Theorem is often more streamlined and less error-prone for these types of remainder problems, especially when the base is not very close to a multiple of the divisor.

Summary and Key Takeaway:

To find the remainder of a large power $A^B$ when divided by $N$:

1. Reduce the base: Find $R_A = A \pmod{N}$. The problem becomes finding $R_A^B \pmod{N}$.
2. Find the cycle length: Determine the cycle length ($L$) of the powers of $R_A$ modulo $N$. If $N$ is prime and $R_A$ is not $0 \pmod N$, Fermat's Little Theorem states $R_A^{N-1} \equiv 1 \pmod N$, so the cycle length is a divisor of $N-1$.
3. Reduce the exponent: Find $R_B = B \pmod{L}$ (if $R_B = 0$, use $L$). The problem becomes finding $R_A^{R_B} \pmod{N}$.
4. Calculate the final remainder: Compute the simplified power.

This systematic approach makes complex remainder problems accessible by breaking them down into manageable modular operations. The final remainder obtained through this method is $5$.