Key Concept: Modular Arithmetic and Binomial Theorem

This problem requires a strong understanding of modular arithmetic, which deals with remainders from division. The core idea is to simplify numbers and expressions by taking their remainders with respect to a given modulus. The Binomial Theorem, which states that $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$, is particularly useful when dealing with large powers, as it allows us to expand terms and often identify parts that are divisible by the modulus, thereby simplifying the remainder calculation. A critical property is that if $A \equiv B \pmod M$, then $A^n \equiv B^n \pmod M$.

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Step-by-Step Solution

1. Express $x$ using the given remainder information: We are informed that when $x$ is divided by 4, the remainder is 3. Mathematically, this can be written as: $x \equiv 3 \pmod{4}$ This means $x$ can be represented in the form $x = 4k + 3$ for some integer $k$. Explanation: This step translates the word problem into an algebraic expression, which is the starting point for further calculations.

2. Simplify the base of the expression modulo 8: We need to find the remainder when $(2020 + x)^{2022}$ is divided by 8. First, substitute the expression for $x$ into the base $(2020+x)$: $2020 + x = 2020 + (4k + 3) = 2023 + 4k$ Now, we need to determine the value of $(2023 + 4k)$ modulo 8. Let's find $2023 \pmod{8}$: $2023 = 8 \times 252 + 7$ So, $2023 \equiv 7 \pmod{8}$. Therefore, $2023 + 4k \equiv 7 + 4k \pmod{8}$.

The term $4k \pmod{8}$ depends on the parity (even or odd) of $k$:

• Case A: $k$ is an even integer. If $k = 2m$ for some integer $m$, then $4k = 4(2m) = 8m$. In this scenario, $4k \equiv 0 \pmod{8}$. So, $2023 + 4k \equiv 7 + 0 \pmod{8} \equiv 7 \pmod{8}$.

• Case B: $k$ is an odd integer. If $k = 2m + 1$ for some integer $m$, then $4k = 4(2m + 1) = 8m + 4$. In this scenario, $4k \equiv 4 \pmod{8}$. So, $2023 + 4k \equiv 7 + 4 \pmod{8} \equiv 11 \pmod{8} \equiv 3 \pmod{8}$.

Explanation: We are simplifying the base of the exponentiation $(2020+x)$ before dealing with the large power. We first find the remainder of 2023 when divided by 8. The term $4k$ requires special attention because its remainder modulo 8 changes based on whether $k$ is even or odd. This necessitates splitting the problem into two cases to ensure all possibilities are covered.

3. Evaluate the exponentiated expression for each case: Now we raise the simplified base to the power of 2022 for each case.

• Case A: When $2020 + x \equiv 7 \pmod{8}$ We need to calculate $7^{2022} \pmod{8}$. Since $7 \equiv -1 \pmod{8}$, we can substitute this: $7^{2022} \equiv (-1)^{2022} \pmod{8}$ Because 2022 is an even number, $(-1)^{2022}$ equals 1. Thus, $7^{2022} \equiv 1 \pmod{8}$.

• Case B: When $2020 + x \equiv 3 \pmod{8}$ We need to calculate $3^{2022} \pmod{8}$. Let's observe the pattern of powers of $3 \pmod{8}$: $3^1 \equiv 3 \pmod{8}$ $3^2 \equiv 9 \equiv 1 \pmod{8}$ Since $2022$ is an even number, we can express it as $2022 = 2 \times 1011$. $3^{2022} = (3^2)^{1011} \equiv 1^{1011} \pmod{8} \equiv 1 \pmod{8}$ Thus, $3^{2022} \equiv 1 \pmod{8}$.

Explanation: In both distinct cases for the base, the exponentiation results in a remainder of 1. This demonstrates a common pattern in modular arithmetic where a number raised to an even power often simplifies to 1 if its square is 1 modulo the divisor, or if it is congruent to $-1$ modulo the divisor. The binomial theorem is implicitly used in understanding that terms like $(8m \pm 1)^n$ or $(8m \pm 3)^n$ simplify nicely modulo 8, as higher-order terms become multiples of 8.

4. Final Conclusion: In both possible scenarios arising from the initial condition on $x$, the remainder when $(2020 + x)^{2022}$ is divided by 8 is consistently 1.

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Discrepancy Note: The calculated remainder based on the problem statement and standard mathematical principles, including the implicit application of the Binomial Theorem as suggested by the topic, is 1. The provided "Correct Answer" is 4. This suggests a potential typo in the question, the options, or the stated correct answer.

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Tips and Common Mistakes to Avoid:

• Don't assume base parity: Always thoroughly analyze the base modulo the divisor. If a term like $Nk$ appears where $N$ is a factor of the modulus but not the modulus itself (e.g., $4k \pmod{8}$), you must consider the parity of $k$.
• Look for patterns in powers: When dealing with modular exponentiation, cycle lengths of powers are very useful. For example, $3^1 \equiv 3$, $3^2 \equiv 1$, $3^3 \equiv 3$, $3^4 \equiv 1 \pmod 8$.
• Use negative congruences: Sometimes, expressing a number $A \pmod M$ as $A \equiv -r \pmod M$ (e.g., $7 \equiv -1 \pmod 8$) can greatly simplify exponentiation, especially for even or odd powers.
• Binomial Theorem for specific forms: For expressions of the form $(NM \pm R)^P \pmod M$, only the term involving $R^P$ (or a few initial terms) often contributes to the remainder, as terms containing $NM$ will be multiples of $M$. For example, $(8m-1)^{2022} \equiv (-1)^{2022} \pmod 8 \equiv 1 \pmod 8$.

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Summary/Key Takeaway: When solving remainder problems involving expressions with variables and large exponents, systematically reduce the base modulo the divisor. Pay close attention to how variable terms (like $4k$) interact with the modulus, often requiring a case-by-case analysis. Utilizing properties like $A \equiv -1 \pmod M$ or $A^2 \equiv 1 \pmod M$ can significantly simplify large power calculations, often leading to a consistent remainder across all valid cases.