Key Concept: Modular Arithmetic

This problem requires finding the remainder of a large sum of powers when divided by 7. The most efficient way to solve this is by using modular arithmetic. Modular arithmetic deals with remainders after division.

The key properties of modular arithmetic that we will use are:

1. Congruence: $a \equiv b \pmod n$ means that $a$ and $b$ have the same remainder when divided by $n$, or equivalently, $a-b$ is a multiple of $n$.
2. Powers: If $a \equiv b \pmod n$, then $a^k \equiv b^k \pmod n$ for any positive integer $k$. This allows us to reduce the base of a power before evaluating it.
3. Sum: If $a \equiv b \pmod n$ and $c \equiv d \pmod n$, then $a+c \equiv b+d \pmod n$. We can sum the remainders.

Step-by-step Solution

Our goal is to find the remainder of $(2021)^{2022} + (2022)^{2021}$ when divided by 7.

1. Simplify the bases using modular arithmetic First, we find the remainder of each base ($2021$ and $2022$) when divided by 7.

• For the base 2021: We divide 2021 by 7: $2021 \div 7 = 288 \text{ with a remainder of } 5$ So, $2021 \equiv 5 \pmod 7$. For easier calculation with powers, it's often beneficial to use a negative remainder if it's smaller in magnitude. Since $5-7 = -2$, we can also write: $2021 \equiv -2 \pmod 7$ This means $(2021)^{2022} \equiv (-2)^{2022} \pmod 7$.

• For the base 2022: We divide 2022 by 7: $2022 \div 7 = 288 \text{ with a remainder of } 6$ So, $2022 \equiv 6 \pmod 7$. Using a negative remainder for simplicity ($6-7 = -1$): $2022 \equiv -1 \pmod 7$ This means $(2022)^{2021} \equiv (-1)^{2021} \pmod 7$.

2. Evaluate each term modulo 7

Now we substitute these simplified bases into the original expression: $(2021)^{2022} + (2022)^{2021} \equiv (-2)^{2022} + (-1)^{2021} \pmod 7$

Let's evaluate each term:

• First Term: $(-2)^{2022} \pmod 7$ Since the exponent 2022 is an even number, $(-2)^{2022}$ is positive: $(-2)^{2022} = 2^{2022}$ Now, we need to find $2^{2022} \pmod 7$. Let's look for a pattern in the powers of 2 modulo 7: $2^1 \equiv 2 \pmod 7$ $2^2 \equiv 4 \pmod 7$ $2^3 \equiv 8 \equiv 1 \pmod 7$ The powers of 2 modulo 7 cycle with a length of 3 (i.e., $2^k \pmod 7$ repeats every 3 powers). This is a very useful property. To use this, we find the remainder of the exponent $2022$ when divided by the cycle length $3$: $2022 \div 3 = 674 \text{ with a remainder of } 0$ Since the remainder is 0, this means $2022$ is a multiple of 3. We can write $2022 = 3 \times 674$. Therefore, $2^{2022} = (2^3)^{674} \equiv (1)^{674} \pmod 7$ $2^{2022} \equiv 1 \pmod 7$

• Second Term: $(-1)^{2021} \pmod 7$ Since the exponent 2021 is an odd number, $(-1)^{2021}$ is negative: $(-1)^{2021} = -1$ In modular arithmetic, a negative remainder can be converted to a positive one by adding the modulus. $-1 \equiv 6 \pmod 7$ So, $(-1)^{2021} \equiv -1 \pmod 7$.

3. Combine the results

Finally, we add the remainders of the two terms: $(2021)^{2022} + (2022)^{2021} \equiv 1 + (-1) \pmod 7$ $\equiv 0 \pmod 7$

Therefore, the remainder when $(2021)^{2022} + (2022)^{2021}$ is divided by 7 is $\boxed{0}$.

Relevant Tips and Common Mistakes to Avoid

• Choosing Remainders: When finding $a \pmod n$, you can choose any integer $b$ such that $a \equiv b \pmod n$. Often, choosing $b$ to be the smallest non-negative integer (e.g., $5 \pmod 7$) or the integer with the smallest absolute value (e.g., $-2 \pmod 7$) can simplify calculations, especially with large exponents. Negative remainders are particularly useful when the exponent is odd or even, as seen with $(-1)^{odd} = -1$ and $(-1)^{even} = 1$.
• Cyclicity of Powers (Order of an Element): For powers like $a^k \pmod n$, the sequence of remainders $a^1, a^2, a^3, \dots \pmod n$ will eventually repeat. Finding this cycle length (or the order of $a$ modulo $n$) is crucial. For a prime modulus $p$, Fermat's Little Theorem states that $a^{p-1} \equiv 1 \pmod p$ for any integer $a$ not divisible by $p$. In our case, $2^{7-1} = 2^6 \equiv 1 \pmod 7$. However, we found an even shorter cycle: $2^3 \equiv 1 \pmod 7$. Always look for the smallest cycle.
• Binomial Expansion: While binomial expansion could technically be used, e.g., $(7k-2)^{2022} = \sum_{j=0}^{2022} \binom{2022}{j} (7k)^j (-2)^{2022-j}$, all terms except the last one (when $j=0$) are multiples of 7. Thus, $(7k-2)^{2022} \equiv (-2)^{2022} \pmod 7$. This confirms that reducing the base first using modular arithmetic is a direct and efficient shortcut, avoiding lengthy expansions.

Summary

By skillfully applying modular arithmetic, we first simplified the bases of the powers to their equivalent remainders modulo 7 ($2021 \equiv -2 \pmod 7$ and $2022 \equiv -1 \pmod 7$). Then, by leveraging the cyclic nature of powers modulo 7, particularly $2^3 \equiv 1 \pmod 7$, and the properties of negative numbers raised to odd/even powers, we reduced the expression to $1 + (-1) \pmod 7$, which ultimately results in a remainder of 0. This method demonstrates the power of modular arithmetic in simplifying complex remainder problems.