Key Concept: Binomial Theorem and Modular Arithmetic

This problem requires us to find the remainder of a large sum when divided by a specific number, which is a classic application of Modular Arithmetic often simplified by properties derived from the Binomial Theorem.

Modular Arithmetic deals with remainders after division. We say that $a$ is congruent to $b$ modulo $m$, written as $a \equiv b \pmod m$, if $a$ and $b$ have the same remainder when divided by $m$, or equivalently, if $a-b$ is a multiple of $m$.

The Binomial Theorem states the algebraic expansion of $(x+y)^n$ for any non-negative integer $n$: $(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k = \binom{n}{0}x^n y^0 + \binom{n}{1}x^{n-1}y^1 + \dots + \binom{n}{n-1}x^1 y^{n-1} + \binom{n}{n}x^0 y^n$ A very useful property for remainder problems arises when one of the terms in the binomial is a multiple of the modulus. If we have an expression of the form $(am \pm b)^n$, where $am$ is a multiple of $m$, then almost all terms in its binomial expansion will also be multiples of $m$. Specifically: $(am+b)^n = (am)^n + \binom{n}{1}(am)^{n-1}b + \dots + \binom{n}{n-1}(am)b^{n-1} + b^n$ Since every term from $(am)^n$ down to $\binom{n}{n-1}(am)b^{n-1}$ contains $am$ as a factor, they are all multiples of $m$. Therefore, when considering the expression modulo $m$: $(am+b)^n \equiv b^n \pmod m$ This powerful simplification allows us to drastically reduce complex power calculations to simply evaluating the remainder of the non-multiple term raised to the power.

Problem Analysis and Strategy

Our goal is to find the remainder when $7^{2022}+3^{2022}$ is divided by $5$. The strategy will involve:

1. Transforming the bases $7$ and $3$ into forms that are easier to handle modulo $5$.
2. Utilizing the even exponent $2022$ to simplify the bases.
3. Applying the modular property of the Binomial Theorem to each term.
4. Combining the individual remainders.
5. Ensuring the final remainder is a positive integer.

Step-by-Step Solution

Step 1: Express the terms with simplified bases using the even exponent

The given expression is $7^{2022}+3^{2022}$. Notice that the exponent $2022$ is an even number. This is a crucial observation because we can rewrite $a^{2022}$ as $(a^2)^{1011}$. $7^{2022}+3^{2022} = (7^2)^{1011}+(3^2)^{1011}$ Why this step? When dealing with modular arithmetic, we often look for patterns that make the base congruent to $1$, $-1$, or $0$ modulo the divisor. Let's look at the powers of $7$ and $3$ modulo $5$:

• $7 \equiv 2 \pmod 5$
• $7^2 = 49 \equiv 4 \pmod 5 \equiv -1 \pmod 5$
• $3 \equiv 3 \pmod 5$
• $3^2 = 9 \equiv 4 \pmod 5 \equiv -1 \pmod 5$ By squaring the bases, both $7^2$ and $3^2$ become congruent to $-1 \pmod 5$. This will greatly simplify the subsequent calculations.

Step 2: Rewrite the squared bases in the form $(multiple\_of\_5 - 1)$

Now, we evaluate the squared bases and express them in a form suitable for applying the Binomial Theorem for remainders.

• $7^2 = 49$. We can write $49$ as $50 - 1$.
• $3^2 = 9$. We can write $9$ as $10 - 1$.

Substituting these into our expression: $(7^2)^{1011}+(3^2)^{1011} = (50-1)^{1011}+(10-1)^{1011}$ Why this step? This transformation aligns perfectly with the Binomial Theorem property for modular arithmetic discussed earlier. Both $50$ and $10$ are multiples of $5$. By expressing the bases as $(multiple\_of\_5 \pm 1)$, all terms in the binomial expansion containing a factor of $50$ or $10$ will be multiples of $5$, leaving only the last term to determine the remainder.

Step 3: Apply the Binomial Theorem (modular property) to each term

Let's find the remainder for $(50-1)^{1011}$ when divided by $5$: Using the property $(am+b)^n \equiv b^n \pmod m$, with $am=50$, $b=-1$, and $m=5$: $(50-1)^{1011} \equiv (-1)^{1011} \pmod 5$ Since $1011$ is an odd number, $(-1)^{1011} = -1$. Thus, $(50-1)^{1011} \equiv -1 \pmod 5$

Similarly, for the second term, $(10-1)^{1011}$: Using the same property with $am=10$, $b=-1$, and $m=5$: $(10-1)^{1011} \equiv (-1)^{1011} \pmod 5$ Again, since $1011$ is odd, $(-1)^{1011} = -1$. Thus, $(10-1)^{1011} \equiv -1 \pmod 5$ Why this step? This is the core of the solution. By isolating the $(-1)$ term, we avoid expanding the entire binomial expression, which would be computationally intensive. The property allows us to quickly find the remainder based on the power of the non-multiple term.

Step 4: Combine the individual remainders

Now we add the remainders of the two parts of the expression: $7^{2022}+3^{2022} \equiv (50-1)^{1011}+(10-1)^{1011} \pmod 5$ $7^{2022}+3^{2022} \equiv (-1) + (-1) \pmod 5$ $7^{2022}+3^{2022} \equiv -2 \pmod 5$ Why this step? A fundamental rule of modular arithmetic states that if $A \equiv a \pmod m$ and $B \equiv b \pmod m$, then $A+B \equiv a+b \pmod m$. We simply add the remainders found for each term.

Step 5: Convert the negative remainder to a positive remainder

The remainder obtained is $-2$. While mathematically correct, remainders are conventionally expressed as non-negative integers less than the divisor. To convert a negative remainder to its positive equivalent, we add the modulus ($5$ in this case) to it: $-2 \pmod 5 \equiv -2 + 5 \pmod 5$ $-2 \pmod 5 \equiv 3 \pmod 5$ Why this step? By definition, the remainder $r$ when $N$ is divided by $D$ satisfies $0 \le r < D$. Adding the modulus to a negative remainder shifts it into this standard range without changing its congruence.

Therefore, the remainder when $7^{2022}+3^{2022}$ is divided by 5 is $3$.

Important Tips and Common Mistakes

• Prioritize modular reduction: Always try to reduce the base modulo the divisor as early as possible. For instance, $7 \equiv 2 \pmod 5$ and $3 \equiv 3 \pmod 5$. Then $2^{2022} + 3^{2022} \pmod 5$.
• Look for $1$ or $-1$ remainders: Powers that yield $1$ or $-1$ modulo $m$ are incredibly useful because they repeat in cycles. $a^k \equiv 1 \pmod m$ or $a^k \equiv -1 \pmod m$ (which leads to $a^{2k} \equiv 1 \pmod m$) are golden. Here, $7^2 \equiv -1 \pmod 5$ and $3^2 \equiv -1 \pmod 5$ were key insights.
• Negative remainders are powerful: Don't be afraid to use negative remainders (e.g., $4 \equiv -1 \pmod 5$). They can often simplify calculations, especially when dealing with large exponents and parity (even/odd). Just remember to convert to a positive remainder at the very end.
• Binomial property shortcut: Remember the shortcut $(am \pm b)^n \equiv (\pm b)^n \pmod m$. This saves you from writing out the full binomial expansion.

Summary and Key Takeaway

This problem demonstrates an elegant application of modular arithmetic combined with the Binomial Theorem. By recognizing that both $7^2$ and $3^2$ are congruent to $-1 \pmod 5$, we transformed the original expression into a much simpler form. The property $(multiple\_of\_m \pm 1)^n \equiv (\pm 1)^n \pmod m$ then allowed us to quickly determine the individual remainders for each term. Finally, summing these remainders and adjusting for a positive result yielded the remainder of $3$. This approach avoids cumbersome direct calculations of large powers.

The final answer is $\boxed{3}$.