Key Concepts and Formulas

• Modular Arithmetic: The core concept used here. When we say $a \equiv b \pmod n$, it means that $a$ and $b$ have the same remainder when divided by $n$. This also implies that $n$ divides $(a-b)$.
  • Property of Exponents: 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 before calculating large powers.
• Binomial Theorem: $(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$. We often use this to expand expressions like $(An \pm B)^k$ and observe terms that are multiples of $n$.
• Remainder Property: The remainder $r$ when an integer $D$ is divided by an integer $d$ must satisfy $0 \le r < |d|$. If a calculation yields a negative remainder, add the divisor until a positive remainder is obtained. For example, $-28 \equiv -28+35 \equiv 7 \pmod{35}$.

Step-by-Step Solution

1. Reduce the base modulo 35: Our goal is to find the remainder when $(2023)^{2023}$ is divided by 35. First, we simplify the base, 2023, modulo 35. Dividing 2023 by 35: $2023 = 35 \times 57 + 28$ This means $2023 \equiv 28 \pmod{35}$. However, using a negative remainder often simplifies calculations, especially with the Binomial Theorem. We can write: $2023 = 35 \times 58 - 7$ So, $2023 \equiv -7 \pmod{35}$. Using the property $a^k \equiv b^k \pmod n$ if $a \equiv b \pmod n$, we can rewrite the original problem: $(2023)^{2023} \equiv (-7)^{2023} \pmod{35}$ Explanation: By reducing the base first, we work with smaller numbers, which makes subsequent calculations more manageable. Using $-7$ instead of $28$ often leads to simpler binomial expansions.

2. Simplify using the parity of the exponent: The exponent is $2023$, which is an odd number. For any negative number $x$ and an odd exponent $k$, we know that $x^k = -( |x|^k )$. $(-7)^{2023} = -(7^{2023})$ Therefore, the problem reduces to finding the remainder of $- (7^{2023})$ when divided by 35. $(2023)^{2023} \equiv -7^{2023} \pmod{35}$ Explanation: This step eliminates the negative sign from the base, making it easier to work with. We will find $7^{2023} \pmod{35}$ first, and then apply the negative sign.

3. Manipulate the power of 7: To simplify $7^{2023}$, we can factor out $7$ and work with powers of $7^2$: $7^{2023} = 7 \times 7^{2022} = 7 \times (7^2)^{1011}$ Since $7^2 = 49$, we have: $7^{2023} = 7 \times (49)^{1011}$ Now we need to evaluate $7 \times (49)^{1011} \pmod{35}$. Explanation: Factoring out a $7$ is crucial here because the modulus is $35 = 5 \times 7$. This allows us to implicitly handle the factor of $7$ and focus on the other part of the expression.

4. Apply Binomial Theorem to $(49)^{1011}$: We need to evaluate $(49)^{1011}$ in a way that helps with modulo 35. Notice that $49$ is close to a multiple of $5$. We can write $49$ as $50-1$. $(49)^{1011} = (50-1)^{1011}$ Let's expand $(50-1)^{1011}$ using the Binomial Theorem: $(50-1)^{1011} = \binom{1011}{0} (50)^{1011} (-1)^0 + \binom{1011}{1} (50)^{1010} (-1)^1 + \ldots + \binom{1011}{1010} (50)^1 (-1)^{1010} + \binom{1011}{1011} (50)^0 (-1)^{1011}$ Consider this expression modulo 5. Since $50$ is a multiple of $5$ (i.e., $50 \equiv 0 \pmod 5$), all terms containing $50$ as a factor will be multiples of $5$, and thus equivalent to $0 \pmod 5$. The only term that does not contain a factor of $50$ is the very last term: $\binom{1011}{1011} (50)^0 (-1)^{1011} = 1 \times 1 \times (-1) = -1$ So, when $(50-1)^{1011}$ is divided by 5, the remainder is $-1$. $(50-1)^{1011} \equiv -1 \pmod 5$ This means $(50-1)^{1011}$ can be expressed in the form $5\lambda - 1$ for some integer $\lambda$. $(49)^{1011} = 5\lambda - 1$ Explanation: The Binomial Theorem allows us to systematically expand the expression. By observing that $50$ is a multiple of $5$, we can quickly determine the remainder modulo $5$ for the entire expansion, as all terms except the last one become $0 \pmod 5$. This form $5\lambda - 1$ is key to relating it back to the modulus 35.

5. Substitute back and find the final remainder: Now substitute $(49)^{1011} = 5\lambda - 1$ back into the expression from Step 3: $-7^{2023} = -7 \times (49)^{1011} = -7 \times (5\lambda - 1)$ Distribute the $-7$: $-7 \times (5\lambda - 1) = -35\lambda + 7$ Now we evaluate this expression modulo 35: $-35\lambda + 7 \equiv 0 + 7 \pmod{35}$ $-35\lambda + 7 \equiv 7 \pmod{35}$ Thus, the remainder when $(2023)^{2023}$ is divided by 35 is 7. Explanation: The term $-35\lambda$ is an exact multiple of 35, so its remainder is 0. The remaining term, 7, is the desired remainder.

Final Answer: The remainder when $(2023)^{2023}$ is divided by 35 is $\boxed{7}$.

Tips and Common Mistakes

• Negative Remainders: While using negative remainders like $-7 \pmod{35}$ can simplify calculations, always ensure the final answer is a positive remainder within the range $[0, \text{divisor}-1]$.
• Binomial Theorem Application: When using the Binomial Theorem for modular arithmetic, carefully identify which terms become zero modulo the divisor. In $(A \cdot n \pm B)^k \pmod n$, terms with $A \cdot n$ will often vanish.
• Composite Moduli: For composite moduli (like $35 = 5 \times 7$), sometimes the Chinese Remainder Theorem can be used, or as shown here, factoring out common terms can simplify the problem. Be cautious if the base shares a factor with the modulus, as properties like Euler's Totient Theorem or Fermat's Little Theorem might need careful application or an alternative approach.
• Sign Errors: Double-check signs, especially when dealing with negative bases raised to odd or even powers.

Summary and Key Takeaway

This problem demonstrates an effective strategy for finding remainders of large powers:

1. Reduce the base modulo the divisor. Opt for negative remainders if they simplify expressions.
2. Simplify powers of negative bases based on the exponent's parity.
3. Factor out common terms if the base shares a factor with the modulus.
4. Utilize the Binomial Theorem to expand powers, focusing on terms that are not multiples of the relevant smaller modulus (e.g., modulo 5 in this case). By systematically applying these principles, complex remainder problems can be broken down into manageable steps.