Key Concepts: Modular Arithmetic and the Binomial Theorem

This problem involves finding the remainder of a large number raised to a large power when divided by a smaller number. This is best solved using modular arithmetic.

• Modular Arithmetic: We say that two integers $a$ and $b$ are congruent modulo $m$, written as $a \equiv b \pmod m$, if they have the same remainder when divided by $m$. Equivalently, $a-b$ is a multiple of $m$.
• Properties of Congruence:
  • If $a \equiv b \pmod m$, then $a^n \equiv b^n \pmod m$ for any positive integer $n$. This allows us to simplify the base of an exponent before computing the power.
  • If $a \equiv b \pmod m$ and $c \equiv d \pmod m$, then $a \times c \equiv b \times d \pmod m$.
• Binomial Theorem in Modular Arithmetic: A crucial application here is for expressions of the form $(km + r)^n$. When expanded using the Binomial Theorem, every term except the last one (if $r$ is the isolated term) will contain a factor of $m$. Thus, $(km + r)^n \equiv r^n \pmod m$. This is particularly useful when $r=1$ or $r=-1$.

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Step-by-Step Solution

1. Simplify the Base Modulo the Divisor

• Goal: The first step is to reduce the base of the exponent, $428$, to its equivalent remainder when divided by $21$. Working with smaller numbers in modular arithmetic is always more manageable.
• Calculation: Divide $428$ by $21$: $428 = 21 \times 20 + 8$
• Explanation: This means that $428$ leaves a remainder of $8$ when divided by $21$. In modular arithmetic notation, this is expressed as: $428 \equiv 8 \pmod{21}$
• Implication: Using the property $a \equiv b \pmod m \implies a^n \equiv b^n \pmod m$, we can substitute $428$ with $8$ in our expression: $428^{2024} \equiv 8^{2024} \pmod{21}$ Now, our task is to find the remainder of $8^{2024}$ when divided by $21$.

2. Find a Cyclical Pattern for Powers of the Simplified Base

• Goal: To simplify the large exponent $2024$, we look for a smaller power of $8$ that gives a remainder of $1$ (or $-1$, which is $20 \pmod{21}$) when divided by $21$. This often creates a useful cycle.
• Calculation: Let's compute the first few powers of $8$ modulo $21$: $8^1 \equiv 8 \pmod{21}$ $8^2 = 64$ Now, find the remainder of $64$ when divided by $21$: $64 = 21 \times 3 + 1$
• Explanation: We found that $8^2 \equiv 1 \pmod{21}$. This is a very convenient result because any power of $1$ is $1$. This effectively breaks down the large exponent.

3. Apply the Pattern to the Exponent using Congruence and Binomial Principle

• Goal: Use the identity $8^2 \equiv 1 \pmod{21}$ to simplify $8^{2024} \pmod{21}$.
• Calculation: We can rewrite the exponent $2024$ in terms of $2$: $2024 = 2 \times 1012$ So, we can express $8^{2024}$ as: $8^{2024} = (8^2)^{1012}$ Now, substitute the congruence $8^2 \equiv 1 \pmod{21}$ into this expression: $(8^2)^{1012} \equiv (1)^{1012} \pmod{21}$ Since $1$ raised to any positive integer power is $1$: $(1)^{1012} = 1$
• Explanation: This step leverages the property that if $a \equiv b \pmod m$, then $a^n \equiv b^n \pmod m$. Here, $8^2 \equiv 1 \pmod{21}$, so raising both sides to the power of $1012$ gives $(8^2)^{1012} \equiv 1^{1012} \pmod{21}$, which simplifies to $1 \pmod{21}$. Alternatively, we can see $8^2 = 64 = 3 \times 21 + 1$. So, $8^{2024} = (3 \times 21 + 1)^{1012}$. By the Binomial Theorem principle, $(km+r)^n \equiv r^n \pmod m$, so $(3 \times 21 + 1)^{1012} \equiv 1^{1012} \pmod{21}$.

4. Final Remainder

• Conclusion: We established in Step 1 that $428^{2024} \equiv 8^{2024} \pmod{21}$. We established in Step 3 that $8^{2024} \equiv 1 \pmod{21}$. Combining these, we conclude that: $428^{2024} \equiv 1 \pmod{21}$
• Answer: The remainder when $428^{2024}$ is divided by $21$ is $\mathbf{1}$.

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Tips and Common Mistakes to Avoid

• Always simplify the base first: Before dealing with large exponents, always reduce the base modulo the divisor. This dramatically simplifies subsequent calculations.
• Look for powers congruent to 1 or -1: When finding patterns, try to find $a^k \equiv 1 \pmod m$ or $a^k \equiv -1 \pmod m$ (which is equivalent to $a^k \equiv m-1 \pmod m$). These are the most powerful simplifications.
• Ensure the remainder is in the correct range: A remainder when divided by $m$ must be an integer between $0$ and $m-1$ (inclusive). If your calculation yields a negative remainder (e.g., $-5 \pmod{21}$), add the modulus to get the correct positive remainder (e.g., $-5+21 = 16 \pmod{21}$).
• Careful with exponent splitting: Ensure that when you split an exponent (e.g., $a^{mn} = (a^m)^n$), the modular equivalence applies correctly.

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Summary

This problem effectively demonstrates the power of modular arithmetic for simplifying calculations involving large numbers and exponents. By first reducing the base modulo the divisor, and then finding a small power of the simplified base that is congruent to $1$ (or $-1$), we can efficiently determine the final remainder using the properties of congruence and the binomial theorem principle.