Key Concept: Modular Arithmetic and Congruence Properties

This problem involves finding the remainder of a large number raised to a power when divided by another number. The core concept used is modular arithmetic, which deals with remainders. Specifically, we leverage the property of congruences: If $a \equiv b \pmod n$, then for any positive integer $k$, $a^k \equiv b^k \pmod n$. We also utilize the idea that $(X - Y)^N \equiv (-Y)^N \pmod X$ and $(X + Y)^N \equiv Y^N \pmod X$, which stems from the Binomial Theorem, to simplify expressions.

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Step 1: Simplify the Base Modulo the Divisor

• Goal: Our first step is to reduce the base number, $2021$, to its equivalent remainder when divided by the divisor, $17$. Working with smaller numbers is essential for simplifying subsequent calculations involving the large exponent.
• Working: We perform the division of $2021$ by $17$: $2021 \div 17 = 118 \text{ with a remainder of } 15$ This can be written in congruence notation as: $2021 \equiv 15 \pmod{17}$ To make further calculations easier, we can express $15$ using a negative remainder. Since $15$ is $2$ less than $17$, we can write: $15 \equiv -2 \pmod{17}$ Therefore, $2021 \equiv -2 \pmod{17}$
• Explanation: Using a negative remainder, like $-2$ instead of $15$, often simplifies computations, especially when dealing with large exponents. This is because raising a small negative number to a power can sometimes be less cumbersome than a larger positive number, particularly if it leads to powers of $-1$.

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Step 2: Apply the Congruence Property to the Power

• Goal: Replace the original base with its simplified modular equivalent. This allows us to work with a much smaller base while preserving the remainder property.
• Working: Since $2021 \equiv -2 \pmod{17}$, we can substitute this into the original expression $(2021)^{3762}$: $(2021)^{3762} \equiv (-2)^{3762} \pmod{17}$ As the exponent $3762$ is an even number, any negative base raised to an even power becomes positive. Therefore, $(-2)^{3762}$ simplifies to $2^{3762}$. So, the problem transforms into finding the remainder of $2^{3762}$ when divided by $17$.
• Explanation: This step is a direct application of the fundamental property of modular arithmetic: if two numbers are congruent modulo $n$, then their respective powers will also be congruent modulo $n$. The even exponent simplifies the negative base, which is a common trick in these types of problems.

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Step 3: Simplify the Power of 2 Modulo 17

• Goal: We need to efficiently calculate $2^{3762} \pmod{17}$. The key here is to find a power of $2$ that has a simple remainder (like $1$ or $-1$) when divided by $17$.
• Working: We notice that $2^4 = 16$. When $16$ is divided by $17$, the remainder is $16$, or more conveniently, $-1$. $2^4 = 16 \equiv -1 \pmod{17}$ Now, we want to express the exponent $3762$ in terms of powers of $4$. We divide $3762$ by $4$: $3762 = 4 \times 940 + 2$ This allows us to rewrite $2^{3762}$ as: $2^{3762} = 2^{(4 \times 940 + 2)} = (2^4)^{940} \times 2^2$ Substituting $2^4 \equiv -1 \pmod{17}$ and $2^2 = 4$: $(2^4)^{940} \times 2^2 \equiv (-1)^{940} \times 4 \pmod{17}$
• Explanation: By identifying $2^4 \equiv -1 \pmod{17}$, we've found a way to significantly reduce the complexity. Working with powers of $-1$ is much simpler than working with powers of $2$. We broke down the original exponent into a multiple of $4$ (since $2^4$ gives us the $-1$ congruence) and a remaining smaller exponent. This is a strategic application of exponent rules.

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Step 4: Final Remainder Calculation

• Goal: Complete the calculation to find the final remainder.
• Working: Since $940$ is an even number, $(-1)^{940}$ evaluates to $1$. Therefore, our expression simplifies to: $1 \times 4 \pmod{17}$ $4 \pmod{17}$ The remainder is $4$.
• Explanation: The even exponent of $-1$ is critical here. A positive result directly gives us the final remainder, as $4$ is less than $17$.

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Tips and Common Mistakes to Avoid

• Choosing the Right Remainder: Always consider both positive and negative remainders. For example, $15 \pmod{17}$ and $-2 \pmod{17}$ are equivalent. The choice that leads to $1$ or $-1$ for a small power of the base is usually the most efficient.
• Fermat's Little Theorem: For a prime modulus $p$, if $a$ is not divisible by $p$, then $a^{p-1} \equiv 1 \pmod p$. Here, $2^{16} \equiv 1 \pmod{17}$ could also be used. Dividing $3762$ by $16$ ($3762 = 16 \times 235 + 2$) would lead to $2^{3762} \equiv (2^{16})^{235} \times 2^2 \equiv 1^{235} \times 4 \equiv 4 \pmod{17}$. This is an equally valid and often powerful alternative method.
• Exponent Parity: Always pay attention to whether the exponent is even or odd when dealing with negative bases, as $(-x)^{\text{even}} = x^{\text{even}}$ and $(-x)^{\text{odd}} = -x^{\text{odd}}$.
• Final Remainder Must Be Positive: The final remainder must always be a positive integer between $0$ and $n-1$ (inclusive), where $n$ is the divisor. If your calculation yields a negative remainder, add the divisor until it's positive. For example, if you got $-13 \pmod{17}$, you would add $17$ to get $4 \pmod{17}$.

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Summary and Key Takeaway

To solve problems involving large powers and remainders:

1. Simplify the base modulo the divisor.
2. Look for a small power of the new base that results in a remainder of $1$ or $-1$ (modulo the divisor). This is the key to breaking down large exponents.
3. Use exponent rules to rewrite the original exponent in terms of multiples of the power found in step 2.
4. Substitute the congruences and simplify to arrive at the final remainder. This systematic application of modular arithmetic properties allows us to solve seemingly complex problems without dealing with astronomically large numbers.