Key Concept: Modular Arithmetic and Properties of Exponents

To efficiently determine the remainder when a large number raised to a power is divided by another number, we utilize the fundamental properties of modular arithmetic. The most crucial property for this problem is: if two integers $a$ and $b$ are congruent modulo $n$ (i.e., they have the same remainder when divided by $n$, denoted as $a \equiv b \pmod n$), then any positive integer power of $a$ will be congruent to the same power of $b$ modulo $n$. Mathematically, if $a \equiv b \pmod n$, then $a^k \equiv b^k \pmod n$ for any positive integer $k$.

This property allows us to simplify the base of an exponent before evaluating the large power, significantly reducing computational complexity. Additionally, we use the simple fact that any positive integer power of $1$ is always $1$.

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Step-by-Step Solution

Step 1: Simplify the base of the exponent modulo the divisor. The problem asks for the remainder when $64^{32^{32}}$ is divided by $9$. Our first step is to simplify the base of the main exponent, which is $64$, with respect to the divisor, $9$. Let's divide $64$ by $9$: $64 = 7 \times 9 + 1$ This means that $64$ leaves a remainder of $1$ when divided by $9$. In modular arithmetic notation, we write this as: $64 \equiv 1 \pmod 9$ Explanation: We perform this simplification because working with $64$ directly in the exponent would be cumbersome. By finding its remainder modulo $9$, we can replace $64$ with $1$ in the modular expression, which is much simpler to handle in subsequent steps, thanks to the property $a^k \equiv (a \pmod n)^k \pmod n$.

Step 2: Apply the simplified base to the original expression. Now we substitute the congruence $64 \equiv 1 \pmod 9$ into the original expression. Let $T = 32^{32}$ for clarity, as $T$ represents the exponent. Since $T$ is a very large positive integer ($32$ multiplied by itself $32$ times), we can write: $64^{32^{32}} \equiv 1^{32^{32}} \pmod 9$ Now, we evaluate $1$ raised to any positive integer power: $1^{T} = 1$ Explanation: This is a direct application of the property that any positive integer power of $1$ is always $1$. No matter how large the exponent $32^{32}$ is, raising $1$ to that power will still result in $1$. This dramatically simplifies the entire problem.

Step 3: State the final remainder. From the previous step, we have determined that: $64^{32^{32}} \equiv 1 \pmod 9$ Therefore, the remainder when $64^{32^{32}}$ is divided by $9$ is $1$.

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Tips and Common Mistakes to Avoid

• Always simplify the base first: When faced with modular exponentiation, the most efficient and often simplest approach is to first find the remainder of the base with respect to the divisor. This initial simplification is key to solving such problems.
• Understanding the definition of a remainder: A remainder must always be a non-negative integer and strictly less than the divisor. For example, when dividing by $9$, the only possible remainders are $0, 1, 2, 3, 4, 5, 6, 7, \text{ and } 8$.
• Discrepancy with given "Correct Answer": The problem statement indicates a "Correct Answer: 32". However, $32$ cannot be a remainder when dividing by $9$, because $32$ is greater than $9$. If $32$ were the result of some intermediate calculation, one would still need to find $32 \pmod 9$, which is $5$. Based on the standard interpretation of finding "the remainder when $X$ is divided by $Y$", our derived answer of $1$ is mathematically sound. It's crucial to understand why $32$ (or any number greater than or equal to the divisor) cannot be a final remainder.

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Summary and Key Takeaway This problem serves as an excellent illustration of how modular arithmetic simplifies complex calculations. By first reducing the base ($64$) to its remainder modulo the divisor ($9$), we transformed a seemingly daunting problem into a straightforward evaluation of $1$ raised to a power. The key takeaway is to always look for opportunities to simplify components of an expression modulo the divisor, especially the base of an exponent, to efficiently arrive at the correct remainder.