Key Concept: Modular Arithmetic and Properties of Exponents

This problem involves finding the remainder of a highly nested exponential expression when divided by a prime number. The key concepts to apply are:

1. Modular Congruence: If two integers $a$ and $b$ have the same remainder when divided by $n$, we write $a \equiv b \pmod n$.
2. Properties of Modular Exponentiation:
   • If $a \equiv b \pmod n$, then $a^k \equiv b^k \pmod n$ for any positive integer $k$.
   • Any positive integer power of $1$ is $1$, i.e., $1^k = 1$.
3. Order of Operations for Exponents: An expression like $(a^b)^c$ is equivalent to $a^{b \times c}$, while $a^{b^c}$ (a power tower) means $a^{(b^c)}$. The parentheses in the given problem are crucial.

Problem: Find the remainder when $\left((64)^{(64)}\right)^{(64)}$ is divided by 7.

Step-by-step working:

1. Simplify the structure of the expression The given expression is $\left((64)^{(64)}\right)^{(64)}$. This is of the form $(X^Y)^Z$, which by exponent rules simplifies to $X^{Y \times Z}$. Here, $X = 64$, $Y = 64$, and $Z = 64$. So, $\left((64)^{(64)}\right)^{(64)} = (64)^{64 \times 64}$.

Now, let's calculate the exponent: $64 \times 64 = 4096$. Therefore, the expression simplifies to $64^{4096}$.

$\left((64)^{(64)}\right)^{(64)} = (64)^{64 \times 64} = 64^{4096}$

2. Simplify the base modulo 7 To find the remainder of $64^{4096}$ when divided by 7, we first find the remainder of the base, $64$, when divided by 7. Divide 64 by 7: $64 = 9 \times 7 + 1$ This means $64$ leaves a remainder of $1$ when divided by $7$. In modular arithmetic notation, we write this as: $64 \equiv 1 \pmod{7}$ Explanation: This step is crucial because it significantly simplifies the problem. Instead of working with large numbers like 64, we can work with its remainder modulo 7, which is 1.

3. Apply modular exponentiation property Since we have simplified the base $64$ to $1 \pmod 7$, we can use the property that if $a \equiv b \pmod n$, then $a^k \equiv b^k \pmod n$. Substituting $64 \equiv 1 \pmod 7$ into our simplified expression $64^{4096}$: $64^{4096} \equiv 1^{4096} \pmod{7}$ Explanation: This step allows us to replace the original base with its modular equivalent without changing the final remainder of the entire expression.

4. Evaluate the simplified power Now we need to calculate $1^{4096}$. Any positive integer power of $1$ is always $1$. So, $1^{4096} = 1$. Therefore: $1^{4096} \equiv 1 \pmod{7}$ Explanation: This is the final calculation. Since the base reduced to 1, any power of it will also be 1.

5. State the final remainder The remainder when $\left((64)^{(64)}\right)^{(64)}$ is divided by 7 is 1.

Tips & Common Mistakes:

• Understanding Power Notation: Always pay close attention to parentheses in exponential expressions. $\left((A)^B\right)^C$ is fundamentally different from $A^{\left(B^C\right)}$. The former simplifies to $A^{B \times C}$, while the latter is a power tower where the exponent $B^C$ is calculated first. In this problem, the explicit parentheses make the interpretation unambiguous as $(64^{64})^{64}$.
• Early Simplification of Base: For modular exponentiation problems, the first step should almost always be to simplify the base modulo $n$. If the base simplifies to $1 \pmod n$ or $-1 \pmod n$, the problem becomes much easier.
• Fermat's Little Theorem/Euler's Totient Theorem: While not strictly needed here because the base simplifies to $1 \pmod 7$, remember these theorems for cases where the base does not simplify to $1$ or $-1$. Fermat's Little Theorem states that if $p$ is a prime number, then for any integer $a$ not divisible by $p$, $a^{p-1} \equiv 1 \pmod p$. Euler's totient theorem generalizes this for composite moduli.

Summary: This problem demonstrates how simplifying the base of a large exponentiation using modular arithmetic can drastically reduce complexity. By recognizing that $64 \equiv 1 \pmod 7$, the entire expression simplifies to $1$ raised to a large power, which always results in $1$. Thus, the remainder is $1$.