What is the remainder when `32^(32^(32))` is divided by 7 ?

Answer

To find the remainder when \( 32^{32^{32}} \) is divided by 7, we can follow these steps:

### Step 1: Simplify the Base
First, we need to find the remainder of 32 when divided by 7:
\[ 32 \div 7 = 4 \quad \text{(remainder)} \]
So, we can rewrite the expression as:
\[ 32^{32^{32}} \equiv 4^{32^{32}} \mod 7 \]

**Hint:** When dealing with large powers, reducing the base modulo the divisor can simplify calculations.

### Step 2: Identify the Pattern of Powers of 4 Modulo 7
Next, we will find the pattern of \( 4^n \mod 7 \):
- \( 4^1 \equiv 4 \mod 7 \)
- \( 4^2 \equiv 16 \equiv 2 \mod 7 \)
- \( 4^3 \equiv 64 \equiv 1 \mod 7 \)
- \( 4^4 \equiv 256 \equiv 4 \mod 7 \)
- \( 4^5 \equiv 1024 \equiv 2 \mod 7 \)
- \( 4^6 \equiv 4096 \equiv 1 \mod 7 \)

From this, we see a repeating cycle every 3 terms:
\[ 4, 2, 1 \]

**Hint:** Identifying cycles in modular arithmetic can help simplify the problem significantly.

### Step 3: Determine the Exponent Modulo 3
Now, we need to determine which term in the cycle \( 4, 2, 1 \) corresponds to \( 4^{32^{32}} \). To do this, we need to find \( 32^{32} \mod 3 \):
\[ 32 \div 3 = 2 \quad \text{(remainder)} \]
Thus, \( 32 \equiv 2 \mod 3 \).

Now we need to calculate \( 2^{32} \mod 3 \):
- \( 2^1 \equiv 2 \mod 3 \)
- \( 2^2 \equiv 1 \mod 3 \)

Since \( 2^2 \equiv 1 \mod 3 \), we see that \( 2^{32} \) is:
\[ 2^{32} = (2^2)^{16} \equiv 1^{16} \equiv 1 \mod 3 \]

**Hint:** Reducing the exponent modulo the cycle length can help identify which term in the cycle to use.

### Step 4: Find the Corresponding Remainder
Now that we have \( 32^{32} \equiv 1 \mod 3 \), we can use this to find the corresponding term in the cycle of \( 4^n \mod 7 \):
- Since \( 1 \mod 3 \) corresponds to the first term in our cycle, we have:
\[ 4^{32^{32}} \equiv 4^1 \equiv 4 \mod 7 \]

### Final Answer
Thus, the remainder when \( 32^{32^{32}} \) is divided by 7 is:
\[ \boxed{4} \]