`(32^(32^(32)))/(7)` find Remainder.

Answer

To find the remainder of \( \frac{32^{32^{32}}}{7} \), we can use modular arithmetic, specifically Fermat's Little Theorem.

### Step-by-Step Solution:

1. **Identify the Base and Modulus**:
We need to find \( 32^{32^{32}} \mod 7 \).

2. **Reduce the Base Modulo 7**:
First, we simplify \( 32 \mod 7 \):
\[ 32 \div 7 = 4 \quad \text{(remainder 4)} \]
Thus, \( 32 \equiv 4 \mod 7 \).

3. **Rewrite the Expression**:
Now we can rewrite the expression:
\[ 32^{32^{32}} \mod 7 \equiv 4^{32^{32}} \mod 7 \]

4. **Apply Fermat's Little Theorem**:
Since 7 is a prime number, we can use Fermat's Little Theorem which states that if \( p \) is a prime and \( a \) is not divisible by \( p \), then:
\[ a^{p-1} \equiv 1 \mod p \]
Here, \( a = 4 \) and \( p = 7 \), so:
\[ 4^{6} \equiv 1 \mod 7 \]
Therefore, we need to find \( 32^{32} \mod 6 \) (since \( 6 = 7 - 1 \)).

5. **Calculate \( 32 \mod 6 \)**:
Now, we simplify \( 32 \mod 6 \):
\[ 32 \div 6 = 5 \quad \text{(remainder 2)} \]
Thus, \( 32 \equiv 2 \mod 6 \).

6. **Calculate \( 32^{32} \mod 6 \)**:
We now need to find \( 2^{32} \mod 6 \). Notice that:
\[ 2^1 \equiv 2 \mod 6 \]
\[ 2^2 \equiv 4 \mod 6 \]
\[ 2^3 \equiv 8 \equiv 2 \mod 6 \]
\[ 2^4 \equiv 16 \equiv 4 \mod 6 \]
The powers of 2 modulo 6 alternate between 2 and 4. Since \( 32 \) is even, we have:
\[ 2^{32} \equiv 4 \mod 6 \]

7. **Substitute Back**:
Now we substitute back into our expression:
\[ 4^{32^{32}} \mod 7 \equiv 4^{4} \mod 7 \]

8. **Calculate \( 4^4 \mod 7 \)**:
Now we compute \( 4^4 \):
\[ 4^2 = 16 \equiv 2 \mod 7 \]
\[ 4^4 = (4^2)^2 \equiv 2^2 = 4 \mod 7 \]

9. **Final Remainder**:
Therefore, the remainder when \( 32^{32^{32}} \) is divided by 7 is:
\[ \boxed{4} \]