Remainder when `64^(32^(32)`is divided by 9 is equal to ____.

Answer

To find the remainder when \( 64^{32^{32}} \) is divided by 9, we can use modular arithmetic.

### Step 1: Simplify the base modulo 9
First, we simplify \( 64 \) modulo \( 9 \):
\[ 64 \div 9 = 7 \quad \text{(since } 9 \times 7 = 63\text{)} \]
So,
\[ 64 \equiv 1 \mod 9 \]

### Step 2: Rewrite the expression
Now we can rewrite the original expression:
\[ 64^{32^{32}} \equiv 1^{32^{32}} \mod 9 \]

### Step 3: Evaluate the power
Since \( 1 \) raised to any power is still \( 1 \):
\[ 1^{32^{32}} = 1 \]

### Step 4: Find the remainder
Thus, we have:
\[ 64^{32^{32}} \equiv 1 \mod 9 \]

### Conclusion
The remainder when \( 64^{32^{32}} \) is divided by \( 9 \) is:
\[ \boxed{1} \]

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