Question

Remainder when \( 64^{32^{32}} \) is divided by 9 is equal to ______.

Answer 1

Correct Answer: 1

Let \( 32^{32} = t \).  

Then,  

\(64^{32^{32}} = 64^t = 8^{2t} = (9 - 1)^{2t}\)

Expanding using the binomial theorem, we get:  
\((9 - 1)^{2t} = 9k + 1,\)
for some integer \( k \).  

Thus, the remainder when \( 64^{32^{32}} \) is divided by 9 is 1.

The Correct Answer is: 1

Answer 2

We are given the expression \( 64^{32^{32}} \) and are asked to find the remainder when it is divided by 9.

Step 1: Simplify using modular arithmetic. 

We start by simplifying \( 64 \mod 9 \): \[ 64 \equiv 1 \, (\text{mod} \, 9) \] This is because: \[ 64 \div 9 = 7 \text{ remainder } 1 \]

Step 2: Apply this result to the expression \( 64^{32^{32}} \).

Since \( 64 \equiv 1 \, (\text{mod} \, 9) \), we can simplify the expression \( 64^{32^{32}} \mod 9 \) as: \[ 64^{32^{32}} \equiv 1^{32^{32}} \, (\text{mod} \, 9) \] This simplifies further to: \[ 1^{32^{32}} = 1 \]

Step 3: Conclusion.

Therefore, the remainder when \( 64^{32^{32}} \) is divided by 9 is: \[ \boxed{1} \]