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