Jee Main 2024 29 01 2024 Shift 2 - Question11
Question 11
The remainder when $64^{32^{32}}$ is divided by 9 is
Show Answer
Answer: (1)
Solution:
$64 \equiv 1(\bmod 9)$
$$ \begin{aligned} & 64^{32^{32}} \equiv 1^{32^{32}}(\bmod 9) \\ & \Rightarrow \text { Remainder }=1 \end{aligned} $$





