KVPY #17

$\begin{aligned} 32^{32^{32}} & \equiv 32^{\color{#3D99F6}32^{32} \text{ mod }\phi(9)} \text{ (mod 9)} & \small \color{#3D99F6} \text{Since } \gcd(32, 9) = 1 \text{, Euler's theorem applies.} \\ & \equiv 32^{\color{#3D99F6}32^{32} \text{ mod }6} \text{ (mod 9)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(9) = 6 \\ & \equiv 32^{(30+2)^{32} \text{ mod }6} \text{ (mod 9)} \\ & \equiv 32^{2^{32} \text{ mod }6} \text{ (mod 9)} \\ & \equiv 32^{2^{5\times 6 + 2} \text{ mod }6} \text{ (mod 9)} \\ & \equiv 32^{32^6\times 2^2 \text{ mod }6} \text{ (mod 9)} \\ & \equiv 32^{2^6\times 2^2 \text{ mod }6} \text{ (mod 9)} \\ & \equiv 32^{32\times 2^3 \text{ mod }6} \text{ (mod 9)} \\ & \equiv 32^{2^4 \text{ mod }6} \text{ (mod 9)} \\ & \equiv 32^4 \text{ (mod 9)} \\ & \equiv (36-4)^4 \text{ (mod 9)} \\ & \equiv 4^4 \text{ (mod 9)} \\ & \equiv 16^2 \text{ (mod 9)} \\ & \equiv (18-2)^2 \text{ (mod 9)} \\ & \equiv \boxed{4} \text{ (mod 9)} \end{aligned}$

$32^{32^{32}}$ $= 32^{2^{5*32}}$ . We know $2 \equiv -1 \pmod 3$ , therefore, $2^{32*5} \equiv (-1)^{32*5} \equiv 1 \pmod 3$ . So, let $2^{5*32} - 1 = 3k$ , where $k$ is an odd positive number. $32^{2^{5*32}} = 32^{3k} * 32 \equiv (2^{3})^{5k} * (-4) \equiv (-1)^{5k} * (-4) \equiv (-1).(-4) \equiv \boxed {4} \pmod 9$

We know that remainder when 2^3(8) divided by 9 is -1.

So, remainder when 2^4 divided by 9 is -2.

Extending this logic we can say remainder when 32 divided by 9 as 4, since powers are even.

Let us see inner powers----> 4^32---->2^64---->(2^3)^21*2---->-2(outer power is 32).

Let us see outer power---->2^32---->(2^3)^10*4---->4

So, our answer is 4.