NMTC Junior Level Stage 1

$\huge {32}^{{32}^{32}} \equiv (3 \times 11-1)^{{32}^{32}} \equiv (-1)^{{32}^{32}} \equiv \boxed{1} \mod 11$

$32^n \equiv (3\dot{}11-1)^n \equiv (-1)^n \pmod {11}$

This part is true because if you expand.

$(3\dot{}11-1)^n = 3^n11^n - n\dot{}3^{n-1}11^{n-1}+\frac {n(n-1)}{2} n\dot{}3^{n-2}11^{n-2}-... (-1)^n$

We note that all terms before the last $(-1)^n$ have a factor of $11$ , therefore the remainder is $(-1)^n$ .

A remainder of $(-1)^n$ means $+1$ when $n$ is even and $-1$ when $n$ is odd.

A $(-1)$ remainder means a remainder of $11-1=10$ . This can be seen from the first few $32^n$ as below.

$\begin{matrix} n & 32^n & x \mod {11} \\ 1& 32 & 10 \\ 2 &1024& 1 \\ 3 &32768& 10\\ 4 &1048576& 1\\ 5 &33554432& 10 \\ 6 &1073741824& 1 \end{matrix}$

Now since $n = 32^{32}$ is even therefore, $32^{32^32} \equiv \boxed{1} \pmod{11}$ .

(33-1)^32^32 when divided by 11 .-1 is not divisible thus -1 is the remainder

Satvik Golechha's solution is very clear and correct. Quite obvious, $(33-1)^{32^{32}}$ gives all terms that are divisible by 11 but $-1^{32^{32}}$ .

I present alternative solution nonetheless. The following when divided by 11 give the corresponding remainder. $2^0$ 1

$2^1$ 2

$2^2$ 4

$2^3$ 8

$2^4$ 5

$2^5$ 10

$2^6$ 9

$2^7$ 7

$2^8$ 3

$2^9$ 6

$2^{10}$ 1

So,

$32^{32^{32}}=(2^5)^{32^{32}}$

$32^{32}$ is even and when multiplied by 5 makes it divisible by 10. Thus $32^{32^{32}}$ divided by 11 gives a remainder of 1.

Or similarly, $32=32^1=2^5=(2^5)^1$ and this gives a remainder of 10 when divided by 11. $32^2=(2^5)^2$ gives a remainder of 1 when divided by 11.

I liked all solutions.

We can find $32^{32}\pmod{10}$ and let the remainder be $x$ .

Then to find the original remainder, find the remainder $32^{x}\pmod{11}$