It's too big...

By Fermat's little theorem states that if $p$ is prime and $m$ and $n$ are positive integers such that $m\equiv n\pmod{p-1}$ then for every integer $a$ we have $a^m\equiv a^n\pmod{p}$ .

Since $7$ is prime , to solve the problem, $32^{32^{32}} \equiv x \pmod{7}$ we first solve for $n$ in $32^{32} \equiv n \pmod {6}$ .

$32^{32} \equiv (30+2)^{32} \equiv 2^{32} \equiv 256^4 \equiv 4^4 \equiv 256 \equiv 4 \pmod {6}$

$32^{32^{32}} \equiv 32^4 \pmod{7} \quad \Rightarrow 32^4 \equiv (28+4)^4 \equiv 4^4 \equiv 256 \equiv \boxed {4} \pmod {7}$

Notation: $R(\frac{a}{b})$ =Remainder when $a$ is divided by $b$ .

$R(\frac{32^{32^{32}}}{7})$ $=R(\frac{4^{32^{32}}}{7})$ Now $R(\frac{32^{32}}{3})=R(\frac{2^{32}}{3})$ $=R(\frac{{2^{2}}^{16}}{3})=1$ $\Rightarrow {32^{32}}=3n+1$ $\Rightarrow R(\frac{4^{32^{32}}}{7})=R(\frac{4^{3n}•4}{7})=4$ $\Rightarrow R(\frac{32^{32^{32}}}{7})=\boxed 4$

2 x 1 = 2 (2^1) 2 x 2 = 4 (2^2) 4 x 2 = 8 (2^3) 8 x 2 = 16 (2^4) 16 x 2 = 32 32 x 2 = 64 64 x 2 = 128 128 x 2 = 256 256 x 2 = 512 (2^9)

the end digits are 2,4,8,6,2,4,8,6...

take note of the pattern of the remainders at the above sequence when divided by 7 when exponents are increased

2, 4, 1, 2, 4, 1, 2, 4, 1, 2, ...

at 2^32 the remainder, count 32 starting from 2,4,1,2,4,1,... you will arrive ar 4

same when 32 is raised to the power of 1 to 32, the above patterns goes the same

so the answer will be 4

4 x 7 = 28, 32 - 28 = 4

I dont understand others solution. I used this logic. Is there any wrong?

(32 x 32 x 32 x 32 x ........32^32 times)/7 can be written as

(32/7) x (32 x 32 x 32 x........(32^32 - 1) times).

32/7 leaves remainder as 4. The second part has no remainder. That's what I understood.

32^32^32 -- 2^5^2^5^2^5

=> 2^800

(2^x)/7 has a cyclicity or repeatability of 3. That is (2^1)/7 - 2, (2^2)/7- 4, (2^3)/7 - 1 .....

Hence (2^800)/7 boils down to the same as (2^2)/7 which is 4.

Every 4th power of 32 leaves remainder 4 when divided by 7

32^32^32 looks quite daunting

Rof 32^32^32 when divided by 7 = Rof 4^32^32 by 7 as 28+4 = 32 4^32^32 =Rof 4 ^ (3k + 1) = Rof 4 ^3k * Rof 4 ^1 = 1 * 4 = 4 Hence Answer

I used Binomial