Time for 7

By using Fermat's little theorem and the prime factorisation of 100: $7^4 \equiv 1 \pmod 5 \Rightarrow\ 7^4 \equiv 1 \pmod {100} \Rightarrow\ 7^7 \equiv 7^3 \pmod{100}$ So this means that: $7^{7^{7^{7^{7^{7^7}}}}} \equiv 7^{3^{3^{3^{3^{3^3}}}}} \pmod{100}$ Earlier we saw that the exponents produce the same remainder on division of 100 , in cycles of 4 so we need to evaluate: $3^{3^{3^{3^{3^3}}}} \pmod{4}$ Now by testing powers of 3 we can see that: $3^3 \equiv 3\pmod{4}$ $\therefore\ 7^{3^{3^{3^{3^{3^3}}}}} \equiv 7^3 = 343 \equiv 43 \pmod{100}$ Hence the answer is $\boxed{43}$ .

By Euler's theorem we need to determine the value of exponent $\large{\color{#20A900}7^{\color{#D61F06}7^{\color{#624F41}7^{\color{magenta}7^{\color{grey}7^\color{#BA33D6}7}}}}}$ modulo $\phi(100)=100\times\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{5}\right)=40$ .

Again, by Euler's theorem we need to determine the value of exponent $\large{\color{#D61F06}7^{\color{#624F41}7^{\color{magenta}7^{\color{grey}7^\color{#BA33D6}7}}}}$ modulo $\phi(40)=40\times\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{5}\right)=16$ .

Again, by Euler's theorem we need to determine the value of exponent $\large{\color{#624F41}7^{\color{magenta}7^{\color{grey}7^\color{#BA33D6}7}}}$ modulo $\phi(16)=16\times\left(1-\frac{1}{2}\right)=8$ .

Again, by Euler's theorem we need to determine the value of exponent $\large{\color{magenta}7^{\color{grey}7^\color{#BA33D6}7}}$ modulo $\phi(8)=8\times\left(1-\frac{1}{2}\right)=4$ .

Again, by Euler's theorem we need to determine the value of exponent $\large{\color{grey}7^\color{#BA33D6}7}$ modulo $\phi(4)=4\times\left(1-\frac{1}{2}\right)=2$ .

And, now we have : $\begin{aligned}\large{\color{grey}7^\color{#BA33D6}7}&\equiv\left(2\times3+1\right)^7\equiv1\pmod{2}\\ \large{\color{magenta}7^{\color{grey}7^\color{#BA33D6}7}}&\equiv\color{magenta}7^{1}\equiv3\pmod{4}\\ \large{\color{#624F41}7^{\color{magenta}7^{\color{grey}7^\color{#BA33D6}7}}}&\equiv\color{#624F41}7^{3}\equiv\left(8\times{6}+1\right)\times{7}\equiv7\pmod{8}\\ \large{\color{#D61F06}7^{\color{#624F41}7^{\color{magenta}7^{\color{grey}7^\color{#BA33D6}7}}}}&\equiv\color{#D61F06}7^{7}\equiv\left(16\times{3}+1\right)^{3}\times{7}\equiv7\pmod{16}\\ \large{\color{#20A900}7^{\color{#D61F06}7^{\color{#624F41}7^{\color{magenta}7^{\color{grey}7^\color{#BA33D6}7}}}}}&\equiv\color{#20A900}7^{7}\equiv{49}^{3}\times{7}\equiv81\times{9}\times{7}\equiv1\times{63}\equiv23\pmod{40}\\ \large{\color{#3D99F6}7^{\color{#20A900}7^{\color{#D61F06}7^{\color{#624F41}7^{\color{magenta}7^{\color{grey}7^\color{#BA33D6}7}}}}}}&\equiv\large\color{#3D99F6}7^{23}\equiv\left(100\times{24}+1\right)^{5}\times{7^3}\equiv43\pmod{100}\end{aligned}$ So, the last two digits of $\LARGE\color{#3D99F6}7^{\color{#20A900}7^{\color{#D61F06}7^{\color{#624F41}7^{\color{magenta}7^{\color{grey}7^\color{#BA33D6}7}}}}}$ is $43$

The last two digits of 7 make up a series such that it repeats these two last digits all the time: 1) 07 2) 49 3) 43 4) 01 So, to know which one of these pairs appears, you have to only know the last two digits of the number 7^7, so as we know 7 has the following sum:4+3, 4 means that it crossed all the terms and the 3 means that it stopped at the third option which is 43, and since the last two digits of 7^7 are 43, the number shown in the question will have the same two last digits as 43mod4=3, so the last two digits keep coming up, this means that the last two digits of the number shown in the question are 43.

We observe that the order of $7$ in the multiplicative group $\mathbb{Z}_{100}^{\ast}$ is $4$ . That is, $7^4 \equiv 1 (mod\ 100)$ .

With that fact, the original question $\text{What is the remainder when }7^{7^{7^{7^{7^{7^7}}}}} \text{ is divided by }100?$ is now reduced to $\text{What is the remainder when }7^{7^{7^{7^{7^7}}}} \text{is divided by }4?$

Since $7 \equiv -1 (mod\ 4)$ and $7^{7^{7^{7^7}}}$ is odd, we have $7^{7^{7^{7^{7^7}}}} \equiv (-1)^{7^{7^{7^{7^7}}}} \equiv -1 \equiv 3\ (mod\ 4)$

Let us call the quotient of the above operation $q$ . That is, $7^{7^{7^{7^{7^7}}}} = 4q + 3$ .

So now, $7^{7^{7^{7^{7^{7^7}}}}} = 7^{4q+3} = (7^4)^q \times 7^3 \equiv (1)^q \times 7^3 \equiv 7^3 \ (mod\ 100)$ .

Hence the last two digits of $7^{7^{7^{7^{7^{7^7}}}}}$ is the same as the last two digits of $\textit{a less scary looking number}$ $7^3$ which is $\boxed{43}$ .

May be simple way to just do 7^7 = 823543 and derive the last two digits.

All congruences will be modulo 100. We will freely use the results $7^3\equiv43$ and $7^4\equiv1$ . We will show, by induction, that $7^{7^{7^{....}}}\equiv43$ for any power tower of at least two 7s.

First, $7^7\equiv7^3\equiv43$ . If $n\equiv43$ , then $7^n=7^{100m+43}=(7^4)^{25m+10}7^3\equiv43$ .

I have a question, why did I get 7^(7 7 7 7 7*7) on my calculator and the last two digits are 07

Use the Clock model and begin with 7^1. 7^1 ends in 7, 7^2 ends in 49.
7^3 ends in 43 So the. 3rd 7 ends in 43. Now, there are six sevens. Since the 6th seven is a multiple of 3, then according to the Clock model , the 6th 7 will also end in 43. Done !

I think it should be 07

7^3^3^3^3^3^3 = 7^9^9^9

7^9 equiv to 07 (mod 100)

7^9^9^9 = 7^9 (mod 100)

7^9 \equiv 07 \ (mod 100)

Hence answer is 07

in other word

7^7 equiv to 7^3 equiv to 43 (mod 100)

7^7^7 equiv to 7^3^3 = 7^9 equiv to 07 (mod 100)

7^7^7^7 equiv to 43 (mod 100)

and so on

so 7^7^7^7^7^7^7 is equiv to 07

Pls correct the answer

We observe that $7^{4}\equiv1\pmod{100}$

Now, $7\equiv-1\pmod4$ .

$\therefore 7^{7^{7^{7^{7^{7}}}}}\equiv-1\equiv3\pmod4$ .

$\Rightarrow 7^{7^{7^{7^{7^{7^{7}}}}}}\equiv{7^{(4k + 3)}}\equiv{7^{4k}}\times{7^3}\equiv{1 \cdot 7^3}\equiv7^3\equiv43\pmod{100}$ .

Always there are common pattern that can be observed while writing the answer for the exponent series of 7. The common last 2 digits were found as 43. hence the answer is 43

2 lines of Ruby

a=(1..7).to a.permutation.to a.map(&:join).map(&:to_i)

puts a.select{|l|l % 5 != 0}.sort[1999]