Three years, back and forth #4

The values in question are exponential factorials. Let $t_n$ denote the $n$ th exponential factorial. We wish to compute $t_{2015}$ , $t_{2018}$ , and $t_{2021}$ modulo $10000$ .

For $t_{2015}$ , we will consider the congruences $t_{2015} \equiv 2015^{t_{2014}} \pmod {16} \ \ \ \ \ \text{and}\ \ \ \ \ t_{2015} \equiv 2015^{t_{2014}} \pmod {625}.$ Notice that $2015^2 \equiv 1 \pmod{16}$ and $t_{2014}$ is even. Also notice that $5$ is a factor of $2015$ and $t_{2014}$ is sufficiently large so that $5^4 = 625$ is a factor of $2015^{t_{2014}}$ . These observations imply $t_{2015} \equiv 1 \pmod{16} \ \ \ \ \ \text{and}\ \ \ \ \ t_{2015} \equiv 0 \pmod{625}.$ By the Chinese remainder theorem, $t_{2015} \equiv 625 \pmod{10000},$ so the last four digits of $t_{2015}$ are $0625$ .

For $t_{2018}$ , we will consider the congruences $t_{2018} \equiv 2018^{t_{2017}} \pmod {16} \ \ \ \ \ \text{and} \ \ \ \ \ t_{2018} \equiv 2018^{t_{2017}} \pmod {625}.$ In this case, $2$ is a factor of $2018$ , and $t_{2017}$ is sufficiently large so that $2^4 = 16$ is a factor of $2018^{t_{2017}}$ . Furthermore, we can check that $2018^{100} \equiv 1 \pmod{625}$ . These observations imply $t_{2018} \equiv 0 \pmod {16} \ \ \ \ \ \text{and}\ \ \ \ \ t_{2018} \equiv 2018^r \pmod {625} \tag{1}$ for any value $r$ satisfying $r \equiv t_{2017} \equiv 2017^{t_{2016}} \pmod{100}.$ To find a suitable value of $r$ , we can consider the congruences $r \equiv 2017^{t_{2016}} \pmod 4 \ \ \ \ \ \text{and}\ \ \ \ \ r \equiv 2017^{t_{2016}} \pmod {25}.$ Notice that $2017^2 \equiv 1 \pmod 4$ and $t_{2016}$ is an even number. Also check that $2017^{20} \equiv 1 \pmod {25}$ . This means $r \equiv 2017^{2k} \equiv 1^k \equiv 1 \pmod 4 \ \ \ \ \ \text{and}\ \ \ \ \ r \equiv 2017^{r'} \pmod{25} \tag{2}$ for any value $r'$ satisfying $r' \equiv t_{2016} \equiv 2016^{t_{2015}} \pmod {20}.$ Since $2016 \equiv 0 \pmod 4$ and $2016 \equiv 1 \pmod 5$ , we can put $r' \equiv 0^{t_{2015}} \equiv 0 \pmod 4 \ \ \ \ \ \text{and}\ \ \ \ \ r' \equiv 1^{t_{2015}} \equiv 1 \pmod 5. \tag{3}$ When we apply the Chinese remainder theorem to $(3)$ , $(2)$ , and $(1)$ , we find $r' \equiv 16 \pmod{20}$ and thus $r \equiv 1 \pmod 4 \ \ \ \ \ \text{and}\ \ \ \ \ r \equiv 2017^{16} \equiv 6 \pmod {25},$ so $r \equiv 81 \pmod{100}$ and $t_{2018} \equiv 0 \pmod {16} \ \ \ \ \ \text{and}\ \ \ \ \ t_{2018} \equiv 2018^{81} \equiv 518 \pmod {625}.$ This means $t_{2018} \equiv 6768 \pmod{10000},$ so the last four digits of $t_{2018}$ are $6768$ .

Finally, for $t_{2021} = 2021^{t_{2020}}$ , we can apply Euler's theorem because $2021$ and $10000$ are relatively prime. Since $\phi(10000) = 4000$ , $t_{2020} \equiv r \pmod{4000} \implies t_{2021} \equiv 2021^{t_{2020}} \equiv 2021^r \pmod{10000}.$ Notice that $4000 = 2^5 \cdot 5^3$ , and both $2$ and $5$ are prime factors of $2020$ . Since $t_{2019}$ is quite large, $4000$ must be a factor of $t_{2020} = 2020^{t_{2019}}$ and thus $t_{2020} \equiv 0 \pmod{4000} \implies t_{2021} \equiv 2021^0 \equiv 1 \pmod{10000}.$ This means the last four digits of $t_{2021}$ are $0001$ .

Therefore, the desired value is $625 + 6768 + 1 = \boxed{7394}.$