For Ramanujan

$\begin{aligned}26^{4^{1920}}&=(50-24)^{4^{1920}}\\ (50-24)^{4^{1920}}\pmod{100}&\equiv24^{4^{1920}}\pmod{100}\\ 24^n\pmod{100}&\equiv\begin{cases}76 : \text{ if n is even}\\ 24: \text{ if n is odd}\end{cases}\\ \implies24^{4^{1920}}\pmod{100}&\equiv 76\\ \overline{mn}&=76\\ m\times n-10&=7\times6-10=\color{#EC7300}\boxed{\color{#333333}32}\end{aligned}$

Let $N = 26^{4^{1920}}$ . We need to find $N \bmod 100$ . Using Chinese remainder theorem on factors 4 and 25, we have:

Factor 4: $N \equiv 26^{4^{1920}} \equiv 0 \text{ (mod 4)}$

Factor 25: $N \equiv 26^{4^{1920}} \equiv (25+1)^{4^{1920}} \equiv 1 \text{ (mod 25)}$ , $\implies N \equiv 25n + 1$ , where $n$ is an integer. Then $25n + 1 \equiv 0 \text{ (mod 4)}$ , $\implies n \equiv 3$ and $N \equiv 3\times 25 + 1 \equiv 76 \text{ (mod 100)}$ . Therefore, $mn-10 = \boxed {32}$ .

We want $26^{4^{1920}} \pmod{100}$ .

Write $n=4^{1920}$ .

We have $26^n \equiv 1 \pmod{25}$ and $26^n \equiv 0 \pmod4$ . By the Chinese Remainder Theorem, these two congruences have a unique solution modulo $100$ ; by inspection this is $76$ , so these are the last two digits we need. The answer is then $7 \times 6-10=\boxed{32}$ .

Since $26^2\Rightarrow 676$ , and $26^3=17576$ , the last two digits are 76. Therefore $7\times6-10\Rightarrow 32$ .