Seven, Seven, Seven

$2801= 7^4 + 7^3 +7^2 +7 +1$

When written in a polynomial form, $x^5 = (x-1)(x^4+x^3+x^2+x+1) +1$ .

Thus, when $7^5$ is divided by $2801$ , the remainder is $1$ .

Finally, $7^{777} \equiv 7^{5\times 155 + 2} \equiv 7^2 \equiv \boxed{49} \pmod {2801}$

We know, $7^5\equiv 1\pmod{2801}$

So, $7^{775}\equiv 1\pmod{2801}$

or, $7^{777}\equiv 49\pmod{2801}$