The answer lies somewhere near the question?

Note that $4096={ 2 }^{ 12 }\equiv 1(mod\quad 91)$ , then ${ 2 }^{ 80 }={ ({ 2 }^{ 12 }) }^{ 6 }\cdot { 2 }^{ 8 }\equiv { 1 }^{ 6 }\cdot 256(mod\quad 91)$ , thus, ${ 2 }^{ 80 }\equiv 74(mod\quad 91)$ , and the desired answer is 91.

Relevant wiki: Euler's Theorem

Since (2,91) = 1, we can apply Euler's Theorem,

$a^{\varphi(p)} \equiv 1$ mod p, where a & p are coprimes.

$2^{\varphi(91)} \equiv 1$ mod 91

$\Rightarrow 2^{72} \equiv 1$ mod 91 ............(1) (since $\varphi(91) = 91\times\frac{12}{13} \times \frac{6}{7} = 72$ )

Now we know that $2^{8} \equiv$ 74 mod 91 ................(2)

Multiplying (1) & (2), we get

$2^{80} \equiv$ 74 mod 91. Hence, the remainder is 74. Adding 17 to it, we get $\boxed{91}$ as our answer.