Modulus 13 - #1

Since $13$ is a prime and $13\not\mid 2$ , Fermat's Little Theorem comes to our rescue and we reduce the exponents on the given expression. For those who don't know, Fermat's Little Theorem (FLT) states,

$a^{p-1}\equiv 1\pmod{p}~,~\textrm{where }p\textrm{ is a prime and }a\in\mathbb{Z}~,~p\not\mid a$

So, from FLT, we have $2^{12}\equiv 1\pmod{13}$ . Coming onto the calculations,

$12^{111}\equiv (-1)^{111}\equiv (-1)\pmod{13}~\ldots (i)$

$2^{111}\equiv \left(2^{12}\right)^9 \times 2^3\equiv 1^9\times 8\equiv 8\pmod{13}~\ldots (ii)$

Adding the congruencies $(i)$ and $(ii)$ , we have,

$12^{111}+2^{111}\equiv (-1)+8\equiv 7\pmod{13}$

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Note: $x\not\mid y$ means " $x$ does not divide $y$ "