Residue of the (next,next) year

First, note that $7^{2} \equiv -3 (mod 13)$ . This implies $7^{2016} mod 13= (-3)^{1008} mod 13 = 3^{1008} mod 13$ .

Now, since, $3^{3} \equiv 1 (mod 13)$ and $1008 \equiv 0 (mod 3)$ , thus $7^{2016} \equiv 1 (mod 13)$

For Fermat's Little Theorem, $n\cdot (n^{p-1})=kp$ where $p$ is prime. In this case, $7\cdot (7^{13-1}-1)=13k$

So, $7^{12} \equiv 1 (mod 13)$

Then, $7^{2016}=(7^{12})^{168}$ $7^{2016} \equiv 1(mod 13)$ And so r=1