Check whether $36^{36}+41^{41}$ a multiple of $77$ Let $$a=\frac{36^{36}+41^{41}}{77}.$$

Is $a$ an integer?

I know that:


The last digit ofr $41^{41}$ is $1$.
The last digit of $36^{36}=6^{72}$ is $6$.


How can I use this information to answer my question, or do I need another thing to help me?
Hint: $36^{36}+41^{41} = 36^{36}+(77-36)^{41} = 77m+36^{36}-36^{41} = 77m-36^{36}(6^5+1)(6^5-1)$ and $6^5+1=(6+1)(6^3\cdot5+6\cdot5+1)$.
$\displaystyle{\large%
a = {36^{36} + 41^{41} \over 77}.\quad
\mbox{Is}\quad a\quad \mbox{an integer ?}\\[3mm]}$


$\large\mbox{Hint: Try these tests}$


$$
7\ |\ 7777\ \mbox{because}\
777 - 14=763\,,\quad 76 - 6 = 70\,,\quad 7 - 0 = 7\quad\mbox{and}\quad 7\ |\ 7
$$


$$
11\ |\ 8096\ \mbox{because}\
809 - 6=803\,,\quad 80 - 3 = 77\,,\quad 7 - 7 = 0\quad\mbox{and}\quad 11\ |\ 0
$$



WA yields

$a = 17284125074533757891391113841182593918679814596127517531883081101$