Absurdly large primes

Given a palindrome with an even number of digits, such as $\overline{{ a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }{ a }_{ 4 }{ a }_{ 5 }......{ a }_{ 5 }{ a }_{ 4 }{ a }_{ 3 }{ a }_{ 2 }{ a }_{ 1 }}$ , we can prove that it will always be divisible by 11.

We can group common terms together to get $({ 10 }^{ 2017 }+1){ a }_{ 1 }+({ 10 }^{ 2016 }+{ 10 }^{ 1 }){ a }_{ 2 }+({ 10 }^{ 2015 }+{ 10 }^{ 2 }){ a }_{ 3 }+......({ 10 }^{ 1009 }+{ 10 }^{ 1008 }){ a }_{ 1009 }$

Notice that for $k>0$ , we have $({ x }^{ 2k+1 }+1)=(x+1)({ x }^{ 2k }-{ x }^{ 2k-1 }+......-x+1)$

Plugging in $x=10$ , we get $(10^{ 2k+1 }+1)=(11)({ 10 }^{ 2k }-{ 10 }^{ 2k-1 }+......-10+1)$ which has a factor of 11.

Great! Now, the only thing we have to do is to factor 11 out of each term of palindrome.

$({ 10 }^{ 2017 }+1){ a }_{ 1 }+({ 10 }^{ 2016 }+{ 10 }^{ 1 }){ a }_{ 2 }+......({ 10 }^{ 1009 }+{ 10 }^{ 1008 }){ a }_{ 1009 }\\ =({ 10 }^{ 2017 }+1){ a }_{ 1 }+{ 10 }({ 10 }^{ 2015 }+1){ a }_{ 2 }+......{ 10 }^{ 1008 }({ 10 }+1){ a }_{ 1009 }$

It is obvious that $({ 10 }^{ 2017 }+1)$ , $({ 10 }^{ 2015 }+1)$ and so on will contain a factor of 11.

Thus, the whole palindrome will be divisible by 11 and therefore not a prime.

I'm sorry for the messy solution as I am clearly not very good at $LaTeX$ .