Long Palindrome Primes

Relevant wiki: Application of Divisibility Rules

We will show, that all 100 digit palindrome numbers are divisible by 11.

The divisibility rule for 11:

The number is divisible by 11 iff the (sum of the odd numbered digits (O)) - (sum of the even numbered digits (E)) is divisible by 11.

Now, our 100 digit palindrome number looks like the following:

$N = \overline { a_1a_2...a_{49}a_{50}a_{50}a_{49}..a_2a_1 }$

As we can see, the 1st and the 100th, the 2nd and the 99th, ... , the 52nd and the 49th, and finally, the 50th and the 51st digits are equal (pairwise). This means, that each of this pair contains 1 odd and 1 even numbered digits, therefore the O and E sums are equal:

$O_N = E_N \iff O_N - E_N = 0$

$11 | 0 \iff 11 | N$

Hence (as they are all divisible by 11, and therefore they cannot be prime), our answer should be:

$\boxed { \text {None} }$

It is easy to see now, that all palindrome numbers having an even number (> 2) of digits are divisible by 11 as well.