A palindromic number is one which reads the same way forwards
and backward, such as 14599541.
There are no 100digit palindromic primes; all 100digit palindromic
numbers are divisible by 11.
To see this, first observe that 10^{2k+1} + 1 is divisible
by 11, for k = 0, 1, 2, ..., since:
10^{2k+1} + 1 
= 
(11  1)^{2k+1} + 1 

= 
11^{2k+1}  11.11^{2k} + 55.11^{2k1} + ... + 11.11  1 + 1 
by the binomial theorem, and the quantity on the right is clearly a
multiple of 11.
Now, if N is a 100digit palindromic number, we have:
N 
= 
a_{0}a_{1}...a_{49}a_{49}...a_{1}a_{0} 

= 
a_{0}(10^{99} + 1) + a_{1}(10)(10^{97} + 1) + ... + a_{49}(10^{49})(10^{1}+1) 
All of the expressions in brackets are divisible by 11, so N is divisible
by 11 as well.