A palindromic number is one which reads the same way forwards
and backward, such as 14599541.
There are no 100-digit palindromic primes; all 100-digit palindromic
numbers are divisible by 11.
To see this, first observe that 102k+1 + 1 is divisible
by 11, for k = 0, 1, 2, ..., since:
| 102k+1 + 1 |
= |
(11 - 1)2k+1 + 1 |
|
= |
112k+1 - 11.112k + 55.112k-1 + ... + 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 100-digit palindromic number, we have:
| N |
= |
a0a1...a49a49...a1a0 |
| |
= |
a0(1099 + 1) + a1(10)(1097 + 1) + ... + a49(1049)(101+1) |
All of the expressions in brackets are divisible by 11, so N is divisible
by 11 as well.