10) The answer is $(\mathbf{A})$. Let $a_{0}, a_{1}, \ldots, a_{2 n-1}$ be the base 10 digits of a palindrome number $N$ with an even number of digits (i.e., $N=a_{0}+a_{1} \cdot 10+\cdots+a_{2 n-1} 10^{2 n-1}$). We then have $a_{i}=a_{2 n-1-i}$ and thus the digits in even positions are the same as the digits in odd positions. The alternating sum of the digits is equal to zero, so $N$ is divisible by 11. If $N$ is prime, necessarily $N=11$.
10) The largest prime palindrome with an even number of digits has $\begin{array}{lll}\text { (A) } 2 \text { digits } & \text { (B) } 4 \text { digits } & \text { (C) } 10 \text { digits }\end{array}$

(D) there do not exist numbers with these properties

(E) there exist numbers of any size with these properties.

Note: a number is called a palindrome if it can be read the same way from left to right or from right to left. For example, 141 and 2552 are palindromes, while 1231 is not.