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.
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 $N=11$.