A positive integer $N$ is a palindrome if the integer obtained by reversing the sequence of digits of $N$ is equal to $N$. The year 1991 is the only year in the current century with the following 2 properties:
(a) It is a palindrome
(b) It factors as a product of a 2-digit prime palindrome and a 3-digit prime palindrome. 
How many years in the millenium between 1000 and 2000 have properties (a) and (b)?
$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$
Solution by e_power_pi_times_i
Notice that all four-digit palindromes are divisible by $11$, so that is our two-digit prime. Because the other factor is a three-digit number, we are looking at palindromes between $1100$ and $2000$, which also means that the last digit of the three-digit number is $1$. Checking through the three-digit numbers $101, 111, 121,\dots, 181$, we find out that there are $\boxed{\textbf{(D) } 4}$ three-digit prime numbers, which when multiplied by $11$, result in palindromes.