$32 \cdot 44$ For a positive integer $N$, if the number obtained by arranging its digits from right to left is exactly equal to $N$, then $N$ is called a palindrome. The year 1991 is the only year in this century that has the following two properties:
(a) It is a palindrome;
(b) It can be factored into the product of a two-digit palindrome prime and a three-digit palindrome prime. The number of years from 1000 to 2000 (inclusive) that have properties (a) and (b) (including 1991) is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(E) 5.
(42nd American High School Mathematics Examination, 1991)
［Solution］From the problem, the palindromic years with property $(a)$ are:
$$
1001,1111,1221,1331,1441,1551,1661,1771,1881,1991 \text {. }
$$

Also, $1001=11 \cdot 91, \quad 1111=11 \cdot 101, \quad 1221=11 \cdot 111$,
$1331=11 \cdot 121, \quad 1441=11 \cdot 131, \quad 1551=11 \cdot 141$,
$1661=11 \cdot 151, \quad 1771=11 \cdot 161, \quad 1881=11 \cdot 171$,
$1991=11 \cdot 181$.

Notice that 3 | 111, 3 | 141, 3 | 171, and $11|121, 7| 161$, while 101, 131, 151, 181 are prime numbers.
Thus, the number of years with both properties $(a)$ and $(b)$ is 4.
Therefore, the answer is $(D)$.