Two different 2-digit positive integers are called a reversal pair if the position of the digits in the first integer is switched in the second integer. For example, 52 and 25 are a reversal pair. The integer 2015 has the property that it is equal to the product of three different prime numbers, two of which are a reversal pair. Including 2015, how many positive integers less than 10000 have this same property?
(A) 18
(B) 14
(C) 20
(D) 17
(E) 19
All 2-digit prime numbers are odd numbers, so to create a reversal pair, both digits of each prime must be odd (so that both the original number and its reversal are odd numbers).

We also note that the digit 5 cannot appear in either prime number of the reversal pair since any 2-digit number ending in 5 is not prime.

Combining these two facts together leaves only the following list of prime numbers from which to search for reversal pairs: $11,13,17,19,31,37,71,73,79$, and 97 .

This allows us to determine that the only reversal pairs are: 13 and 31, 17 and 71 , 37 and 73 , and 79 and 97 .

(Note that the reversal of 11 does not produce a different prime number and the reversal of 19 is 91 , which is not prime since $7 \times 13=91$.)

Given a reversal pair, we must determine the prime numbers (different than each prime of the reversal pair) whose product with the reversal pair is a positive integer less than 10000.

The product of the reversal pair 79 and 97 is $79 \times 97=7663$.

Since the smallest prime number is 2 and $2 \times 7663=15326$, which is greater than 10000 , then the reversal pair 79 and 97 gives no possibilities that satisfy the given conditions.

We continue in this way, analyzing the other 3 reversal pairs, and summarize our results in the table below.

| Prime | Product of the Prime Number with the Reversal Pair |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: |
|  | 13 and 31 | 17 and 71 | 37 and 73 | 79 and 97 |  |
| 2 | $2 \times 13 \times 31=806$ | $2 \times 17 \times 71=2414$ | $2 \times 37 \times 73=5402$ | greater than 10000 |  |
| 3 | $3 \times 13 \times 31=1209$ | $3 \times 17 \times 71=3621$ | $3 \times 37 \times 73=8103$ |  |  |
| 5 | $5 \times 13 \times 31=2015$ | $5 \times 17 \times 71=6035$ | greater than 10000 |  |  |
| 7 | $7 \times 13 \times 31=2821$ | $7 \times 17 \times 71=8449$ |  |  |  |
| 11 | $11 \times 13 \times 31=4433$ | greater than 10000 |  |  |  |
| 13 | can't use 13 twice |  |  |  |  |
| 17 | $17 \times 13 \times 31=6851$ |  |  |  |  |
| 19 | $19 \times 13 \times 31=7657$ |  |  | 0 |  |
| 23 | $23 \times 13 \times 31=9269$ |  |  |  |  |
| 29 | greater than 10000 |  | 2 |  |  |
| Total | 8 | 4 |  |  |  |

In any column, once we obtain a product that is greater than 10000 , we may stop evaluating subsequent products since they use a larger prime number and thus will exceed the previous product.

In total, there are $8+4+2=14$ positive integers less than 10000 which have the required property.