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The Boulevard Hotel has a peculiar way of numbering its rooms. The first room on each floor is numbered as the product of all floor number below it starting from floor number on ( the ground floor is not counted). For example , the first room on the fourth floor would be numbered as `1 xx 2 xx 3 = 6`. All successive room numbers would be numbered three more than the previous room number. Thus , on the fourth floor, rooms are numbered as `6,6+3 = 9,6 + 2 xx 3 = 12, 6+3 xx 3 = 15` etc. If it is known that there are six rooms on each floor, how many numbers on the fifteenth floor are prime numbers ?

Options

A. 0
B. 2
C. 4
D. 5

Answer

To solve the problem, we need to determine the room numbers on the 15th floor of the Boulevard Hotel and check how many of those numbers are prime.

### Step 1: Determine the first room number on the 15th floor.
The first room number on any floor is given by the product of all floor numbers below it. For the 15th floor, we need to calculate the product of the floor numbers from 1 to 14.

\[ \text{First room number on 15th floor} = 1 \times 2 \times 3 \times \ldots \times 14 = 14! \]

### Step 2: Calculate the room numbers on the 15th floor.
The room numbers on the 15th floor are derived from the first room number, with each successive room number increasing by 3. Therefore, the room numbers are:

1. First room: \( 14! \)
2. Second room: \( 14! + 3 \)
3. Third room: \( 14! + 6 \)
4. Fourth room: \( 14! + 9 \)
5. Fifth room: \( 14! + 12 \)
6. Sixth room: \( 14! + 15 \)

So, the room numbers are:
- \( 14! \)
- \( 14! + 3 \)
- \( 14! + 6 \)
- \( 14! + 9 \)
- \( 14! + 12 \)
- \( 14! + 15 \)

### Step 3: Check if each room number is prime.
A prime number is defined as a number greater than 1 that has no positive divisors other than 1 and itself. We will check each room number:

1. **Room 1: \( 14! \)**
- \( 14! \) is the product of all integers from 1 to 14, hence it has multiple factors (1, 2, 3, ..., 14). Therefore, it is **not prime**.

2. **Room 2: \( 14! + 3 \)**
- Since \( 14! \) is even (as it includes the factor 2), \( 14! + 3 \) is odd. However, \( 14! \) is divisible by 3 (as it includes the factor 3), thus \( 14! + 3 \) is also divisible by 3. Therefore, it is **not prime**.

3. **Room 3: \( 14! + 6 \)**
- \( 14! + 6 \) can be expressed as \( 14! + 2 \times 3 \). Since \( 14! \) is even, \( 14! + 6 \) is even and greater than 2, thus it is **not prime**.

4. **Room 4: \( 14! + 9 \)**
- \( 14! + 9 \) can be expressed as \( 14! + 3 \times 3 \). Like before, \( 14! \) is divisible by 3, hence \( 14! + 9 \) is also divisible by 3. Therefore, it is **not prime**.

5. **Room 5: \( 14! + 12 \)**
- \( 14! + 12 \) can be expressed as \( 14! + 3 \times 4 \). Again, since \( 14! \) is divisible by 3, \( 14! + 12 \) is also divisible by 3. Therefore, it is **not prime**.

6. **Room 6: \( 14! + 15 \)**
- \( 14! + 15 \) can be expressed as \( 14! + 3 \times 5 \). Since \( 14! \) is divisible by 3, \( 14! + 15 \) is also divisible by 3. Therefore, it is **not prime**.

### Conclusion:
After checking all six room numbers on the 15th floor, we find that none of them are prime numbers.

Thus, the answer to the question is:
**0 prime numbers on the 15th floor.**

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