Let p be a prime number such that `3 lt p lt 50`, then `p^2-1` is:

Options

A. always divisible by 8
B. always divisible by 24
C. always divisible by 12
D. all of a, b, c

Answer

To solve the problem, we need to find the value of \( p^2 - 1 \) where \( p \) is a prime number such that \( 3 < p < 50 \).

### Step-by-Step Solution:

1. **Identify the Range of Prime Numbers**:
We need to find all prime numbers \( p \) such that \( 3 < p < 50 \). The prime numbers in this range are:
- 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

2. **Calculate \( p^2 - 1 \) for Each Prime**:
We will calculate \( p^2 - 1 \) for each of the identified prime numbers:
- For \( p = 5 \):
\[ p^2 - 1 = 5^2 - 1 = 25 - 1 = 24 \]
- For \( p = 7 \):
\[ p^2 - 1 = 7^2 - 1 = 49 - 1 = 48 \]
- For \( p = 11 \):
\[ p^2 - 1 = 11^2 - 1 = 121 - 1 = 120 \]
- For \( p = 13 \):
\[ p^2 - 1 = 13^2 - 1 = 169 - 1 = 168 \]
- For \( p = 17 \):
\[ p^2 - 1 = 17^2 - 1 = 289 - 1 = 288 \]
- For \( p = 19 \):
\[ p^2 - 1 = 19^2 - 1 = 361 - 1 = 360 \]
- For \( p = 23 \):
\[ p^2 - 1 = 23^2 - 1 = 529 - 1 = 528 \]
- For \( p = 29 \):
\[ p^2 - 1 = 29^2 - 1 = 841 - 1 = 840 \]
- For \( p = 31 \):
\[ p^2 - 1 = 31^2 - 1 = 961 - 1 = 960 \]
- For \( p = 37 \):
\[ p^2 - 1 = 37^2 - 1 = 1369 - 1 = 1368 \]
- For \( p = 41 \):
\[ p^2 - 1 = 41^2 - 1 = 1681 - 1 = 1680 \]
- For \( p = 43 \):
\[ p^2 - 1 = 43^2 - 1 = 1849 - 1 = 1848 \]
- For \( p = 47 \):
\[ p^2 - 1 = 47^2 - 1 = 2209 - 1 = 2208 \]

3. **Identify Common Factors**:
Now we observe the results:
- \( 24, 48, 120, 168, 288, 360, 528, 840, 960, 1368, 1680, 1848, 2208 \)

We can see that all these numbers are divisible by 24.

4. **Conclusion**:
Therefore, \( p^2 - 1 \) for any prime number \( p \) in the specified range will always be divisible by 24.

### Final Answer:
The value of \( p^2 - 1 \) is always divisible by 24.