Consider the sets T = {n, n+1, n+2, n +3, n+4}, where n=1, 2, 3, ...............,96. How many of these sets contains 6 or any integral multiple thereof (i e., any one of the numbers 6, 12, 18.............)?

Options

A. 80
B. 81
C. 82
D. 83

Answer

To solve the problem, we need to determine how many sets \( T = \{ n, n+1, n+2, n+3, n+4 \} \) contain the number 6 or any integral multiple of 6 (i.e., 6, 12, 18, etc.) for values of \( n \) ranging from 1 to 96.

### Step-by-Step Solution:

1. **Identify the Range of \( n \)**:
The values of \( n \) can be from 1 to 96. Therefore, we have:
\[ n = 1, 2, 3, \ldots, 96 \]

2. **Determine the Condition for Inclusion of 6**:
The set \( T \) will contain the number 6 if:
\[ n \leq 6 \quad \text{and} \quad n + 4 \geq 6 \]
This means \( n \) must be in the range where 6 is included in the set \( T \).

3. **Finding the Range for \( n \)**:
The above condition can be rewritten as:
\[ n \leq 6 \quad \text{and} \quad n \geq 2 \]
Thus, the possible values of \( n \) that satisfy this condition are:
\[ n = 2, 3, 4, 5, 6 \]
This gives us 5 values of \( n \).

4. **Determine the Condition for Inclusion of Multiples of 6**:
Next, we need to consider the multiples of 6. The multiples of 6 up to 96 are:
\[ 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96 \]
This gives us 16 multiples of 6.

5. **Finding the Range for Each Multiple**:
For each multiple \( m \) of 6, we need to find \( n \) such that:
\[ m - 4 \leq n \leq m \]
This ensures that \( m \) is included in the set \( T \).

6. **Calculate Valid \( n \) for Each Multiple**:
- For \( m = 6 \): \( n \) can be \( 2, 3, 4, 5, 6 \) (5 values)
- For \( m = 12 \): \( n \) can be \( 8, 9, 10, 11, 12 \) (5 values)
- For \( m = 18 \): \( n \) can be \( 14, 15, 16, 17, 18 \) (5 values)
- For \( m = 24 \): \( n \) can be \( 20, 21, 22, 23, 24 \) (5 values)
- For \( m = 30 \): \( n \) can be \( 26, 27, 28, 29, 30 \) (5 values)
- For \( m = 36 \): \( n \) can be \( 32, 33, 34, 35, 36 \) (5 values)
- For \( m = 42 \): \( n \) can be \( 38, 39, 40, 41, 42 \) (5 values)
- For \( m = 48 \): \( n \) can be \( 44, 45, 46, 47, 48 \) (5 values)
- For \( m = 54 \): \( n \) can be \( 50, 51, 52, 53, 54 \) (5 values)
- For \( m = 60 \): \( n \) can be \( 56, 57, 58, 59, 60 \) (5 values)
- For \( m = 66 \): \( n \) can be \( 62, 63, 64, 65, 66 \) (5 values)
- For \( m = 72 \): \( n \) can be \( 68, 69, 70, 71, 72 \) (5 values)
- For \( m = 78 \): \( n \) can be \( 74, 75, 76, 77, 78 \) (5 values)
- For \( m = 84 \): \( n \) can be \( 80, 81, 82, 83, 84 \) (5 values)
- For \( m = 90 \): \( n \) can be \( 86, 87, 88, 89, 90 \) (5 values)
- For \( m = 96 \): \( n \) can be \( 92, 93, 94, 95, 96 \) (5 values)

7. **Total Count of Valid Sets**:
Since there are 16 multiples of 6, and for each multiple, there are 5 corresponding values of \( n \):
\[ \text{Total sets} = 5 \times 16 = 80 \]

### Final Answer:
Thus, the total number of sets \( T \) that contain 6 or any integral multiple thereof is **80**.