HCF of 3!, 4! and 5! is k!, then k =

Options

A. 3
B. 4
C. 6
D. 60

Answer

To find the value of \( k \) such that the HCF of \( 3! \), \( 4! \), and \( 5! \) is \( k! \), we can follow these steps:

### Step 1: Calculate the factorials
First, we need to calculate the factorials of 3, 4, and 5.

- \( 3! = 3 \times 2 \times 1 = 6 \)
- \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
- \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

### Step 2: List the values
Now we have:
- \( 3! = 6 \)
- \( 4! = 24 \)
- \( 5! = 120 \)

### Step 3: Find the HCF
Next, we need to find the HCF of these three numbers: 6, 24, and 120.

- The factors of 6 are: \( 1, 2, 3, 6 \)
- The factors of 24 are: \( 1, 2, 3, 4, 6, 8, 12, 24 \)
- The factors of 120 are: \( 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 \)

The common factors are \( 1, 2, 3, 6 \). The greatest of these is \( 6 \).

### Step 4: Relate HCF to \( k! \)
We have found that the HCF of \( 3! \), \( 4! \), and \( 5! \) is \( 6 \). We can express this as:
\[ HCF(3!, 4!, 5!) = 6 = 3! \]
This means that \( k! = 3! \).

### Step 5: Solve for \( k \)
Since \( k! = 3! \), we can equate:
\[ k = 3 \]

Thus, the value of \( k \) is \( \boxed{3} \).

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