The largest natural number by which the product of three consecutive even natural number is always divisible is -----

Options

A. A)16
B. B)24
C. C)48
D. D)96

Answer

To find the largest natural number by which the product of three consecutive even natural numbers is always divisible, we can follow these steps:

### Step 1: Define the three consecutive even natural numbers
Let the three consecutive even natural numbers be represented as:
- \( n \) (the first even number)
- \( n + 2 \) (the second even number)
- \( n + 4 \) (the third even number)

### Step 2: Write the product of these numbers
The product of these three consecutive even natural numbers can be expressed as:
\[ P = n \times (n + 2) \times (n + 4) \]

### Step 3: Factor out the common terms
Since \( n \) is even, we can express it as \( n = 2k \) for some integer \( k \). Thus, the three consecutive even numbers can be rewritten as:
- \( 2k \)
- \( 2k + 2 = 2(k + 1) \)
- \( 2k + 4 = 2(k + 2) \)

Now, the product becomes:
\[ P = 2k \times 2(k + 1) \times 2(k + 2) = 2^3 \times k \times (k + 1) \times (k + 2) \]
This simplifies to:
\[ P = 8k(k + 1)(k + 2) \]

### Step 4: Analyze the factors
The expression \( k(k + 1)(k + 2) \) represents the product of three consecutive integers. The product of any three consecutive integers is always divisible by \( 3! = 6 \). Therefore, we have:
\[ P = 8 \times 6 = 48 \]

### Step 5: Conclusion
The largest natural number by which the product of three consecutive even natural numbers is always divisible is:
\[ \boxed{48} \]