For each `n in N`, the greatest positive integer which divides `(n + 1) (n + 2) (n + 3) `is :

Options

A. 2
B. 6
C. 24
D. 120

Answer

To find the greatest positive integer that divides \((n + 1)(n + 2)(n + 3)\) for each \(n \in \mathbb{N}\), we can follow these steps:

### Step 1: Expand the expression
We start with the expression \((n + 1)(n + 2)(n + 3)\).
\[ (n + 1)(n + 2)(n + 3) = (n + 1)((n + 2)(n + 3)) \]
Now, we can expand \((n + 2)(n + 3)\):
\[ (n + 2)(n + 3) = n^2 + 5n + 6 \]
So, we have:
\[ (n + 1)(n^2 + 5n + 6) = n^3 + 6n^2 + 11n + 6 \]

### Step 2: Identify the divisibility
We need to find the greatest positive integer that divides this expression for all natural numbers \(n\).

### Step 3: Evaluate for specific values of \(n\)
Let's evaluate the expression for the first few natural numbers:

- For \(n = 1\):
\[ (1 + 1)(1 + 2)(1 + 3) = 2 \cdot 3 \cdot 4 = 24 \]

- For \(n = 2\):
\[ (2 + 1)(2 + 2)(2 + 3) = 3 \cdot 4 \cdot 5 = 60 \]

- For \(n = 3\):
\[ (3 + 1)(3 + 2)(3 + 3) = 4 \cdot 5 \cdot 6 = 120 \]

### Step 4: Find the GCD
Now we need to find the greatest common divisor (GCD) of these results:
- GCD of \(24\), \(60\), and \(120\).

Calculating the GCD:
- The prime factorization of \(24 = 2^3 \cdot 3\)
- The prime factorization of \(60 = 2^2 \cdot 3 \cdot 5\)
- The prime factorization of \(120 = 2^3 \cdot 3 \cdot 5\)

The GCD is determined by taking the lowest power of each prime:
- For \(2\), the minimum power is \(2^2\).
- For \(3\), the minimum power is \(3^1\).
- For \(5\), it does not appear in all three numbers.

Thus, the GCD is:
\[ 2^2 \cdot 3^1 = 4 \cdot 3 = 12 \]

### Conclusion
The greatest positive integer that divides \((n + 1)(n + 2)(n + 3)\) for all \(n \in \mathbb{N}\) is \(12\).