The function `Delta (m)` is defined for all positive integers m as the product of `m + 4, m + 5 and m + 6`. If m is a positive integer, then `Delta(n)` must be divisible by which of the following numbers?

Options

A. 2
B. 3
C. 6
D. 7

Answer

To solve the problem, we need to analyze the function \( \Delta(m) \) defined as the product of three consecutive integers:

\[ \Delta(m) = (m + 4)(m + 5)(m + 6) \]

We need to determine by which numbers this product is guaranteed to be divisible.

### Step 1: Understand the function
The function \( \Delta(m) \) consists of three consecutive integers: \( m + 4 \), \( m + 5 \), and \( m + 6 \).

### Step 2: Identify divisibility by 3
Among any three consecutive integers, at least one of them is always divisible by 3. Therefore, \( \Delta(m) \) is divisible by 3.

### Step 3: Identify divisibility by 2
Among three consecutive integers, at least one of them is also always divisible by 2. Therefore, \( \Delta(m) \) is divisible by 2.

### Step 4: Combine the results
Since \( \Delta(m) \) is divisible by both 2 and 3, we can conclude that it is divisible by their product:

\[ 2 \times 3 = 6 \]

### Step 5: Conclusion
Thus, \( \Delta(m) \) must be divisible by 6.

### Final Answer
The function \( \Delta(m) \) is divisible by **6**.

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