The product of a non-zero whole number and its successor is always divisible by

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the question "The product of a non-zero whole number and its successor is always divisible by?", we will analyze the problem step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find out what the product of a non-zero whole number (let's call it \( n \)) and its successor (which is \( n + 1 \)) is divisible by.

2. **Formulating the Product**: The product can be expressed mathematically as:
\[ P = n \times (n + 1) \]
Here, \( n \) is a non-zero whole number.

3. **Analyzing the Product**: Notice that \( n \) and \( n + 1 \) are two consecutive integers.

4. **Properties of Consecutive Integers**: One important property of consecutive integers is that one of them is always even. This means that the product \( P \) will always be even.

5. **Divisibility by 2**: Since the product \( P \) contains at least one even number, it is guaranteed that \( P \) is divisible by 2.

6. **Checking Other Options**:
- **Divisibility by 3**: Not all products of two consecutive integers are divisible by 3. For example, if \( n = 1 \) (then \( P = 1 \times 2 = 2 \)), it is not divisible by 3.
- **Divisibility by 4**: Similarly, not all products are divisible by 4. For example, if \( n = 1 \) (then \( P = 1 \times 2 = 2 \)), it is not divisible by 4.
- **Divisibility by 5**: Again, not all products are divisible by 5. For example, if \( n = 1 \) (then \( P = 1 \times 2 = 2 \)), it is not divisible by 5.

7. **Conclusion**: The only option that is always true for the product of a non-zero whole number and its successor is that it is divisible by 2.

### Final Answer:
The product of a non-zero whole number and its successor is always divisible by **2**.