By how much is the greatest of five consecutive even integers greater than the smallest among them ?

Options

A. 1
B. 2
C. 4
D. 8

Answer

To solve the problem of finding how much greater the greatest of five consecutive even integers is than the smallest among them, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Consecutive Even Integers**:
Let's denote the smallest even integer as \( x \). The five consecutive even integers can then be represented as:
- \( x \) (smallest)
- \( x + 2 \)
- \( x + 4 \)
- \( x + 6 \)
- \( x + 8 \) (greatest)

2. **Identify the Greatest and Smallest Integers**:
From our representation:
- The smallest integer is \( x \).
- The greatest integer is \( x + 8 \).

3. **Calculate the Difference**:
To find out how much greater the greatest integer is than the smallest, we calculate the difference:
\[ \text{Difference} = (\text{Greatest Integer}) - (\text{Smallest Integer}) = (x + 8) - x \]

4. **Simplify the Expression**:
Simplifying the expression gives us:
\[ \text{Difference} = x + 8 - x = 8 \]

5. **Conclusion**:
Therefore, the greatest of the five consecutive even integers is 8 greater than the smallest among them.

### Final Answer:
The greatest of the five consecutive even integers is 8 greater than the smallest.