The sum of the 3 consecutive even numbers is always divisible by :

Options

A. 24
B. 48
C. 10
D. none of these

Answer

To solve the problem of finding out by which number the sum of three consecutive even numbers is always divisible, we can follow these steps:

### Step 1: Define the first even number
Let the first even number be represented as \(2n\), where \(n\) is an integer.

**Hint:** Remember that even numbers can be expressed in the form \(2n\), where \(n\) is any whole number.

### Step 2: Write the next two consecutive even numbers
The second consecutive even number will be \(2n + 2\), and the third consecutive even number will be \(2n + 4\).

**Hint:** Consecutive even numbers differ by 2.

### Step 3: Calculate the sum of the three consecutive even numbers
Now, we can find the sum of these three numbers:
\[ \text{Sum} = 2n + (2n + 2) + (2n + 4) \]
Combining like terms, we get:
\[ \text{Sum} = 2n + 2n + 2 + 2n + 4 = 6n + 6 \]

**Hint:** When adding, make sure to combine all like terms correctly.

### Step 4: Factor the sum
We can factor out the common term from the sum:
\[ \text{Sum} = 6(n + 1) \]

**Hint:** Factoring helps to identify the common divisors of the expression.

### Step 5: Determine the divisibility
From the factored form \(6(n + 1)\), we can see that the sum is always divisible by 6, regardless of the value of \(n\).

**Hint:** Focus on the coefficients and factors to determine divisibility.

### Conclusion
Thus, the sum of three consecutive even numbers is always divisible by 6.

**Final Answer:** The sum of the three consecutive even numbers is always divisible by **6**.