The sum of any seven consecutive whole numbers is always divisible by-

Options

A. 2
B. 3
C. 7
D. 11

Answer

To find out what the sum of any seven consecutive whole numbers is always divisible by, we can follow these steps:

### Step 1: Define the consecutive whole numbers
Let the first whole number be \( x \). The seven consecutive whole numbers can then be expressed as:
- \( x, x+1, x+2, x+3, x+4, x+5, x+6 \)

### Step 2: Write the expression for their sum
Now, we can write the sum of these seven numbers:
\[ \text{Sum} = x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) \]

### Step 3: Simplify the sum
Combining like terms, we get:
\[ \text{Sum} = 7x + (0 + 1 + 2 + 3 + 4 + 5 + 6) \]
Calculating the sum of the integers from 0 to 6:
\[ 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21 \]
Thus, the sum can be rewritten as:
\[ \text{Sum} = 7x + 21 \]

### Step 4: Factor the sum
We can factor out a 7 from the expression:
\[ \text{Sum} = 7(x + 3) \]

### Step 5: Conclusion about divisibility
Since the sum \( 7(x + 3) \) is clearly a multiple of 7, we conclude that the sum of any seven consecutive whole numbers is always divisible by 7.

### Final Answer
The sum of any seven consecutive whole numbers is always divisible by **7**.

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