If 604...6 is divisible by 11, then the integer in the blank space is

Options

A. 1
B. 3
C. 7
D. 5

Answer

To determine the integer in the blank space of the number 604...6 such that the entire number is divisible by 11, we will use the divisibility rule for 11.

### Step-by-step Solution:

1. **Identify the digits**: The number is represented as 604x6, where 'x' is the digit in the blank space.

2. **Separate the digits into odd and even positions**:
- Odd position digits: 6 (1st), 4 (3rd), 6 (5th) → Sum of odd position digits = 6 + 4 + 6 = 16
- Even position digits: 0 (2nd), x (4th) → Sum of even position digits = 0 + x = x

3. **Apply the divisibility rule for 11**:
- The rule states that a number is divisible by 11 if the absolute difference between the sum of the odd position digits and the sum of the even position digits is either 0 or divisible by 11.
- Therefore, we need to calculate |Sum of odd position digits - Sum of even position digits| = |16 - x|

4. **Set up the condition for divisibility**:
- We need |16 - x| to be 0 or divisible by 11. This gives us two cases:
- Case 1: 16 - x = 0 → x = 16 (not possible since x must be a single digit)
- Case 2: 16 - x = 11 → x = 5
- Case 3: 16 - x = -11 → x = 27 (not possible since x must be a single digit)

5. **Conclusion**: The only valid solution for x that keeps it as a single digit is x = 5.

Thus, the integer in the blank space is **5**.