In a six digit number, the sum of the digits in the even places is 9 and the sum of the digits in the odd places is 20. All such numbers are divisible by

Options

A. 7
B. 9
C. 6
D. 11

Answer

To solve the problem step by step, we will analyze the information given about the six-digit number and determine what it is divisible by.

### Step 1: Identify the positions of the digits
In a six-digit number, the digits are positioned as follows:
- 1st digit (odd place)
- 2nd digit (even place)
- 3rd digit (odd place)
- 4th digit (even place)
- 5th digit (odd place)
- 6th digit (even place)

### Step 2: Write down the sums given in the problem
According to the problem:
- The sum of the digits in the even places (2nd, 4th, and 6th) is 9.
- The sum of the digits in the odd places (1st, 3rd, and 5th) is 20.

### Step 3: Calculate the difference between the sums
Now, we can find the difference between the sum of the digits in the odd places and the sum of the digits in the even places:
- Difference = Sum of odd place digits - Sum of even place digits
- Difference = 20 - 9 = 11

### Step 4: Determine the divisibility
The difference we calculated (11) is significant because it indicates that any six-digit number that meets the conditions set in the problem will be divisible by this difference.

### Conclusion
Thus, all such six-digit numbers are divisible by **11**.

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