A number of 109 digits is written as follows:
`" 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 ….."`
What is the least possible a positive number is required to add up in the given number to make it divisible by 2 and 5 both ?

Options

A. 0
B. 1
C. 3
D. can't be determined

Answer

To solve the problem, we need to determine the least positive number that can be added to a sequence of numbers from 1 to 59 (which gives us a total of 109 digits) to make the resulting number divisible by both 2 and 5.

### Step-by-Step Solution:

1. **Understanding the Sequence**:
The sequence starts from 1 and goes up to 59. The digits from 1 to 9 contribute 9 digits, and the digits from 10 to 59 contribute 100 digits (since each of these numbers has 2 digits). Therefore, the total number of digits is 109.

2. **Divisibility Conditions**:
To be divisible by both 2 and 5, a number must end in 0 (because 0 is the only digit that satisfies both conditions). Therefore, we need to ensure that the last digit of our number is 0.

3. **Finding the Last Digit**:
The last number in our sequence is 59. The last digit of 59 is 9. To make it end in 0, we need to add a certain number.

4. **Calculating the Required Addition**:
To change the last digit from 9 to 0, we can add 1. Thus:
\[ 59 + 1 = 60 \]
The number 60 ends in 0, making it divisible by both 2 and 5.

5. **Conclusion**:
The least positive number required to add to the sequence to make it divisible by both 2 and 5 is 1.

### Final Answer:
The least positive number required to add is **1**.

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