No. of digits while counting from 1 to 10,000 is N, then digits at unit place of N is

Explanation

To find the number of digits while counting from 1 to 10,000, we will break it down into segments based on the number of digits in each range.

### Step-by-Step Solution:

1. **Count digits from 1 to 9:**
- There are 9 numbers (1 to 9), and each has 1 digit.
- Total digits = \(9 \times 1 = 9\)

2. **Count digits from 10 to 99:**
- There are \(99 - 10 + 1 = 90\) numbers (10 to 99), and each has 2 digits.
- Total digits = \(90 \times 2 = 180\)

3. **Count digits from 100 to 999:**
- There are \(999 - 100 + 1 = 900\) numbers (100 to 999), and each has 3 digits.
- Total digits = \(900 \times 3 = 2700\)

4. **Count digits from 1000 to 9999:**
- There are \(9999 - 1000 + 1 = 9000\) numbers (1000 to 9999), and each has 4 digits.
- Total digits = \(9000 \times 4 = 36000\)

5. **Count digits for the number 10,000:**
- The number 10,000 has 5 digits.
- Total digits = \(5\)

6. **Add all the digits together:**
- Total digits \(N = 9 + 180 + 2700 + 36000 + 5\)
- \(N = 38894\)

7. **Find the unit place of N:**
- The unit place of \(N = 38894\) is \(4\).

### Final Answer:
The digit at the unit place of \(N\) is \(4\).

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