A person starts typing the numbers from 1 to 1999. He press the keys total 'n' number of times. The value of n is L

Options

A. 6889
B. 1000
C. 2888
D. None of these

Answer

To find the total number of key presses (n) when typing the numbers from 1 to 1999, we can break the problem down into steps based on the number of digits in the numbers.

### Step-by-step Solution:

1. **Identify the ranges of numbers**:
- 1-digit numbers: 1 to 9
- 2-digit numbers: 10 to 99
- 3-digit numbers: 100 to 999
- 4-digit numbers: 1000 to 1999

2. **Count the total numbers in each range**:
- **1-digit numbers**:
- From 1 to 9, there are \(9\) numbers.
- **2-digit numbers**:
- From 10 to 99, the count is \(99 - 10 + 1 = 90\) numbers.
- **3-digit numbers**:
- From 100 to 999, the count is \(999 - 100 + 1 = 900\) numbers.
- **4-digit numbers**:
- From 1000 to 1999, the count is \(1999 - 1000 + 1 = 1000\) numbers.

3. **Calculate the total digits pressed for each range**:
- **1-digit numbers**:
- Total key presses = \(9 \times 1 = 9\)
- **2-digit numbers**:
- Total key presses = \(90 \times 2 = 180\)
- **3-digit numbers**:
- Total key presses = \(900 \times 3 = 2700\)
- **4-digit numbers**:
- Total key presses = \(1000 \times 4 = 4000\)

4. **Sum the total key presses**:
- Total key presses = \(9 + 180 + 2700 + 4000\)
- Total key presses = \(6889\)

Thus, the value of \(n\) (the total number of key presses) is **6889**.