For any natural number n, if D (n) is the sum of all the digits of D(n) find the remainder when `sum_1^99` D(n) is divided by 99.

Options

A. 9
B. 0
C. 1
D. 89

Answer

To solve the problem of finding the remainder when the sum of the digits of all natural numbers from 1 to 99 is divided by 99, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding D(n)**:
- D(n) is defined as the sum of the digits of the number n. For example, D(23) = 2 + 3 = 5.

2. **Calculate D(n) for n from 1 to 99**:
- We need to calculate D(n) for each number from 1 to 99 and then sum these values.

3. **Breakdown the Calculation**:
- We can separate the numbers into two ranges: 1 to 9 and 10 to 99.
- For numbers 1 to 9, D(n) is simply n itself.
- For numbers 10 to 99, we can express each number as 10a + b, where a is the tens digit and b is the units digit.
- Thus, D(10a + b) = a + b.

4. **Sum of D(n) from 1 to 9**:
- The sum of D(n) for n = 1 to 9 is:
\[ D(1) + D(2) + D(3) + D(4) + D(5) + D(6) + D(7) + D(8) + D(9) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 \]

5. **Sum of D(n) from 10 to 99**:
- For numbers from 10 to 99, we can calculate the contribution from the tens and units digits separately.
- The tens digit (a) can take values from 1 to 9 (for each complete set of ten numbers).
- For each tens digit a (1 to 9), the units digit (b) can take values from 0 to 9.
- The contribution from the tens digits:
\[ (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) + (2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2) + ... + (9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9) = 10(1 + 2 + 3 + ... + 9) = 10 \times 45 = 450 \]
- The contribution from the units digits (0 to 9) for each tens digit:
\[ 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 \]
- Since there are 9 tens digits (1 to 9), the total contribution from the units digits is:
\[ 9 \times 45 = 405 \]

6. **Total Sum of D(n) from 1 to 99**:
- Now, we can sum the contributions:
\[ \text{Total} = 45 + 450 + 405 = 900 \]

7. **Finding the Remainder**:
- Finally, we need to find the remainder when 900 is divided by 99:
\[ 900 \div 99 = 9 \quad \text{(with a remainder of 9)} \]

### Final Result:
- The remainder when the sum \( \sum_{n=1}^{99} D(n) \) is divided by 99 is **9**.