Let N be a natural number. If its first digit (from the left) deleted, it gets reduced to `(N)/(29)`. The sum of all the digits of N is

Options

A. 14
B. 17
C. 23
D. 29

Explanation

To solve the problem, we need to find a natural number \( N \) such that when its first digit is deleted, the resulting number is equal to \( \frac{N}{29} \). We will denote the digits of \( N \) as \( a_n, a_{n-1}, \ldots, a_1, a_0 \), where \( a_n \) is the first digit.

### Step 1: Express \( N \) in terms of its digits
The number \( N \) can be expressed as:
\[ N = 10^n \cdot a_n + 10^{n-1} \cdot a_{n-1} + 10^{n-2} \cdot a_{n-2} + \ldots + 10^1 \cdot a_1 + a_0 \]

### Step 2: Express the number after deleting the first digit
When we delete the first digit \( a_n \), the resulting number becomes:
\[ M = 10^{n-1} \cdot a_{n-1} + 10^{n-2} \cdot a_{n-2} + \ldots + 10^1 \cdot a_1 + a_0 \]

### Step 3: Set up the equation based on the problem statement
According to the problem, we have:
\[ M = \frac{N}{29} \]
Substituting the expressions for \( N \) and \( M \):
\[ 10^{n-1} \cdot a_{n-1} + 10^{n-2} \cdot a_{n-2} + \ldots + 10^1 \cdot a_1 + a_0 = \frac{1}{29} \left( 10^n \cdot a_n + 10^{n-1} \cdot a_{n-1} + 10^{n-2} \cdot a_{n-2} + \ldots + 10^1 \cdot a_1 + a_0 \right) \]

### Step 4: Multiply both sides by 29
Multiplying both sides by 29 gives:
\[ 29 \left( 10^{n-1} \cdot a_{n-1} + 10^{n-2} \cdot a_{n-2} + \ldots + 10^1 \cdot a_1 + a_0 \right) = 10^n \cdot a_n + 10^{n-1} \cdot a_{n-1} + 10^{n-2} \cdot a_{n-2} + \ldots + 10^1 \cdot a_1 + a_0 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 10^n \cdot a_n = 29 \cdot a_0 + (29 - 1) \cdot (10^{n-1} \cdot a_{n-1} + 10^{n-2} \cdot a_{n-2} + \ldots + 10^1 \cdot a_1) \]
This simplifies to:
\[ 10^n \cdot a_n = 29 \cdot a_0 + 28 \cdot M \]

### Step 6: Analyze the first digit
Since \( a_n \) is the first digit of \( N \), it must be between 1 and 9. We can assume \( a_n = 7 \) (as indicated in the video transcript) since it is a common digit that satisfies the equation.

### Step 7: Substitute \( a_n = 7 \)
Substituting \( a_n = 7 \):
\[ 10^n \cdot 7 = 29 \cdot a_0 + 28 \cdot M \]

### Step 8: Solve for \( n = 2 \) and \( n = 3 \)
1. For \( n = 2 \):
- We find \( a_0 + 10a_1 = 25 \) (from the video).
- Possible values are \( a_0 = 5 \) and \( a_1 = 2 \).
- Thus, \( N = 725 \).
- Sum of digits = \( 7 + 2 + 5 = 14 \).

2. For \( n = 3 \):
- We find \( a_0 + 10a_1 + 100a_2 = 250 \).
- Possible values are \( a_0 = 0, a_1 = 5, a_2 = 2 \).
- Thus, \( N = 7250 \).
- Sum of digits = \( 7 + 2 + 5 + 0 = 14 \).

### Conclusion
In both cases, the sum of the digits of \( N \) is:
\[ \text{Sum of digits} = 14 \]