How many natural numbers are there which give a remainder of 41 after dividing 1997?

Options

A. 2
B. 4
C. 6
D. None of these

Answer

To solve the problem of how many natural numbers give a remainder of 41 when dividing 1997, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Division Concept**:
When a number \( d \) (in this case, 1997) is divided by a natural number \( n \), it can be expressed as:
\[ d = q \cdot n + r \]
where \( q \) is the quotient, \( n \) is the divisor, and \( r \) is the remainder. Here, we know \( r = 41 \).

2. **Set Up the Equation**:
From the equation, we can rearrange it to find \( n \):
\[ 1997 = q \cdot n + 41 \]
Rearranging gives:
\[ q \cdot n = 1997 - 41 \]
Thus,
\[ q \cdot n = 1956 \]

3. **Identify the Divisor**:
The natural number \( n \) must be a divisor of 1956. Therefore, we need to find the factors of 1956.

4. **Factor 1956**:
We can factor 1956 to find its divisors:
- Start dividing by the smallest prime number, which is 2:
\[ 1956 \div 2 = 978 \]
- Divide again by 2:
\[ 978 \div 2 = 489 \]
- Now divide by 3:
\[ 489 \div 3 = 163 \]
- 163 is a prime number.

Thus, the prime factorization of 1956 is:
\[ 1956 = 2^2 \cdot 3^1 \cdot 163^1 \]

5. **Calculate the Total Number of Divisors**:
To find the total number of divisors, we use the formula:
\[ (e_1 + 1)(e_2 + 1)(e_3 + 1) \]
where \( e_1, e_2, e_3 \) are the powers of the prime factors.
- For \( 2^2 \): \( e_1 = 2 \) → contributes \( 2 + 1 = 3 \)
- For \( 3^1 \): \( e_2 = 1 \) → contributes \( 1 + 1 = 2 \)
- For \( 163^1 \): \( e_3 = 1 \) → contributes \( 1 + 1 = 2 \)

Therefore, the total number of divisors is:
\[ 3 \cdot 2 \cdot 2 = 12 \]

6. **Filter Divisors Greater than 41**:
Since we need natural numbers that give a remainder of 41, the divisor \( n \) must be greater than 41. The divisors of 1956 are:
- 1, 2, 3, 4, 6, 12, 163, 326, 489, 652, 978, 1956

From this list, the divisors greater than 41 are:
- 163, 326, 489, 652, 978, 1956

This gives us a total of 6 valid divisors.

### Final Answer:
Thus, the total number of natural numbers that give a remainder of 41 when dividing 1997 is **6**.