How many natural numbers are there between 1000 to 2000, which when divided by 341 leaves remainder 5?

Options

A. 3
B. 2
C. 4
D. 1

Answer

To find how many natural numbers between 1000 and 2000 leave a remainder of 5 when divided by 341, we can follow these steps:

### Step 1: Identify the range of multiples of 341
We need to find the multiples of 341 that fall within the range of 1000 to 2000.

### Step 2: Calculate the first multiple of 341 greater than 1000
To find the first multiple of 341 that is greater than 1000, we can divide 1000 by 341 and round up to the nearest whole number:
\[ \text{First multiple} = 341 \times \lceil \frac{1000}{341} \rceil \]
Calculating:
\[ \frac{1000}{341} \approx 2.93 \quad \Rightarrow \quad \lceil 2.93 \rceil = 3 \]
Thus,
\[ 341 \times 3 = 1023 \]

### Step 3: Calculate the next multiples of 341
Now we calculate the next multiples of 341:
- \(341 \times 4 = 1364\)
- \(341 \times 5 = 1705\)
- \(341 \times 6 = 2046\) (This is greater than 2000, so we stop here)

### Step 4: Add 5 to each multiple to find the required numbers
Since we need numbers that leave a remainder of 5 when divided by 341, we add 5 to each of the valid multiples:
- For \(1023\): \(1023 + 5 = 1028\)
- For \(1364\): \(1364 + 5 = 1369\)
- For \(1705\): \(1705 + 5 = 1710\)

### Step 5: List the valid numbers
The valid numbers between 1000 and 2000 that leave a remainder of 5 when divided by 341 are:
- 1028
- 1369
- 1710

### Step 6: Count the valid numbers
Counting these, we find there are a total of 3 natural numbers.

### Final Answer
Thus, the total number of natural numbers between 1000 and 2000 that leave a remainder of 5 when divided by 341 is **3**.

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