How many natural numbers between 1000 and 2000 (inclusive) leave a remainder of 5 when divided by 341?

Introduction / Context:
This question tests understanding of arithmetic progressions and modular arithmetic. We are asked to count how many numbers in a given range satisfy a specific remainder condition with respect to a divisor. This skill is frequently used in number theory and aptitude tests where direct listing of all numbers is not practical, and a pattern based approach is needed instead.

Given Data / Assumptions:

• We consider natural numbers between 1000 and 2000, including 1000 and 2000.
• The divisor is 341.
• We want numbers that give remainder 5 when divided by 341.
• All numbers are positive integers.

Concept / Approach:
If a number N leaves remainder 5 when divided by 341, it can be written as N = 341k + 5 for some integer k. So the set of such numbers forms an arithmetic progression with first term 5 and common difference 341. Our job is to find all terms of this form that lie between 1000 and 2000. That reduces to finding suitable integer values of k and counting how many such values exist.

Step-by-Step Solution:
Write N in the form N = 341k + 5.We need 1000 ≤ 341k + 5 ≤ 2000.Subtract 5 from all parts: 995 ≤ 341k ≤ 1995.Now divide the inequalities by 341.For the lower bound: 995 / 341 is slightly less than 3 because 341 * 3 = 1023.So k must be at least 3.For the upper bound: 1995 / 341 is slightly more than 5 because 341 * 5 = 1705 and 341 * 6 = 2046, which is greater than 1995.So k can be at most 5.Possible integer values of k are 3, 4, and 5.Compute the corresponding numbers: for k = 3, N = 341 * 3 + 5 = 1028.For k = 4, N = 341 * 4 + 5 = 1369.For k = 5, N = 341 * 5 + 5 = 1710.All three numbers lie between 1000 and 2000.

Verification / Alternative check:
Check remainders: 1028 divided by 341 gives remainder 5, 1369 divided by 341 gives remainder 5, and 1710 divided by 341 gives remainder 5.Next terms in the progression would be for k = 2 or k = 6. For k = 2, N = 687, which is less than 1000. For k = 6, N = 2051, which is greater than 2000.So only three numbers in the required interval satisfy the condition.

Why Other Options Are Wrong:
Options 2, 4, and 1 are incorrect because they do not match the count of valid k values found through the inequality approach.Any different count would imply either missing a valid term or including a term that lies outside the interval from 1000 to 2000.

Common Pitfalls:
Learners sometimes treat 1000 and 2000 as exclusive and mistakenly ignore boundary checks, but here they are inclusive.Another common mistake is calculating only one or two terms and guessing the answer without checking all integer values of k within the bounds.Errors in multiplication such as 341 * 3 or 341 * 5 can also lead to incorrect counts.

Final Answer:
The number of natural numbers between 1000 and 2000 that leave remainder 5 when divided by 341 is 3.