What is the sum of all natural numbers between 100 and 400 (inclusive) that are divisible by 13?

Introduction / Context:
This problem involves arithmetic progressions and divisibility. We are asked to find the sum of all natural numbers in a given range that are multiples of 13. Instead of listing each number and adding them individually, we treat the set of multiples as an arithmetic progression and use the formula for the sum of an arithmetic series. This approach is standard in aptitude tests and helps develop efficient calculation skills.

Given Data / Assumptions:

• We consider natural numbers from 100 to 400 inclusive.
• We are interested only in those numbers that are divisible by 13.
• We must compute the sum of all such numbers.

Concept / Approach:
First, find the smallest and largest multiples of 13 within the interval from 100 to 400. Once we know the first term, last term, and common difference, we can determine how many terms are in this arithmetic progression. Then we use the sum formula for an arithmetic progression: S = n / 2 * (first term + last term), where n is the number of terms.

Step-by-Step Solution:
Find the smallest multiple of 13 that is at least 100.Compute 100 / 13 ≈ 7.69, so the next integer multiple is for k = 8.First multiple = 13 * 8 = 104.Find the largest multiple of 13 that is at most 400.Compute 400 / 13 ≈ 30.76, so the largest integer k is 30.Last multiple = 13 * 30 = 390.So the sequence of multiples is 104, 117, 130, ..., 390 with common difference 13.Number of terms n is given by n = ((last - first) / difference) + 1.Compute n = ((390 - 104) / 13) + 1 = (286 / 13) + 1 = 22 + 1 = 23.Now use the sum formula S = n / 2 * (first + last).S = 23 / 2 * (104 + 390) = 23 / 2 * 494.Compute 494 / 2 = 247.Therefore S = 23 * 247 = 5681.

Verification / Alternative check:
We can quickly verify by computing 13 times the sum of the integers from 8 to 30.Sum of integers 8 to 30 is equal to sum from 1 to 30 minus sum from 1 to 7.Sum from 1 to 30 = 30 * 31 / 2 = 465.Sum from 1 to 7 = 7 * 8 / 2 = 28.So sum from 8 to 30 = 465 - 28 = 437.The sum of multiples is 13 * 437 = 5681, which matches the earlier calculation.

Why Other Options Are Wrong:
5334, 5434, and 5761 do not equal the sum produced by the arithmetic progression method.They may arise from incorrect counts of terms or from using the wrong first or last multiple within the interval.

Common Pitfalls:
Some learners mistakenly use 91 or 403 as the first or last multiple, which fall outside the interval 100 to 400.Another error is miscalculating n, the number of terms, leading to a wrong sum.Incorrect application of the sum formula, such as using only first plus last without multiplying by n / 2, can also cause mistakes.

Final Answer:
The sum of all natural numbers between 100 and 400 that are divisible by 13 is 5681.