In this article, we'll dive into the mathematical problem of finding consecutive natural numbers whose sum equals 243. We'll explore the process step-by-step and provide a detailed solution.

Introduction to Consecutive Natural Number Sums

Consecutive natural numbers are numbers that follow each other in order, such as 1, 2, 3, 4, and so on. The challenge here is to find a sequence of these numbers that add up to a specific target, in this case, 243.

Formulating the Problem

Let's denote the first number in the sequence as $n$ and the length of the sequence as $k$. The sum of $k$ consecutive natural numbers starting from $n$ can be expressed as:

Sum of k consecutive natural numbers starting from n.

To find the sum equal to 243, we set up the equation:

$knspace;plus;space;frac{k(kspace;-space;1)}{2}space;space;243$

Multiplying the entire equation by 2 to eliminate the fraction, we get:

$2knspace;plus;space;k(kspace;-space;1)space;space;486$

Further simplification gives:

$k^2space;-space;kspace;plus;space;2knspace;space;486$

This can be rearranged to:

$k(2nspace;plus;space;kspace;-space;1)space;space;486$

Factoring 486

Next, we need to factor 486 to find suitable values of $k$, which are its divisors. The factors of 486 are:

1 2 3 6 9 18 27 54 81 162 243 486

We will then solve for $n$ for each factor of 486 that provides a natural number sequence.

Solving for Each k

For $k 1$

$1(2nspace;plus;space;1space;-space;1)space;space;486$ implies $2nspace;space;486$ implies $nspace;space;243$.

The sequence is 243.

For $k 2$

$2(2nspace;plus;space;2space;-space;1)space;space;486$ implies $22nspace; space;1space;space;486$ implies $22nspace;space;243$ implies $nspace;space;121$:

The sequence is 121, 122.

For $k 3$

$3(2nspace;plus;space;3space;-space;1)space;space;486$ implies $3(2nspace;plus;space;2)space;space;486$ implies $2nspace;plus;space;2space;space;162$ implies $2nspace;space;160$ implies $nspace;space;80$.

The sequence is 80, 81, 82.

For $k 6$

$6(2nspace;plus;space;6space;-space;1)space;space;486$ implies $6(2nspace;plus;space;5)space;space;486$ implies $2nspace;plus;space;5space;space;81$ implies $2nspace;space;76$ implies $nspace;space;38$.

The sequence is 38, 39, 40, 41, 42, 43.

For $k 9$

$9(2nspace;plus;space;9space;-space;1)space;space;486$ implies $9(2nspace;plus;space;8)space;space;486$ implies $2nspace;plus;space;8space;space;54$ implies $2nspace;space;46$ implies $nspace;space;23$.

The sequence is 23, 24, 25, 26, 27, 28, 29, 30, 31.

For $k 18$

$18(2nspace;plus;space;18space;-space;1)space;space;486$ implies $18(2nspace;plus;space;17)space;space;486$ implies $2nspace;plus;space;17space;space;27$ implies $2nspace;space;10$ implies $nspace;space;5$.

The sequence is 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.

For $k 27$, we find that $n$ is not a natural number, so it is not a valid solution.

The Final Answer

The smallest sequence of consecutive natural numbers that sums to 243 is:

$5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20$, starting from $n 5$ and ending at $20$.

Therefore, the consecutive natural numbers that sum to 243 start from $5$ and include up to $20$.