Question

I am owed 3240 florins. The debtor pays me 1 florin the first day, 2 the second day, 3 the third day, and so on. How many days does it take to pay off the debt?

Short answer

Answer: It takes at least 80 days to pay off the debt.

Step-by-step solution

Understanding the problem

We want to find out how many days it takes for the debtor to pay off a debt of 3240 florins, given the payments are increasing by 1 florin every day (e.g., 1, 2, 3...).

Setup the sum of an arithmetic series

We can represent the total amount of florins paid after 'n' days as the sum of the arithmetic series: S = 1 + 2 + 3 + ... + n S = (n * (n + 1)) / 2 We know that the total sum S needs to be at least 3240 florins, therefore: ( n * (n + 1) ) / 2 ≥ 3240

Solve the inequality for n

We now have to solve the inequality for n: ( n * (n + 1) ) / 2 ≥ 3240 Multiplying both sides by 2 to remove the fraction yields n * (n + 1) ≥ 6480 Expanding the brackets results in: n^2 + n ≥ 6480 Now, rearrange the equation to a quadratic inequality and solve for n: n^2 + n - 6480 ≥ 0 As this is a quadratic inequality, we will find the zeros (roots) of the quadratic equation: n^2 + n - 6480 = 0

Finding the roots

Now, we'll find the roots of the quadratic equation by either factoring, using the quadratic formula, or estimation. In this case, we'll use estimation with some educated guesswork. n^2 + n - 6480 = 0 This factors into: (n - 80)(n + 81) = 0 So, the roots of the equation are n = 80 and n = -81.

Analyze the roots

Since n represents the number of days, negative values don't have any meaning. Additionally, setting n=80 results in the sum being exactly 3240. Therefore, it takes at least 80 days to pay off the debt.

Key concepts

Arithmetic Progression

Understanding an arithmetic progression is key to solving problems like the debt repayment scenario described. An arithmetic progression, or arithmetic sequence, is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the common difference, denoted as 'd'. In our exercise, each day's payment amount represents a term in the sequence, and the common difference is 1, since the payment increases by 1 florin each day.

In general, the nth term of an arithmetic progression can be found using the formula: \( a_n = a_1 + (n - 1)d \), where \( a_1 \) is the first term and \( n \) is the term number. Consequently, the sum of the first n terms, denoted by S, is given by \( S_n = \frac{n}{2}(a_1 + a_n) \), or alternatively, \( S_n = \frac{n}{2}(2a_1 + (n - 1)d) \). This formula is fundamental as it allows us to calculate the total amount paid over a certain number of days.

This concept of arithmetic progression not only simplifies the problem but also offers a clear pathway to finding a solution through its inherent pattern and predictability.

Quadratic Inequality

Quadratic inequalities such as the one we encounter in our debt repayment problem require some manipulative work to reach a solution. These inequalities are similar to quadratic equations but instead of finding a specific value of 'n', we are looking for a range of values. Our original inequality, after summing the arithmetic progression, is \( \frac{n(n + 1)}{2} \geq 3240 \), which simplifies to \( n^2 + n - 6480 \geq 0 \).

The process of solving this inequality involves transforming it into a standard form, much like a quadratic equation, and then finding its roots. If we let the inequality be represented by the quadratic equation \( n^2 + n - 6480 = 0 \), we could find the roots by factoring, completing the square, or using the quadratic formula.

Once we determine the roots of the associated quadratic equation, we use these values to ascertain the solution to the inequality. We evaluate the intervals determined by the roots to see where the inequality holds true, keeping in mind that we are looking for positive values since negative days would not make sense in our context.

Mathematical Induction

While not directly used in the provided solution, mathematical induction is a powerful technique that can confirm the validity of formulas over a sequence of numbers, such as the sum of an arithmetic progression. Induction consists of two main steps: the base case and the inductive step.

For the base case, we validate that the formula is true for the first term of our sequence. When dealing with our arithmetic series, we'd verify that \( S = \frac{n(n + 1)}{2} \) holds true when \( n = 1 \). Afterwards, the inductive step involves proving that if the formula holds for some arbitrary positive integer \( k \), then it must also hold true for the next integer, \( k+1 \).

To apply induction to our series sum formula, we would assume that \( S_k = \frac{k(k + 1)}{2} \) is true, and then show that \( S_{k+1} = \frac{(k+1)(k + 2)}{2} \) is also true by incorporating \( S_k \) into our working out.

This process essentially builds a chain of correctness from the first term through all positive integers and is a cornerstone of many proofs in mathematics, reinforcing our understanding of the concepts involved in sequences and sums.