Consecutive Integers: Sum Equals Product - Find The Possibilities!

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Consecutive Integers: Sum Equals Product - Find the Possibilities!

Hey guys! Let's dive into a fascinating math problem today: How many sets of three consecutive integers have a sum that is equal to their product? This might sound a bit tricky at first, but we'll break it down step by step. We'll explore the concept of consecutive integers, understand how to represent them algebraically, and then use that knowledge to find the solutions. So, grab your thinking caps, and let's get started!

Understanding Consecutive Integers

First, let's make sure we're all on the same page about what consecutive integers are. Simply put, they are integers that follow each other in order, each differing from the previous one by 1. Think of it like counting on your fingers: 1, 2, 3, 4, 5... Those are consecutive integers. Similarly, -3, -2, -1, 0, 1 are also consecutive integers. The key is that they form a sequence where you add 1 to get the next number.

To tackle this problem algebraically, we need a way to represent these consecutive integers using variables. Let's say we have three consecutive integers. We can represent them as follows:

  • The first integer: n
  • The second integer: n + 1
  • The third integer: n + 2

Here, n can be any integer, positive, negative, or zero. This representation allows us to create an equation based on the problem's condition: the sum of these integers equals their product. This is a crucial step in solving the problem, as it transforms the word problem into a manageable algebraic equation. By using this representation, we can explore different values of n and see which sets of consecutive integers satisfy the given condition. This algebraic approach is a powerful tool for solving various mathematical problems, and it's especially useful when dealing with sequences and patterns.

Setting Up the Equation

Now that we know how to represent consecutive integers, let's translate the problem into an equation. The problem states that the sum of three consecutive integers is equal to their product. Using our representation from above, we can write this as:

n + (n + 1) + (n + 2) = n * (n + 1) * (n + 2)

This equation is the heart of our problem. It captures the relationship between the sum and the product of the three consecutive integers. The left side of the equation represents the sum, and the right side represents the product. Our goal is to find the values of n that make this equation true. To do this, we'll need to simplify the equation and solve for n. This involves algebraic manipulation, such as combining like terms and potentially factoring. Once we find the values of n, we can then determine the sets of consecutive integers that satisfy the condition.

Solving the Equation

The next step is to solve the equation we set up. Let's simplify and rearrange it:

  1. Expand both sides:

    n + (n + 1) + (n + 2) = n * (n + 1) * (n + 2)

    3n + 3 = n * (n^2 + 3n + 2)

    3n + 3 = n^3 + 3n^2 + 2n

  2. Move everything to one side to set the equation to zero:

    0 = n^3 + 3n^2 + 2n - 3n - 3

    0 = n^3 + 3n^2 - n - 3

Now we have a cubic equation. Solving cubic equations can sometimes be tricky, but in this case, we can try factoring by grouping. This technique involves grouping terms together and factoring out common factors.

  1. Factor by grouping:

    0 = (n^3 + 3n^2) + (-n - 3)

    0 = n^2(n + 3) - 1(n + 3)

    0 = (n^2 - 1)(n + 3)

    0 = (n - 1)(n + 1)(n + 3)

Now we have factored the cubic equation into a product of three linear factors. This makes it much easier to find the solutions for n. The solutions are the values of n that make each factor equal to zero.

  1. Set each factor to zero and solve for n:

    • n - 1 = 0 => n = 1
    • n + 1 = 0 => n = -1
    • n + 3 = 0 => n = -3

So, we have found three possible values for n: 1, -1, and -3. Each of these values corresponds to a set of three consecutive integers that satisfy the problem's condition. Now, let's find those sets of integers.

Finding the Sets of Integers

We've found three possible values for n: 1, -1, and -3. Now we need to plug each value back into our representation of consecutive integers (n, n+1, n+2) to find the actual sets of numbers.

  1. For n = 1:

    The integers are 1, 1 + 1, 1 + 2, which gives us the set 1, 2, 3.

  2. For n = -1:

    The integers are -1, -1 + 1, -1 + 2, which gives us the set -1, 0, 1.

  3. For n = -3:

    The integers are -3, -3 + 1, -3 + 2, which gives us the set -3, -2, -1.

So, we have identified three sets of consecutive integers that satisfy the condition that their sum is equal to their product: (1, 2, 3), (-1, 0, 1), and (-3, -2, -1). To be absolutely sure, let's quickly verify that these sets indeed work:

  • For 1, 2, 3: 1 + 2 + 3 = 6 and 1 * 2 * 3 = 6. It checks out!
  • For -1, 0, 1: -1 + 0 + 1 = 0 and -1 * 0 * 1 = 0. This one works too!
  • For -3, -2, -1: -3 + (-2) + (-1) = -6 and -3 * -2 * -1 = -6. Perfect!

Conclusion

Alright, guys, we've successfully solved this problem! We started with the question: How many sets of three consecutive integers have a sum equal to their product? Through a combination of algebraic representation, equation solving, and factoring, we discovered that there are exactly three such possibilities. These sets are (1, 2, 3), (-1, 0, 1), and (-3, -2, -1).

This problem is a great example of how algebra can be used to solve number theory problems. By representing the consecutive integers with variables and setting up an equation, we were able to systematically find the solutions. It also highlights the importance of factoring in solving polynomial equations. Factoring allowed us to break down a complex cubic equation into simpler linear factors, making it easier to find the roots.

So, the final answer is C) Only three. I hope you found this explanation helpful and insightful! Keep practicing these types of problems, and you'll become a math whiz in no time! Cheers!