Solving Integer Equations: Finding Non-Negative Solutions

by ADMIN 58 views

Hey math enthusiasts! Today, we're diving into a fascinating problem that combines algebra and number theory: figuring out how many non-negative integer solutions exist for the equation 20x + y + 50z = 82. This isn't just about crunching numbers; it's about understanding the constraints and the elegant ways we can find all possible solutions. Let's break this down, step by step, so you can tackle similar problems with confidence. The equation given, 20x + y + 50z = 82, may seem a bit intimidating at first glance, but it's actually quite approachable. We're looking for whole number (non-negative integer) values for x, y, and z that satisfy this equation. The presence of x, y, and z means this is a Diophantine equation—specifically, a linear Diophantine equation. This type of equation requires integer solutions, which adds an extra layer of complexity. The key to solving this is to isolate variables and systematically explore the possibilities.

First, because we are dealing with non-negative integers, each variable must be greater than or equal to zero. This constraint significantly narrows down the possible values for x, y, and z. Starting with z, the coefficient is the largest which is 50, and since we're looking for solutions that sum up to 82, z can only take on a limited number of values, that is 0 or 1. Let's consider these two scenarios separately. For each possible value of z, we'll then determine the range of values that x can take and, finally, deduce the value of y. This methodical approach will guarantee that we find all solutions without missing any. Think of this process like detectives solving a puzzle, where each variable's value is a clue leading us to the ultimate answer. We'll start with the most restrictive variable, z, to reduce the complexity. Then we will proceed with the systematic enumeration of the solutions, checking all the possibilities. Remember, the goal here is not just to find a solution but to find all possible non-negative integer solutions.

Step-by-Step Solution: Unpacking the Equation

Okay, guys, let's get down to the nitty-gritty and work through the equation 20x + y + 50z = 82 step by step. We'll methodically explore the possible values of z, x, and y to identify all non-negative integer solutions. This systematic approach is crucial to ensure we don't miss any valid combinations. This is a journey of breaking down the problem into smaller, manageable chunks.

Case 1: z = 0

When z equals zero, our equation simplifies to 20x + y = 82. Now, we need to determine the possible values for x and then calculate y. Since x and y must be non-negative integers, we can derive the maximum value for x. If x is greater than 4, then 20x would be greater than 82, and there would be no non-negative value for y to balance the equation. So, x can be 0, 1, 2, 3, or 4.

  • If x = 0, then y = 82 - 20(0) = 82. So, one solution is (0, 82, 0).
  • If x = 1, then y = 82 - 20(1) = 62. The second solution is (1, 62, 0).
  • If x = 2, then y = 82 - 20(2) = 42. Another solution is (2, 42, 0).
  • If x = 3, then y = 82 - 20(3) = 22. So, we find the solution (3, 22, 0).
  • If x = 4, then y = 82 - 20(4) = 2. This gives us the solution (4, 2, 0).

In this first case, where z = 0, we found a total of five solutions. These represent all possible combinations of x and y that satisfy the original equation when z is zero. We'll now examine the second case to see if any more solutions exist.

Case 2: z = 1

Now, let's explore the scenario where z = 1. Our equation becomes 20x + y + 50(1) = 82, which simplifies to 20x + y = 32. Following the same logic as before, we'll determine the possible values for x and calculate the corresponding values for y. Similarly, x can't be more than 1 because 20x would exceed 32 if x is 2 or greater. Therefore, x can only be 0 or 1.

  • If x = 0, then y = 32 - 20(0) = 32. This gives us the solution (0, 32, 1).
  • If x = 1, then y = 32 - 20(1) = 12. So, we get the solution (1, 12, 1).

In this second case, where z = 1, we found two solutions. The solutions found in this case are also non-negative integer solutions that satisfy the original equation when z is equal to 1. This concludes our exploration of all possible values for z. Now, we have successfully identified all solutions by methodically exploring each possible value of z and determining the corresponding values for x and y. We have covered all possible scenarios, ensuring that we didn't miss any valid solutions.

Enumerating the Solutions: Final Results

Alright, folks, we've done the hard work, and now it's time to gather all the solutions we've found and present them in a clear, concise manner. The solutions we've identified are all the possible combinations of non-negative integers that satisfy the equation 20x + y + 50z = 82. Let's recap what we've discovered. Remember, we meticulously went through each possible value of z and then found the corresponding values of x and y. This systematic approach guaranteed that we didn't miss any potential solutions.

From Case 1 (where z = 0), we found five solutions: (0, 82, 0), (1, 62, 0), (2, 42, 0), (3, 22, 0), and (4, 2, 0). These are all the combinations of x and y that work when z is set to zero.

From Case 2 (where z = 1), we found two solutions: (0, 32, 1) and (1, 12, 1). These are the combinations of x and y that satisfy the equation when z is set to one.

So, by combining the solutions from both cases, we have a total of seven solutions that satisfy the original equation. Each of these solutions represents a unique set of non-negative integers that, when plugged into the equation, make the equation true. The complete set of solutions is: (0, 82, 0), (1, 62, 0), (2, 42, 0), (3, 22, 0), (4, 2, 0), (0, 32, 1), and (1, 12, 1). The total number of non-negative integer solutions to the equation 20x + y + 50z = 82 is seven.

Conclusion: Wrapping Up the Puzzle

And there you have it, guys! We've successfully solved the equation 20x + y + 50z = 82 and found all of its non-negative integer solutions. This problem highlights the power of a systematic approach and how breaking down a complex equation into smaller, manageable cases can make it much easier to solve. The ability to methodically consider different scenarios, like the values of z, is a powerful tool in problem-solving. We started with the variable that was most restrictive (z) and worked our way through the possible values, calculating the corresponding values for the other variables. This approach allowed us to ensure that we found all possible solutions without any omissions. This method can be applied to other Diophantine equations or similar problems, allowing you to tackle a wide range of mathematical challenges. The key takeaway from this exercise is not just the answer (seven solutions) but the methodology. Remember, mathematics is about more than just finding answers; it's about the journey of problem-solving. Keep practicing, keep exploring, and you'll find that these types of problems become more and more approachable.

In summary, the equation 20x + y + 50z = 82 has a total of seven non-negative integer solutions. We hope this explanation helped you understand the process and gave you the confidence to approach similar problems. Keep up the great work, and happy solving!