Solve The Mathematical Problem: Xyz = Xy + Yz + Zx
Hey guys! Let's dive into a cool mathematical problem where we need to find a number xyz that satisfies a specific condition. We'll break it down step by step to make it super easy to understand. So, grab your thinking caps, and let's get started!
Understanding the Problem
The problem states that we have a three-digit number xyz where xyz = xy + yz + zx. Our mission is to figure out if the digit x can be equal to 2 and then find the actual number xyz that makes this equation true. Sounds like fun, right?
Part A: Can x be equal to 2?
Let's start by exploring whether the digit x can be 2. If x = 2, our number becomes 2yz. According to the problem, we need to check if 2yz = 2y + yz + 2z is possible.
Breaking Down the Equation
First, let's express 2yz in terms of its place values:
2yz = 200 + 10y + z
Now, let's rewrite the given equation:
200 + 10y + z = (20 + y) + (10y + z) + (20 + z)
Simplify the equation:
200 + 10y + z = 20 + y + 10y + z + 20 + z
Combine like terms:
200 + 10y + z = 40 + 11y + 2z
Now, let's isolate the variables by moving terms around:
200 - 40 = 11y - 10y + 2z - z
160 = y + z
Analyzing the Result
So, we end up with y + z = 160. Here's the catch: y and z are single digits, meaning they can only be numbers from 0 to 9. The maximum value y + z can have is 9 + 9 = 18. Since 160 is way larger than 18, it's impossible for x to be 2.
Conclusion for Part A
Therefore, the answer to the first part of the question is a resounding no. It's not possible for the digit x to be equal to 2 because it leads to a contradiction. The sum of the digits y and z cannot be 160 when they are both single-digit numbers. Mystery solved!
Part B: Finding the Number xyz
Now, let's find the actual number xyz that satisfies the original condition xyz = xy + yz + zx. We'll need to use a bit of algebra and logical reasoning to crack this one.
Setting Up the Equation
Express the three-digit number xyz in terms of its place values:
100x + 10y + z = (10x + y) + (10y + z) + (10z + x)
Simplifying the Equation
Expand and simplify the equation:
100x + 10y + z = 10x + y + 10y + z + 10z + x
Combine like terms:
100x + 10y + z = 11x + 11y + 11z
Move all terms to one side to set the equation to zero:
100x - 11x + 10y - 11y + z - 11z = 0
Simplify further:
89x - y - 10z = 0
Rearranging the Equation
Rearrange the equation to isolate y:
y = 89x - 10z
Finding Possible Values
Since x, y, and z are digits (0-9), we need to find values that satisfy the equation. Let's consider different values for x and see what we get.
If x = 1:
y = 89(1) - 10z
y = 89 - 10z
Now, we need to find a value for z that makes y a single-digit number. If z = 8:
y = 89 - 10(8)
y = 89 - 80
y = 9
So, we have x = 1, y = 9, and z = 8. This gives us the number 198.
If x > 1:
If x is greater than 1, y will be a very large number (greater than 9), which is not possible since y must be a single digit. Thus, x = 1 is the only possible value.
Verifying the Solution
Let's check if 198 satisfies the original equation:
198 = 19 + 98 + 18
198 = 19 + 98 + 18 = 135
Oops! It seems we made a mistake somewhere. Let's backtrack and re-examine our steps.
Correcting the Approach
We have the equation 89x = y + 10z. We need to find integer solutions for x, y, and z between 0 and 9.
Let's try different values for x again, but this time, let’s be more systematic.
If x = 1:
89 = y + 10z
To find suitable values for y and z, we can express y as:
y = 89 - 10z
Now, we test values for z:
- If
z = 0,y = 89(not possible) - If
z = 1,y = 79(not possible) - If
z = 2,y = 69(not possible) - If
z = 3,y = 59(not possible) - If
z = 4,y = 49(not possible) - If
z = 5,y = 39(not possible) - If
z = 6,y = 29(not possible) - If
z = 7,y = 19(not possible) - If
z = 8,y = 9(possible!) - If
z = 9,y = -1(not possible)
So, we have x = 1, y = 9, and z = 8. The number is 198.
Let's verify again:
198 = 19 + 98 + 18
198 = 135
Still not correct! Let's rethink the problem.
The Correct Solution
Let's go back to the original equation: 100x + 10y + z = (10x + y) + (10y + z) + (10z + x). Simplifying, we get:
89x = y + 10z
We need to find digits x, y, and z that satisfy this equation.
Notice that 89x must be a two-digit number since y and z are digits. This implies that x must be 1 because if x were 2 or greater, 89x would be at least 178, which is a three-digit number.
So, x = 1. Now our equation becomes:
89 = y + 10z
Since y and z are digits, the only solution is z = 8 and y = 9. Therefore, the number is 198.
Let's verify:
198 = 19 + 98 + 18
198 = 135. Still incorrect.
Okay, let's try a different approach entirely.
Reassessing the Original Condition
The original condition is xyz = xy + yz + zx. We express this as:
100x + 10y + z = (10x + y) + (10y + z) + (10z + x)
Which simplifies to:
89x = y + 10z
Since x, y, and z are digits, let's test values. We already determined that x must be 1, so:
89 = y + 10z
The only digits that satisfy this are y = 9 and z = 8. So the number is 198.
However, our verification keeps failing. This suggests there might be no solution that perfectly fits the given condition. The problem might have a subtle trick or might not have a valid solution within the constraints.
After numerous attempts and checks, it appears there is no three-digit number xyz that strictly satisfies the condition xyz = xy + yz + zx with x, y, and z being digits from 0 to 9. Our algebraic manipulations and systematic testing have not yielded a valid solution.
Final Conclusion for Part B
Despite our best efforts, we couldn't find a number xyz that fits the criteria. This indicates that either there's a more complex approach needed or the problem, as stated, has no solution.
Final Thoughts
So, there you have it! We tackled a tricky mathematical problem, explored different possibilities, and learned a lot along the way. Even though we couldn't find a perfect solution, the journey was definitely worth it. Keep those thinking caps on, guys, and keep exploring the fascinating world of mathematics! Happy problem-solving! 🚀
