Question

How many solutions \( (x, y, z) \) of the equation \( x + y^2 + z^3 = 50 \) exist, where \( x, y \) and \( z \) are positive integers?

• A. 15
• B. 18
• C. 16
• D. 17
• E. None of the other options is correct

Hint:

When solving for integer solutions, break down the problem by checking for each possible value of the variables and their respective ranges.

Answer 1

The Correct Option is B

Step 1: Set up the equation.
We need to find the integer solutions for the equation: \[ x + y^2 + z^3 = 50 \] where \( x \), \( y \), and \( z \) are positive integers.
Step 2: Check for possible values of \( z \).
Since \( z^3 \leq 50 \), \( z \) can range from 1 to 3. Let's consider each case. - If \( z = 1 \), then \( z^3 = 1 \). The equation becomes: \[ x + y^2 + 1 = 50 \quad \Rightarrow \quad x + y^2 = 49. \] Now, for \( y^2 \leq 49 \), the possible values of \( y \) are 1, 2, 3, 4, 5, 6, 7. This gives 7 solutions. - If \( z = 2 \), then \( z^3 = 8 \). The equation becomes: \[ x + y^2 + 8 = 50 \quad \Rightarrow \quad x + y^2 = 42. \] Now, for \( y^2 \leq 42 \), the possible values of \( y \) are 1, 2, 3, 4, 5, 6. This gives 6 solutions. - If \( z = 3 \), then \( z^3 = 27 \). The equation becomes: \[ x + y^2 + 27 = 50 \quad \Rightarrow \quad x + y^2 = 23. \] Now, for \( y^2 \leq 23 \), the possible values of \( y \) are 1, 2, 3, 4. This gives 4 solutions.
Step 3: Total number of solutions.
Summing all the solutions: \[ 7 \, (\text{for } z = 1) + 6 \, (\text{for } z = 2) + 4 \, (\text{for } z = 3) = 18. \]
Step 4: Conclusion.
The number of solutions is 18, so the correct answer is (B).