Question:medium

For natural numbers x,y, and z, if xy+yz=19 and yz+xz=51, then the minimum possible value of xyz is

Updated On: Jan 15, 2026
Show Solution

Solution and Explanation

1. Given Equations

The provided equations are: \[ y(x + z) = 19 \tag{1} \] \[ z(x + y) = 51 \tag{2} \]

2. Deduction from Equation (1)

From equation (1), \( y(x + z) = 19 \). Since 19 is prime and \( y \) is a positive integer, \( y \) can only be 1 or 19.
If \( y = 1 \), then \( x + z = 19 \). Substituting into equation (2) yields: \[ z(x + 1) = 51 \tag{3} \]

3. Exploring Cases

Case 1: \( y = 1, z = 3 \Rightarrow x = 16 \)

Substituting these values into equation (3): \[ z(x + 1) = 3 \times (16 + 1) = 3 \times 17 = 51 \quad \text{Satisfied} \] The product \( xyz \) is \( 16 \cdot 1 \cdot 3 = 48 \).

Case 2: \( y = 1, z = 17 \Rightarrow x = 2 \)

Checking equation (3): \[ z(x + 1) = 17 \cdot (2 + 1) = 17 \cdot 3 = 51 \quad \text{Satisfied} \] The product \( xyz \) is \( 2 \cdot 1 \cdot 17 = 34 \).

4. Final Answer

The two possible values for \( xyz \) are 48 and 34. The minimum value is: \[ \boxed{34} \]

Was this answer helpful?
5


Questions Asked in CAT exam