Question

Positive integers x, y, z satisfy xy +z = 160. Compute the smallest possible value of x + yz.​

Answer 1

Answer:

the smallest possible value of x + yz.​ = 50

Step-by-step explanation:

Positive integers x, y, z satisfy xy +z = 160. Compute the smallest possible value of x + yz.​

xy +z = 160

x, y, z are Positive integers

=> they can not be zero

158 + 2 = 160

158*1 + 2 = 160

x + yz = 158 + (1*2) = 160

79*2 + 2 = 160

x + yz = 79 + (2*2) = 83

156 + 4 = 160

156*1 + 4 = 160

x + yz = 156 + (1*4) = 160

78*2 + 4 = 160

x + yz = 78 + (2*4) = 86

52*3 + 4 = 160

x + yz =52 + (3*4) = 64

39*4 + 4 = 160

x + yz =39 + (4*4) = 55

26*6 + 4 = 160

x + yz =26 + (6*4) = 50

13*12 + 4 = 160

x + yz =13 + (12*4) = 61

the smallest possible value of x + yz.​ = 50

Answer 2

Given:

xy +z = 160

To find:

x + yz=?

Solution:

The minimum value of x+yz,

$\to x\geq y\geq z\\\\\to 160 =xy +z \geq z^2+z$

$\to z \leq 12,$

The minimum value of x+ yz with Condition xy+z=160 is specific possibilities.

please find the table:

The final answer is "50"