Question

Positive integers X,y,z , satisfy XY +z = 160. Compute the smallest possible value of X+yz

Answer 1

Answer:

Step-by-step explanation:

XY + Z = 160

Now, Positive integers X, Y, Z are +Ve integers. For smallest value of X+YZ

Sc 1:- Z = 1

XY + 1 = 160

XY  = 159, X = 59, Y = 3

X + YZ = 59 + 3 x 2 = 65

Sc 2:- Z = 2

XY + 2 = 160

XY  = 158, X = 1, Y = 159  --> This will be more than Sc 1 for  X+yz.

XY  = 158, X = 2, Y = 79  --> This will be more than Sc 1 for  X+yz.

Sc 3:- Z = 3

XY + 3 = 160

XY = 157

157 being prime

X = 157, Y =1 --> This will be more than Sc 1 for  X+yz.

Sc 4:- Z = 4

XY + 4 = 160

XY = 156, X = 78, Y = 2  --> This will be more than Sc 1 for  X+yz.

XY = 156, X = 52, Y = 3  --> This will be more than Sc 1 for  X+yz.

XY = 156, X = 39, Y = 4  --> X+yz = 39 + 16 = 55 -- This is new low.

XY = 156, X = 26, Y = 6  --> X+yz = 26 + 24 = 50 -- This is new low.

XY = 156, X = 13, Y = 12  --> X+yz = 13 + 12*4 -- This will be more than new low.

Sc 4:- Z = 5

XY + 5 = 160

XY = 155, X = 31, Y = 5 --> X+yz = 31 + 5*5   -- This will be more than new low.

XY = 155, X = 5, Y = 31 --> X+yz = 5 + 31*5   -- This will be more than new low.

So, based on the results we can say  X+yz = 50 is lowest with X = 26, Y = 6, Z = 4