2018 AMC 10B Problem 4

The three distinct face areas are the pairwise products $XY = 24,$ $XZ = 48,$ $YZ = 72$ in some order. Multiply all three: $(XYZ)^2 = 24 \cdot 48 \cdot 72 = 82944,$ so $XYZ = 288.$ Now divide by each face area. We get $Z = \frac{288}{24} = 12,$ $Y = \frac{288}{48} = 6,$ and $X = \frac{288}{72} = 4,$ so $X + Y + Z = 22.$ Therefore, the answer is B.