Question

you have a bag of ping pong balls. you arrange all but 2 of the balls in an equilateral triangle. then you put all the balls back into the bag and create a new equilateral triangle where each side has one more ball than the first arrangement but this time you are 11 balls short. how many balls were in the first arrangement?

Answer 1

The “edge” effect of a tetrahedral container with edge length of 1,000 mm would be 2*r*√3 (r = radius of ball, see diagram below). For a ping pong ball with a diameter of 40 mm, the edge effect would be 2*20*√3 = 69.28 mm. Subtract that from 1,000 mm and you get 930.72 mm of “available space” along the bottom row. Divide 930.72 mm by 40 mm per ball and you get 23.27. So you wouldn’t be able to fit 24 balls along the base, you’d only be able to fit 23 balls. With 23 balls, the total number that a tetrahedron could contain (T) would be T = n(n+1)(n+2)/6 = 2,300 balls, not 2,600.