Friday puzzle: impossible

Last week’s puzzle was to calculate the volume remaining after a hole 6” long is drilled through a sphere. When I was a math nerd in high-school I actually knew how to do the triple integral in spherical polar coordinates to solve this problem properly, and, in particular, verify that the answer does not depend on the radius of the sphere. No way I have a clue how to do that today. However, if you start from the assumption that I wouldn’t be idiotic enough to ask the question if the radius of the sphere was relevant, then the problem is easy to solve. The limiting case is a sphere of diameter 6” so that the hole will remove nothing. The answer must be the volume of a sphere of radius 3” (remember it’s ⁴/₃ π r³) namely 36π cubic inches.

Today’s puzzle also seems not to have enough information. Two numbers not necessarily different are picked from integers greater than 1 that sum to less than 100. Person S gets told only the sum of the two numbers. Person P gets told only the product. The following conversation takes place. P says to S, “I don’t know the two numbers.” S replies, “I knew that you didn’t know.” P then says, “Now I know the two numbers.” S then says, “Now I know the two numbers.” What were they?

Answer next week