The puzzle as previously presented:
The small school that Nora attends has three standard parking spots out front for parking when you drop your child off or pick them. On the Tuesdays and Thursdays mornings last year I got to drop her off last year, I would meet and talk with to the other parents dropping off their children.
If there are 15 different vehicles (including mine) that drop children off, it would take at least 3 1/2 weeks (7 mornings) for me to be parked with the other 14 vehicles, two other vehicles at a time — assuming I met no vehicle twice during that time.
Is it possible that for these 7 mornings that no one would would meet the same vehicle twice dropping their child off? If it is possible, what would this look like (in a diagram, chart, list, etc.)? If it isn’t possible, why not?
This week’s puzzle garnered fewer submissions than normal, and even fewer complete/correct ones. In fact, there were exactly zero correct solutions submitted, with all answers stating “It can’t be done!” Therefore, by decree of my position of “puzzlemaster,” all submissions were entered into the drawing for the weekly fabulous prize, a $50 Gift Certificate from ThinkGeek. Congratulations to Gus Sterneman for being the lucky winner of this week’s prize.
This week’s puzzle was a variant on Kirkman’s Schoolgirl Problem, also known as the Social Golfer Problem, where subsets of people are created over time meeting certain requirements (i.e., they each meet exactly once, they each meet at most once, they each meet exactly twice, etc.)
If we give each of the 15 vehicles a letter from A to O, one solution that has me, and every other parent, meet every other parent over 7 mornings in sets of three appears below:
|Day 1||Day 2||Day 3||Day 4||Day 5||Day 6||Day 7|
Thanks to everyone that submitted a solution, and congratulations to Gus! Happy puzzling!