This week the question is: Can You Solve a High Schooler’s Favorite Puzzle?

A teacher is handing out candy to his students, of which there are at least four. He abides by the following rules:

• He hands out candy to groups of three students (i.e., “trios”) at a time. Each member of the trio gets one piece of candy.
• Each unique trio can ask for candy, but that same trio can’t come back for seconds. If students in the trio want more candy, they must return as part of a different trio.
• When a trio gets candy, the next trio can’t contain any students from that previous trio.

It turns out that every possible trio can get a helping of candy. What is the smallest class size for which this is possible?

And for extra credit:

Instead of trios of students, suppose now that groups of 10 students come up to get candy. This time, there are at least 11 students in the class. As before:

• Each member of the group of 10 gets one piece of candy per visit.
• Each unique group of 10 can ask for candy, but the exact same group of 10 can’t come back for seconds. If students in the group want more candy, they must return as part of a different group.
• When a group of 10 gets candy, the next group of 10 can’t contain any students from the previous group of 10.

Suppose the class size is the minimum that allows every possible group of 10 to get a helping of candy. How many pieces of candy does each student receive?

Highlight to reveal (possibly incorrect) solution:

Program.

The first trio could be ABC. The next trio could be DEF. The next trio can’t contain any of D, E or F, and it can’t contain all of A, B and C, so there has to be 1 more student: G. So we need at least 7 students.

(If you want to, you can name the students Annie, Bob etc.)

I wrote a program to test this theory, and found a solution. There has to be at least 7 students, and a solution with 7 students exists. We need 7 students.

Solution: abc def abg cde abf deg acf bde acg bdf ceg abd cef bdg ace bfg acd efg bcd afg bce dfg abe cfg ade bcf adg bef cdg aef bcg adf beg cdf aeg. But this isn’t the only one, as far as I can tell.

And for extra credit:

I am not sure. I’ve been trying to find a pattern in the trio solution, some way to expand this solution to 10 students in each group. No such luck. So I’ve made a guess: 21.