My child is going into first grade.
The school has asked each child to writea list of 3 friends.
Of these three, at least one is guaranteed to be with him next year in class.
How can a group of (any n number of) friends circumvent this and end up together next year? (who should they write as their list of friends?)

Please try to find a logical answer to this riddle.
Thank you

If it's about math, it's certainly not a 1st grade question.
If's about thinking out of the box, consider having the kids write their 3 friends like this:
1. Me
2. Myself
3. I

Will that work :)

I'm stumped. I don't think it's possible to beat the system for any n, unless you are allowed to put the same friend's name down more than once. Or if a child is allowed to submit only one or two names, then it's possible.

Quote:

---

Originally Posted by ebaines

I'm stumped. I don't think it's possible to beat the system for any n, unless you are allowed to put the same friend's name down more than once. If they only had to submit one or two names, then it's possible.

---

Yeah, the math is not working... and certainly not something that a 1st grader could solve.

... so maybe it's not math but something else. I'm hoping to get some feedback from the asker on this one.

RickJ: I think you misread the riddle - it's not intended to be solved by 1st graders, but rather is a riddle about first graders.

If that's the case, then Me, Myself and I should work to get everyone in the same class.

Quote:

---

Originally Posted by RickJ

If that's the case, then Me, Myself and I should work to get everyone in the same class.

---

I don't think that's it. That solution only guarantees that each child is in a class with himself, not necessarily with any of his friends.