A class of 37 is to be divided into teams, and each student in the class must be a member of exactly one team. However, each student dislikes three of their classmates. The dislike between students need not be mutual. How many teams must be created so that no student is the teammate of someone they dislike?
My attempt: Considering the extreme case when the dislike is never mutual, label the students $a_1, a_2, \cdots a_{37}$ such that $a_i$ dislikes $a_{i+1}, a_{i+2}, a_{i+3}$ for $i = 1,2\cdots 34, a_{35}$ dislikes $a_{36}, a_{37}$ and $a_{1}, a_{36}$ dislikes $a_{37}, a_{1}$ and $a_{2}$ and $a_{37}$ dislikes $a_{1}, a_{2}$ and $a_{3}.$ Thus, each student can be in a team with at most $30$ other students. So, the students can be divided into $4$ teams as follows:
Team $1: \{a_1, a_5, a_9,\cdots a_{33}\}$
Team $2: \{a_2, a_6, a_{10},\cdots a_{34}\}$
Team $3: \{a_3, a_7, a_{11},\cdots a_{35}\}$
Team $4: \{a_4, a_8, a_{12},\cdots a_{36}, a_{37}\}$
Thus, my answer is $4.$ However, the given answer is $7.$ Can someone point out where I am going wrong?
An answer to where you're going wrong, not an answer to the original question per se:
What you're forgetting is the dislikes may not run through such a nice cyclic pattern (i.e. through all $37$ kids). With a smaller such "pattern", you quickly get 4 teams are impossible.
For example, if
a) $a_1$ dislikes $a_2, a_3, a_4$,
b) $a_2$ dislikes $a_3, a_5$,
c) $a_3$ dislikes $a_4, a_5$,
d) $a_4$ dislikes $a_5$,
e) $a_5$ dislikes $a_1$,
then it is clear that you cannot split them into four teams. It may be best to see this by drawing the directed graph of these five kids.
EDIT: If in addition $a_4$ dislikes $a_6$, $a_5$ dislikes $a_6$, and $a_6$ dislikes $a_1, a_2, a_3$, then $6$ is quickly impossible, too.