How to solve this mathematical puzzle? 10 investors bought 3 stocks each and any two of them bought at least one stock in
common. For the stock owned by most people, what is the minimum number of people who have bought it? I tried applying the concepts of set theory as well as pigeon hole principle, but cannot really wrap my head around it. 
 A: You have $45$ pairs of people who must own a stock in common.  If a stock is owned by $n$ people, it covers $\frac 12n(n-1)$ pairs.  If the most people who own the same stock is $4$, there can be seven stocks owned by four people and one stock owned by two.  This accounts for $43$ pairs, so we must have a stock owned by five.  If the most people who own one stock is five, you can have six stocks owned by five people each, which could cover $90$ pairs.  In this case it is natural to assume (I have not justified this) that the stocks come in three pairs with the stocks in each pair belonging to a disjoint set of five people.  This fails bcause of overlap.  Let stock A belong to people $1-5$ and stock B to people $6-10$.  Then one of C,D belongs to at least three of $1-5$ and one of E,F belongs to at least three of $1-5$ as well.  Say these are C,E.  Somebody of $1-5$ owns both C and E by the pigeonhole principle and will therefore only have two matches from each within $6-10$ and will be missing one.  Then the maximum ownership goes up to six and we have five different stocks.  We can demonstrate a solution here.
$$ \begin {array} {c c c c c c c c c c}\\1&2&3&4&5&6&7&8&9&10\\
A&A&A&A&A&A&B&B&B&B\\B&B&C&C&D&D&C&C&C&C\\E&E&E&E&E&E&D&D&D&D\end {array}$$
We know six is sufficient, four is not enough, and it seems five doesn't work but it is not certain.
