41. The class won a prize of a bag with 2014 coins, which initially is with the class president (and the other children in the class have no money). When two classmates meet, they split the money equally between themselves if the total number of coins they have is even. If the total number is odd, they give one coin to the class teacher and then split the rest equally between themselves. After some time, all the coins ended up with the class teacher. What is the smallest number of children that could have been in the class?
$\square$ Answer: 12 children.

We will show by induction on $k$ that $k+1$ people can get rid of $2^{k}-1$ coins initially held by one of them (and from a smaller number of coins). For $k=1$, this is clear, and the inductive step from $k$ to $k+1$ is also clear: after the first meeting, there remain 2 people, each having no more than $2^{k-1}-1$ coins, and $k$ people with no coins. Then $k+1$ people, by the inductive hypothesis, get rid of their coins, after which another set of $k+1$ people do the same.

Now we will show that if one of the people initially had at least $2^{k}$ coins, and at some point in the exchanges, $m$ people participated, $m \leqslant k+1$, then each of them has at least $2^{k+1-m}$ coins. Then, when $m=k+1$, we get that $k+1$ people cannot get rid of all the coins. Induction on the number of exchanges. The base case for 0 exchanges is obvious. If in the next exchange, a new person does not participate, then the participants still have at least $2^{k+1-m}$ coins. If a new person participates, then $m$ increases by 1, and the value $2^{k+1-m}$ is halved, so the statement remains true.