4. Given 2019 indistinguishable coins. All coins have the same weight, except for one, which is lighter. What is the minimum number of weighings required to guarantee finding the lighter coin using a balance scale without weights? (12 points)
Solution. We will prove the following statement by induction on $k$: if there are $N$ visually identical coins, with $3^{k-1}<N \leq 3^{k}$, and one of them is lighter, then it can be found in $k$ weighings. Base case: $k=0, N=1$, no weighing is needed for a single coin. Inductive step: suppose the statement is true for $0,1,2$, $\ldots, k$. Now let $3^{k}<N \leq 3^{k+1}$. Place no less than $N / 3$ coins, but no more than $3^{k}$ coins, on the left pan, and the same number on the right pan. If the left pan is lighter, then the lighter coin is on it; if the right pan is lighter, then the lighter coin is on it; if the scales are balanced, then the lighter coin is among the remaining coins, the number of which is less than or equal to $N / 3 \leq 3^{k}$. As a result, we need to find the lighter coin among no more than $3^{k}$ coins, and it will take no more than $k$ additional weighings. Since $3^{6}<2019 \leq 3^{7}$, the number of weighings $k$ is 7.

Answer: 7.