Brazil

Let's generalize this problem to $2n$ batteries, $n$ of them charged. The number of charged batteries needed is still two. One can see that the order of the attempts doesn't matter and that a set of attempts is always successful if and only if in every set of $n$ batteries there is an attempt with two of these batteries, that is, no set of $n$ batteries is "pairwise untested". Indeed, if there is a set $S$ of $n$ pairwise untested batteries, you can be unlucky enough: the working batteries might be these in $S$. Conversely, if every set of $n$ batteries has two tested batteries, we will choose at some moment two charged batteries and the radio will work. So consider a graph $G$ in which each battery is a vertex and we connect two batteries iff we do not test these batteries. The number of attempts is the number of disconnected pairs, that is $(2^n)-|E|$, where $E$ is the set of edges of $G$. Since we want to minimize the number of attempts, we must maximize $|E|$, that

is, we need to have the maximum number of edges. But the only restriction is that we don't have a set S of n pairwise untested batteries, which speaking "graph-wise", is the same as G not having a n-clique. But, by Túran's Theorem, the graph with the maximum number of edges without a n-clique is the most balanced $(n-1)$-partite graph. In this case, the partition sets of vertices of $G$ must contain $3,3,2,2,\dots ,2$ vertices (where we have $n-3$ twos). So the least number of attempts is the number of disconnected pairs of vertices, that is, $2\cdot 3+(n-3)=n+3$.