Romanian Mathematical Olympiad

For $k \in \mathbb{N}^*$ there exists $n_k \in \mathbb{N}^*$ such that $2^{n_k} < 10^k < 2^{n_k+1}$ ($n_k + 1$ is the smallest element of the set $\{m \in \mathbb{N}^* \mid 10^k < 2^m\}$). As $2^{n_k} < 10^k < 2^{n_k+1}$, we get $10^k < 2^{n_k+1} < 2 \cdot 10^k$, so $p(2^{n_k+1}) = 1$.

Multiplying by $5^{n_k+1}$, we have $10^k \cdot 5^{n_k+1} < 10^{n_k+1} < 2 \cdot 10^k \cdot 5^{n_k+1}$. Dividing the first inequality by $10^k$ and the second by $2 \cdot 10^k$, we get $5 \cdot 10^{n_k-k} < 5^{n_k+1} < 10^{n_k-k+1}$, so $p(5^{n_k+1}) \ge 5 > p(2^{n_k+1})$.

From $2^{n_k} < 10^k < 2^{n_k+1}$, by dividing the second inequality by $2$, we conclude $5 \cdot 10^{k-1} < 2^{n_k} < 10^k$, so $p(2^{n_k}) \ge 5$. Multiplying the first inequality by $5^{n_k}$, we obtain $10^{k-1} \cdot 5^{n_k+1} < 10^{n_k} < 10^k \cdot 5^{n_k}$, so by dividing the first one by $5 \cdot 10^{k-1}$ and the second by $10^k$, we obtain $10^{n_k-k} < 5^{n_k} < 2 \cdot 10^{n_k-k}$, that is $p(5^{n_k}) = 1 < p(2^{n_k})$.

The set $\{n_k \mid k \in \mathbb{N}^*\}$ has an infinity of elements because $n_k > k$. As $\{n_k \mid k \in \mathbb{N}^*\} \subset A$ and $\{n_k + 1 \mid k \in \mathbb{N}^*\} \subset B$, the sets $A$ and $B$ also have infinitely many elements.