Mathematical Reflections 2017, Issue 3 - Problem U411

Problem:
Let $e$ be a positive integer. We say that a positive integer $m$ is awesome if $m$ has $\omega^e(m)$ digits in base ten, where $\omega(m)$ is the number of distinct primes that divide $m$. Prove that there are finitely many awesome numbers.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let $\mathbb{P}$ be the set of primes and let
$$\mathcal{E}_n=\left\{\prod_{j=1}^{n^e} p_j^{k_j} : p_j \in \mathbb{P}, k_j \in \mathbb{N}^{\ast} \right\}$$ where the $p_j$ are distinct primes. A positive integer $m$ is awesome if and only if the two following conditions are simultaneously satisfied:

    (a) $m \in \mathcal{E}_n$
    (b) $10^{n^e-1} \leq m < 10^{n^e}$

for some $n \in \mathbb{N}^{\ast}$.
Let $$\mathcal{A}_n=\{m \in \mathcal{E}_n : 10^{n^e-1} \leq m < 10^{n^e}\}$$ be the set of awesome numbers with $n^e$ digits and let $$\mathcal{A}=\bigcup_{n \in \mathbb{N}^*} \mathcal{A}_n.$$ It's easy to see that $\mathcal{A}$ contains all the awesome numbers. \\
In order to prove that $\mathcal{A}$ is finite, we need to prove that there are finitely many nonempty $\mathcal{A}_n$.
We prove that for any fixed exponent $e$, there exists $k \in \mathbb{N}$ such that $\forall n > k$ we have $\mathcal{A}_n= \emptyset$.\\
Since $\mathcal{E}_n \neq \emptyset$ for all $n \in \mathbb{N}^*$, we can deduce that each $\mathcal{E}_n$ has a minimum by the well-ordering principle.
Let $m_n=\min_{m \in \mathcal{E}_n} \{m\}$ for all $n \in \mathbb{N}^*$ and let $$\{m_n\}_{n \in \mathbb{N}^*}=\{m_1, m_2, \ldots\}$$be the sequence whose terms are the all the minima of each $\mathcal{E}_n$.\\
By definition, $$m_n = \prod_{j=1}^{n^e}p^*_j$$ where $p^*_1=2, p^*_2=3, p^*_3=5, \ldots$. The sequence $\{m_n\}$ is a positive term sequence and $$m_n = \prod_{j=1}^{n^2} p^*_j \geq 2^{n^e} \quad \forall n \in \mathbb{N}^*,$$ so $\{m_n\}_{n \in \mathbb{N}}$ diverges by the Comparison Test, i.e. $$\forall M \in \mathbb{R} \quad \exists k \in \mathbb{N} : m_n > M \quad \forall n>k$$ and this holds in particular for $M=10^{n^e}$, which gives $m_n > 10^{n^e}$ for any $n>k$. Since $m_n=\min_{m \in \mathcal{E}_n} \{m\}$, we have that $\mathcal{A}_n=\emptyset$ for any $n > k$ and for any fixed eponent $e$, which is the desired conclusion.