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Monday, October 9, 2017

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.

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