# Truncatable primes and unavoidable sets of divisors

• Volume: 14, Issue: 1, page 21-25
• ISSN: 1804-1388

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## Abstract

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We are interested whether there is a nonnegative integer ${u}_{0}$ and an infinite sequence of digits ${u}_{1},{u}_{2},{u}_{3},\cdots$ in base $b$ such that the numbers ${u}_{0}{b}^{n}+{u}_{1}{b}^{n-1}+\cdots +{u}_{n-1}b+{u}_{n},$ where $n=0,1,2,\cdots ,$ are all prime or at least do not have prime divisors in a finite set of prime numbers $S.$ If any such sequence contains infinitely many elements divisible by at least one prime number $p\in S,$ then we call the set $S$ unavoidable with respect to $b$. It was proved earlier that unavoidable sets in base $b$ exist if $b\in \left\{2,3,4,6\right\},$ and that no unavoidable set exists in base $b=5.$ Now, we prove that there are no unavoidable sets in base $b⩾3$ if $b-1$ is not square-free. In particular, for $b=10,$ this implies that, for any finite set of prime numbers $\left\{{p}_{1},\cdots ,{p}_{k}\right\},$ there is a nonnegative integer ${u}_{0}$ and ${u}_{1},{u}_{2},\cdots \in \left\{0,1,\cdots ,9\right\}$ such that the number ${u}_{0}{10}^{n}+{u}_{1}{10}^{n-1}+\cdots +{u}_{n}$ is not divisible by ${p}_{1},\cdots ,{p}_{k}$ for each integer $n⩾0.$

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