For any digit d, there are infinitely many primes which do not have d in their decimal expansion

You need to know: Prime numbers

Background: Any positive integer $n$ can be written in a unique way as $n=\sum\limits_{k=1}^{m}a_k10^{m-k}$ , where $m\geq 0$ is an integer, and $a_1, \dots, a_m$ are integers between $0$ and $9$ . This is called decimal expansion of n with digits $a_1, \dots, a_m$ , and this is the usual way we write integers.

The Theorem: On 4th April 2016, James Maynard submitted to arxiv a paper in which he proved that given any digit $d\in\{0,1,\dots,9\}$ , there are exist infinitely many primes p which do not have the digit d in their decimal expansion.

Short context: For $d\in\{0,1,\dots,9\}$ , let ${\cal A}_d$ be the set of positive integers which do not have the digit d in their decimal expansion. For $x \approx 10^k$ , there are about $9^k \approx x^{0.954}$ integers less than x in ${\cal A}_d$ . Sets of integers with at most $x^c$ integers up to x for some $c<1$ are called sparse. Usually, it is hard to prove that there are infinitely many primes in sparse sets. The Theorem achieves this for sets ${\cal A}_d$ , $d=0,1,\dots,9$ . See here for a result about infinitely many primes in (somewhat similar but not sparse) sets of integers with even/odd sum of digits.