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SOLVED - $1000
Can the smallest modulus of a covering system be arbitrarily large?
Described by Erdős as 'perhaps my favourite problem'. Hough [Ho15], building on work of Filaseta, Ford, Konyagin, Pomerance, and Yu [FFKPY07], has shown the answer is no: the smallest modulus must be at most $10^{18}$.

An alternative, simpler, proof was given by Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST22], who improved the bound on the smallest modulus to $616000$.

OPEN
Is there a covering system all of whose moduli are odd?
Asked by Erdős and Selfridge (sometimes also with Schinzel). They also asked whether there can be a covering system such that all the moduli are odd and squarefree. The answer to this stronger question is no, proved by Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST22].

Hough and Nielsen [HoNi19] proved that at least one modulus must be divisible by either $2$ or $3$. A simpler proof of this fact was provided by Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST22].

Selfridge has shown (as reported in [Sc67]) that such a covering system exists if a covering system exists with moduli $n_1,\ldots,n_k$ such that no $n_i$ divides any other $n_j$ (but the latter has been shown not to exist, see [586]).

Additional thanks to: Antonio Girao
SOLVED
For any finite colouring of the integers is there a covering system all of whose moduli are monochromatic?
Conjectured by Erdős and Graham, who also ask about a density-type version: for example, is \[\sum_{\substack{a\in A\\ a>N}}\frac{1}{a}\gg \log N\] a sufficient condition for $A$ to contain the moduli of a covering system? The answer (to both colouring and density versions) is no, due to the result of Hough [Ho15] on the minimum size of a modulus in a covering system - in particular one could colour all integers $<10^{18}$ different colours and all other integers a new colour.
SOLVED - $100
An $\epsilon$-almost covering system is a set of congruences $a_i\pmod{n_i}$ for distinct moduli $n_1<\ldots<n_k$ such that the density of those integers which satisfy none of them is $\leq \epsilon$. Is there a constant $C>1$ such that for every $\epsilon>0$ and $N\geq 1$ there is an $\epsilon$-almost covering system with $N\leq n_1<\cdots <n_k\leq CN$?
By a simple averaging argument the set of moduli $[m_1,m_2]\cap \mathbb{N}$ has a choice of residue classes which form an $\epsilon(m_1,m_2)$-almost covering system with \[\epsilon(m_1,m_2)=\prod_{m_1\leq m\leq m_2}(1-1/m).\] A $0$-covering system is just a covering system, and so by Hough [Ho15] these only exist for $n_1<10^{18}$.

The answer is no, as proved by Filaseta, Ford, Konyagin, Pomerance, and Yu [FFKPY07], who (among other results) prove that if \[1< C \leq N^{\frac{\log\log\log N}{4\log\log N}}\] then, for any $N\leq n_1<\cdots< n_k\leq CN$, the density of integers not covered for any fixed choice of residue classes is at least \[\prod_{i}(1-1/n_i)\] (and this density is achieved for some choice of residue classes as above).

Additional thanks to: Mehtaab Sawhney
OPEN
Let $n_1<\cdots < n_r\leq N$ with associated $a_i\pmod{n_i}$ such that the congruence classes are disjoint (that is, every integer is $\equiv a_i\pmod{n_i}$ for at most one $1\leq i\leq r$). How large can $r$ be in terms of $N$?
Let $f(N)$ be the maximum possible $r$. Erdős and Stein conjectured that $f(N)=o(N)$, which was proved by Erdős and Szemerédi [ErSz68], who showed that, for every $\epsilon>0$, \[\frac{N}{\exp((\log N)^{1/2+\epsilon})} \ll_\epsilon f(N) < \frac{N}{(\log N)^c}\] for some $c>0$. Erdős believed the lower bound is closer to the truth.
OPEN
Is there an integer $m$ with $(m,6)=1$ such that none of $2^k3^\ell m+1$ are prime, for any $k,\ell\geq 0$?
There are odd integers $m$ such that none of $2^km+1$ are prime, which arise from covering systems (i.e. one shows that there exists some $n$ such that $(2^km+1,n)>1$ for all $k\geq 1$). Erdős and Graham also ask whether there is such an $m$ where a covering system is not 'responsible'. The answer is probably no since otherwise this would imply there are infinitely many Fermat primes of the form $2^{2^t}+1$.
OPEN
Do there exist $n$ with an associated covering system $a_d\pmod{d}$ on the divisors of $n$ (so that $d\mid n$ for all $d$ in the system), such that if \[x\equiv a_d\pmod{d}\textrm{ and }x\equiv a_{d'}\pmod{d'}\] then $(d,d')=1$? For a given $n$ what is the density of the integers which do not satisfy any of the congruences?
The density of such $n$ is zero. Erdős and Graham believed that no such $n$ exist.
OPEN
Is there a covering system all of whose moduli are of the form $p-1$ for some primes $p\geq 5$?
Selfridge has found an example using divisors of $360$ if $p=3$ is allowed.
OPEN
If $G$ is an abelian group then can there exist an exact covering of $G$ by more than one cosets of different sizes? (i.e. each element is contained in exactly one of the cosets)
A problem of Herzog and Schönheim.
Additional thanks to: Boris Alexeev and Dustin Mixon
SOLVED
If a finite system of $r$ congruences $\{ a_i\pmod{n_i} : 1\leq i\leq r\}$ covers $2^r$ consecutive integers then it covers all integers.
This is best possible as the system $2^{i-1}\pmod{2^i}$ shows. This was proved indepedently by Selfridge and Crittenden and Vanden Eynden [CrVE70].
OPEN
Is there an infinite Lucas sequence $a_0,a_1,\ldots,$ where $(a_0,a_1)=1$ and $a_{n+2}=a_{n+1}+a_n$ for $n\geq 0$ such that all $a_k$ are composite, and yet no integer has a common factor with every term of the sequence?
Whether such a composite Lucas sequence even exists was open for a while, but using covering systems Graham [Gr64] showed that \[a_0 = 1786772701928802632268715130455793\] \[a_1 = 1059683225053915111058165141686995\] generate such a sequence. This problem asks whether one can have a composite Lucas sequence without 'an underlying system of covering congruences responsible'.
OPEN
For which $n$ is there a covering system $\{ a_d\pmod{d} : d\mid n\}$ such that if $x\equiv a_{d_1}\pmod{d_1}$ and $x\equiv a_{d_2}\pmod{d_2}$ then $(d_1,d_2)=1$?
The density of such $n$ is zero.
OPEN
Let $A=\{n_1<\cdots<n_r\}$ be a finite set of integers. What is the minimum value of the density of integers not hit by a suitable choice of congreunces $a_i\pmod{n_i}$? Is the minimum achieved when all the $a_i$ are equal?
OPEN
Let $k\geq 3$. Is there a choice of congruence classes $a_p\pmod{p}$ for every prime $p$ such that all except finitely many integers can be written as $a_p+tp$ for some prime $p$ and integer $t\geq k$?
Even the case $k=3$ seems difficult. This may be true with the primes replaced by any set $A\subseteq \mathbb{N}$ such that \[\lvert A\cap [1,N]\rvert \gg N/\log N\] and \[\sum_{\substack{n\in A\\ n\leq N}}\frac{1}{n} -\log\log N\to \infty\] as $N\to \infty$.

For $k=1$ or $k=2$ any set $A$ such that $\sum_{n\in A}\frac{1}{n}=\infty$ has this property.

OPEN
Let $n_1<n_2<\cdots $ be an infinite sequence of integers with associated $a_i\pmod{n_i}$, such that for some $\epsilon>0$ we have $n_k>(1+\epsilon)k\log k$ for all $k$. Then \[\#\{ m<n_k : m\not\equiv a_i\pmod{n_i} \textrm{ for }1\leq i\leq k\}\neq o(k).\]
Erdős and Graham [ErGr80] suggest this 'may have to await improvements in current sieve methods' (of which there have been several since 1980).
OPEN
Let $n_1<n_2<\cdots$ be an infinite sequence with associated congruence classes $a_i\pmod{n_i}$ such that the set of integers not satisfying any of the congruences $a_i\pmod{n_i}$ has density $0$.

Is it true that for every $\epsilon>0$ there exists some $k$ such that the density of integers not satisfying any of the congruences $a_i\pmod{n_i}$ for $1\leq i\leq k$ is less than $\epsilon$?

The latter condition is clearly sufficient, the problem is if it's also necessary. The assumption implies $\sum \frac{1}{n_i}=\infty$.
SOLVED
Is there a covering system such that no two of the moduli divide each other?
Asked by Schinzel, motivated by a question of Erdős and Selfridge (see [7]). The answer is no, as proved by Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST22].