d$, then we can discard all congruences whose moduli are multiples of $p^a$ and still have a covering. \end{corollary} \begin{proof} First, since $p^a(p+1) < 2p^{a+1}$, there is at most one multiple of $p^{a+1}$ in $[c,d]$ so by Corollary \ref{cor:1} we can discard the congruence with modulus $p^{a+1}$ (if there is one). This leaves us with at most $p-1$ multiples of $p^a$ in $[c,d]$, namely $p^a \cdot 1, \ldots , p^a \cdot (p-1)$. By applying Corollary \ref{cor:1} again, we can discard all moduli divisible by $p^a$. \end{proof} \begin{corollary} [Krukenberg] \label{cor:3} Let ${\mathcal C}$ be a covering such that $p^a || L$ for some prime $p$ and integer $a \geq 1$. Let ${\mathcal C_1}$ be the subset of ${\mathcal C}$ consisting of congruences whose moduli are divisible by $p^a$. Suppose $|{\mathcal C_1 }| = p$ and the moduli of the congruences in ${\mathcal C_1}$ are $p^a m_1, \ldots, p^a m_p$. Then (i) one can replace the congruences in ${\mathcal C_1}$ by a single congruence with modulus $${\displaystyle p^{a-1} \lcm (m_1,\ldots, m_p)}$$ and the resulting set is still a covering. (ii) if two of the above $p$ congruences are in the same class modulo $p^a$ we can discard all $p$ congruences and the resulting set is still a covering. \end{corollary} \begin{proof} \noindent \emph{(i)} Again, we use Lemma \ref{lem:1}. Now, we split the $p$ congruences into $p$ bins $B_l$, with $l \in \{0, 1, \ldots, p-1\}$, as in Lemma \ref{lem:1}. The only case when $D \neq \emptyset$ is when there is exactly one congruence in each bin and when the system of the $p$ congruences reduced modulo $p$ with moduli $p^{a-1} m_1, \ldots, p^{a-1} m_p$ has a solution. By a generalization of the Chinese remainder theorem, if a finite system of congruences has a solution, the system of congruences is equivalent to a single congruence whose modulus is the least common multiple of the congruences in the finite system. Lemma \ref{lem:1} now implies \emph{(i)}. \bigskip \noindent \emph{(ii)} Here we note that since a bin contains two or more congruences, at least one of the remaining bins is empty, so $D = \emptyset $ in this case. Then the conclusion of Lemma \ref{lem:1} implies \emph{(ii)}. \end{proof} Next, we define a {\it minimal covering system}. A minimal covering system ${\mathcal C}$ is a covering such that no proper subset of ${\mathcal C}$ is a covering system. Clearly, by discarding one by one redundant congruences, after a finite number of steps, any finite covering system can be reduced to a minimal covering system in at least one way. Next, we use Lemma \ref{lem:1} to prove Theorem \ref{8} under the assumption that Theorem \ref{m4} holds. \begin{proof}[Proof of Theorem \ref{8} assuming Theorem \ref{m4}.] Assume that for some integer $m \geq 3$ there is a distinct covering ${\mathcal C}$ with all moduli in the interval $[m,8m]$. Let ${\mathcal C_m}$ be a minimal covering which is a subset of ${\mathcal C}$. Consider the least common multiple $L$ of the moduli of the congruences in ${\mathcal C_m}$. By Corollary \ref{cor:1}, if $p^a | L$ for some prime $p$ and a positive integer $a$, then the interval $[m,8m]$ contains at least $p$ multiples of $p^a$ that are not multiples of $p^{a+1}$. Since one of every $p$ consecutive multiples of $p^a$ is divisible by $p^{a+1}$, we deduce that the interval $[m,8m]$ contains at least $p+1$ multiples of $p^a$. Denote by ${\mathcal M}\subseteq [m, 8m]$ the set of moduli from the congruences in ${\mathcal C_m}$. Let $p \geq \sqrt{7m+1}$ be a prime. The number of multiples of $p$ in the interval $[m, 8m]$ is $$n_p :=\left \lfloor {\frac{8m}{p}} \right \rfloor - \left \lfloor {\frac{m-1}{p}} \right \rfloor = {\frac{7m+1}{p}} - \left \{ {\frac{8m}{p}} \right \} + \left \{ {\frac{m-1}{p}} \right \},$$ where $\{ x \}$ denotes the fractional part of $x$. Since for each $x$, $0 \leq \{ x \} < 1$, we get $$n_p <{\displaystyle \frac{7m+1}{p} } +1 \leq \sqrt{7m+1} + 1 \leq p+1.$$ Thus, for each $p \geq \sqrt{7m+1}$, there are less than $p+1$ multiples of $p$ in the interval $[m, 8m]$. Therefore, if $n$ is a modulus of one of the congruences in ${\mathcal C_m}$ (that is $n \in {\mathcal M}$), then all the prime divisors of $n$ are less than $\sqrt{7m+1}$. Since the density of integers covered by a congruence modulo $n$ is $1/n$ and ${\mathcal C_m}$ is a covering, we get \begin{equation} \label{m3} \sum_{\substack{ m \leq n \leq 8m \\ P(n) < \sqrt{7m+1}}} \frac{1}{n} \geq \sum_{n \in {\mathcal M}} \frac{1 }{n} \geq 1, \end{equation} where $P(n)$ denotes the largest prime divisor of $n$. Let $$S_m = \sum_{n \in {\mathcal M}} \frac{1}{n} \quad \text{ and } \quad T_m = \sum_{ \substack{m \leq n \leq 8m, \\ P(n) < \sqrt{7m+1}}} \frac{1}{n}.$$ We checked by direct computation and by using the inequality $T_{m-1} \leq T_m + \frac{1}{m-1}$ that $T_m < 1$ for all $m \in [51,616000]$. Details on this computation are in Appendix \ref{Appendix 1} to the paper. Since Balister et al.\ \cite{Bal0} showed that the minimum modulus of a distinct covering system does not exceed $616000$, Theorem \ref{8} holds when $m \geq 51$. Also, since Krukenberg showed that there is no distinct covering system with moduli in $[3,35]$, Theorem \ref{8} holds when $m=3$, and $4$. Furthermore, given Theorem \ref{m4}, there is no distinct covering system with moduli in $[4,59]$; therefore Theorem \ref{8} holds when $m=5, 6,$ and $7$. There are nine occasions when $m \in [8,50]$ and $T_m \geq 1$. They are shown below. \begin{center} $\begin{array}{|c|c|c|c|} \hline m & 8 & 9 & 10 \\ \hline T_m & 1.26537840136054\ldots & 1.168553004535147\ldots & 1.08327522675737\ldots \\ \hline \hline m & 11 & 12 & 18 \\ \hline T_m & 1.007525667674477\ldots & 1.029053445452255\ldots & 1.037818616878952\ldots \\ \hline \hline m & 20 & 25 & 26 \\ \hline T_m & 1.008475955572005\ldots & 1.0628503734652\ldots & 1.0276580657728\ldots \\ \hline \end{array}$ \end{center} So far, we have used Corollary \ref{cor:1} only with $a=1$. Next, we use Corollary \ref{cor:2} for all $a \geq 1$. Define $$L_m= \begin{cases} 720 = 2^4\cdot 3^2 \cdot 5, & \text{ if } m=8; \\ 5040 = 2^4\cdot 3^2 \cdot 5 \cdot 7, & \text{ if } m\in \{9, 10, 11\}; \\ 10080 = 2^5\cdot 3^2 \cdot 5 \cdot 7 , &\text{ if } m=12; \\ 332640 = 2^5\cdot 3^3 \cdot 5 \cdot 7 \cdot 11, & \text{ if } m=18; \\ 1663200 = 2^5\cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11, & \text{ if } m=20; \\ 43243200 =2^6\cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13, & \text{ if } m\in \{25, 26\}. \end{cases}$$ Using Corollary \ref{cor:2}, one checks directly that $L$ divides $L_m$ for each $$m\in\{8, 9, 10, 11, 12, 18, 20, 25, 26\}.$$ Then for such $m$ we have \begin{equation} \label{eq3} \sum_{n\in M} \frac{1}{n} \leq \sum_{\substack{d \mid L_m \\ m \leq d \leq 8m}} \frac{1}{d} < 1, \end{equation} where the last inequality is done by a direct computation. As \eqref{eq3} contradicts the second inequality in \eqref{m3}, the proof is complete. \end{proof} A natural question is whether one can replace the constant $8$ in Theorem \ref{8} by a larger constant? If we try to prove Theorem \ref{8} with a constant $9$, the estimate of $T_m$ for large $m$ is similar to what we did above. However, there are $97$ values of $m$ for which $T_m > 1$ and dealing with these would make the proof of the theorem much longer. Moreover, dealing with some of the smaller exceptions $m$ would require more intricate approach than the one in the proof of Theorem~\ref{8}. \section{Reduction of a covering}\label{sec:3} Our second tool is {\it reduction of a covering}. We start with an example. Consider the covering \eqref{eq:m2}. Let $a \in \{ 0,1,2 \}$. Since \eqref{eq:m2} is a covering, the residue class $3m + a$ is covered by the congruences in \eqref{eq:m2}. Substituting $3m + a$ for $x$ in each of the congruences of \eqref{eq:m2} and solving for $m$ we get a new covering system. When $a=0$ we get $m \equiv \modd{1} {2}$, $ m \equiv \modd{2} {4}$, $ m \equiv\modd{0} {1}$, and two congruences have no solution; when $a=1$ we obtain $m \equiv \modd{0} {2}$, $ m \equiv \modd{3} {4}$, $ m \equiv \modd{1} {2}$, and two congruences have no solution; finally, when $a=2$ we have $ m \equiv \modd{1} {2}$, $ m \equiv \modd{0} {4}$, $ m \equiv \modd{2} {4}$, and two congruences have no solution. In general, let ${\mathcal C}$ be a covering system and let $p$ be a prime. Let ${\mathcal C_0}$ be the subset of ${\mathcal C}$ of congruences whose moduli are not divisible by $p$. Let ${\mathcal M_0}$ be the list of the moduli of the congruences in ${\mathcal C_0}$. Similarly, let ${\mathcal C_1}$ be the subset of ${\mathcal C}$ of congruences whose moduli are divisible by $p$, and let ${\mathcal M_1}$ be the list of the moduli of the congruences in ${\mathcal C_1}$. To reduce the covering modulo $p$ for each $a \in \{ 0,1,\ldots, p-1 \}$ we substitute $pm+a$ for $x$ in each of the congruences of ${\mathcal C}$ and solve for $m$ to get a new covering. This way we end up with $p$ coverings in which each modulus in ${\mathcal M_0}$ is used in all $p$ coverings. However, if $m$ is a modulus in ${\mathcal M_1}$ it gets replaced by $m/p$ and it is used in just one of the $p$ coverings. Indeed, if $x \equiv \modd{r} {n}$ is a congruence in ${\mathcal C_0}$ (so $p \nmid n$), substituting $mp+a$ for $x$ and solving for $m$, we get $m \equiv \modd{p^{-1}(r-a)} {n}$. However, if $x \equiv \modd{r} {n}$ is a congruence in ${\mathcal C_1}$ (so $p | n$), substituting $mp+a$ for $x$, we get the congruence $mp+a \equiv \modd{r} {n}$. The last congruence has a solution if and only if $r \equiv \modd{a} {p}$, in which case we get $m \equiv \modd{(r-a)/p} {n/p}$. Next, we say that two congruences are in the same class modulo a positive integer $q$, if the integers covered by the congruences all belong to one class modulo $q$. In other words, if the two congruences are $x \equiv \modd{r_1} {n_1}$ and $x \equiv \modd{r_2} {n_2}$, we say that they are in the same class modulo $q$ if $q | n_1$, $q | n_2$, and $r_1 \equiv \modd{r_2} {q}$. Assume two congruences $x \equiv \modd{r_1} {n_1}$ and $x \equiv \modd{r_2} {n_2}$, both in ${\mathcal C_0}$ are in the same class modulo $q$ with $p \nmid q n_1 n_2$. After reduction modulo $p$, we get $mp \equiv \modd{( r_1-a)} {n_1}$ and $mp \equiv \modd{( r_2-a)} {n_2}$. Suppose that the reduced congruences are $m \equiv \modd{r_1'} {n_1}$ and $m \equiv \modd{r_2'} {n_2}$. Then $mp \equiv r_1'p \equiv \modd{(r_1-a)} {n_1}$ and $mp \equiv r_2'p \equiv \modd{(r_2-a)} {n_2}$. Since $q | n_1$, $q | n_2$, we obtain $r_1'p \equiv \modd{r_2'p} {q}$. Furthermore, $p \nmid q $, so $r_1' \equiv \modd{r_2'} {q}$. Therefore, the reduced congruences are still in the same class modulo $q$. Conversely, arguing in the same way one gets that if the two congruences $x \equiv \modd{r_1} {n_1}$ and $x \equiv \modd{r_2} {n_2}$, both in ${\mathcal C_0}$ are {\it not} in the same class modulo $q$ with $p \nmid q n_1 n_2$, then after reduction, the reduced congruences are {\it not} in the same class modulo $q$. To summarize, we showed that the following lemma holds. \begin{lemma} \label{lem:2} Let ${\mathcal C}$ be a covering system and let $p$ be a prime. Let ${\mathcal C_0}$ be the subset of ${\mathcal C}$ of congruences whose moduli are not divisible by $p$. Let ${\mathcal M_0}$ be the list of the moduli of the congruences in ${\mathcal C_0}$. Similarly, let ${\mathcal C_1}$ be the subset of ${\mathcal C}$ of congruences whose moduli are divisible by $p$, and let ${\mathcal M_1}$ be the list of the moduli of the congruences in ${\mathcal C_1}$. Reducing the covering ${\mathcal C}$ modulo $p$ produces $p$ coverings where (i) each modulus in ${\mathcal M_0}$ is used in each of the $p$ coverings but each modulus $n$ in ${\mathcal M_1}$ is replaced by $n/p$ and is used in just one of the $p$ coverings, and (ii) if two congruences in ${\mathcal C_0}$ are in the same class modulo a positive integer $q$, then after reduction they are in the same class modulo $q$ in each of the $p$ coverings; furthermore, if two congruences in ${\mathcal C_0}$ are not in the same class modulo a positive integer $q$, then after reduction they are not in the same class modulo $q$. \end{lemma} With the risk of stating the obvious and erring on the side of clarity, we state the following lemma. \begin{lemma} \label{obvious} Let $\modd{r_1} {m_1}$ and $\modd{r_2} {m_2}$ be two congruence (residue) classes with $m_{1} \mid m_{2}$. If there is an integer which belongs to both congruence classes, then every integer in the congruence class $\modd{r_2} {m_2}$ is in the congruence class $\modd{r_1} {m_1}$. \end{lemma} \begin{proof} Assume that there is an integer $r$ in both $ \modd{r_1} {m_1}$ and $\modd{ r_2} {m_2}$. Then $ r \equiv \modd{r_1} {m_1}$ and $ r \equiv \modd{r_2} {m_2}$. Since $m_1 \mid m_2$, $ r \equiv \modd{r_2} {m_2}$ implies $ r \equiv \modd{r_2} {m_1}$. Thus, $ r_1 \equiv r \equiv \modd{r_2} {m_1}$. Let $m$ be an integer in the residue class $\modd{r_2} {m_2}$, that is, $ m \equiv \modd{r_2} {m_2}$. Then $ m \equiv \modd{r_2} {m_1}$ and since $ r_1 \equiv \modd{r_2} {m_1}$, we obtain $ m \equiv \modd{r_1} {m_1}$, so $m$ is in the residue class $\modd{r_1} {m_1}$. Therefore, if the residue classes $\modd{r_1} {m_1}$ and $\modd{r_2} {m_2}$ intersect, then every element of $\modd{r_2} {m_2}$ is an element of $\modd{r_1} {m_1}$. \end{proof} We use the above lemma as follows. Suppose $x \equiv \modd{r_1} {m_1}$ and $x \equiv \modd{r_2} {m_2}$ are two congruences in a certain distinct covering ${\mathcal C}$ and $m_1 \mid m_2$. If the sets of integers covered by the two congruences intersect, then by Lemma \ref{obvious}, every integer covered by $x \equiv \modd{r_2} {m_2}$ is covered by $x \equiv \modd{r_1} {m_1}$. Thus, we can discard the congruence $x \equiv \modd{r_2} {m_2}$ from the covering and still have a covering. Now, if we add a congruence to a covering, we still have a covering, so we can change $r_2$ to $r_2'$ so that the two congruences do not intersect. In fact, if the congruences in ${\mathcal C}$ with moduli which are proper divisors of $m_2$ do not form a covering, without loss of generality, we can assume that the congruence $x \equiv \modd{r_2} {m_2}$ does not intersect any of these congruences. For example, in a distinct covering, without loss of generality, we can assume that a congruence modulo $16$ does not cover any integers covered by the congruences modulo $4$ and $8$. Next, we prove that there is no distinct covering system with moduli in the interval $[4,59]$ using a proof by contradiction. Our proof proceeds as follows. First, we use Corollary \ref{cor:2} and Corollary \ref{cor:3} to reduce the list of possible moduli in the covering to $17$ integers. Next, we reduce the covering modulo $3$. We explore all ways in which it is possible to construct the three coverings from Lemma \ref{lem:2} which satisfy condition (i) of the lemma. It turns out, this can be done in two ways. In both cases, we obtain a contradiction by showing condition (ii) of Lemma \ref{lem:2} with $q=5$ is violated. The reason the details of the proof of Theorem \ref{m4} are somewhat complicated is that using moduli in $[4,59]$ one can get very close to a covering; more precisely, one can cover $179$ out of $180$ classes modulo $180$. In Appendix \ref{Appendix 2} we give an example of a distinct covering system using congruences with moduli in $[4,56]$ and a congruence modulo $180$. \begin{proof} [Proof of Theorem \ref{m4}] Assume that there exists a distinct covering ${\mathcal C}$ with all moduli in $[4,59]$. Since every covering contains a subset which is a minimal covering, without loss of generality, we can assume that ${\mathcal C}$ is a minimal covering. Let ${\mathcal M}$ be the set of the moduli of the congruences in ${\mathcal C}$. Also, let $L$ be the least common multiple of the moduli in ${\mathcal M}$. By Corollary \ref{cor:2}, if $p^a(p+1) > 59$ for some prime $p$ and a positive integer $a$, then $p^a$ does not divide any modulus in ${\mathcal M}$, so $p^a \nmid L$. Since, $2^5 \cdot 3 > 59$, $3^3 \cdot 4 > 59$, $7^2 \cdot 8 > 5^2 \cdot 6 > 59$, and $p(p+1) > 59$ for $p \geq 11$, we get $L | \left ( 2^4 \cdot 3^2 \cdot 5 \cdot 7 \right )$. Therefore, $${\mathcal M} \subseteq \{ 4,8,16, 6,12,24,48, 9,18,36, 5,10,20,40, 15,30,45, 7,14,21,28,35,42,56 \}.$$ Without loss of generality, we can assume that $${\mathcal M} = \{ 4,8,16, 6,12,24,48, 9,18,36, 5,10,20,40, 15,30,45, 7,14,21,28,35,42,56 \}.$$ Indeed, for each modulus $m$ is in the displayed set above which is not in ${\mathcal M}$, we simply add a congruence $x \equiv \modd{0} {m}$ to ${\mathcal C}$. When analyzing or constructing a covering using a given set of moduli, following Krukenberg \cite{Krukenberg}, Nielsen \cite{Nielsen}, Balister et al.\ \cite{Bal1}, we use the moduli in increasing order of arithmetic complexity. For example, above we first list powers of $2$, next, powers of $2$ times $3$, etc. and we do not introduce moduli which are multiples of a not yet used prime $p$, until moduli with all prime divisors less than $p$ are used. Next, by Corollary \ref{cor:3} we can replace the seven congruences in ${\mathcal C}$ with moduli divisible by $7$ by a single congruence modulo $120 =\mbox{ lcm}(1,2,3,4,5,6,8)$ and still have a covering. Also, by Corollary \ref{cor:3} we can replace the two congruences with moduli $16,48$ by a single congruence modulo $24$ and still have a covering. Denote the resulting covering by ${\mathcal C'}$ and denote the {\it list} of the moduli of the congruences in ${\mathcal C'}$ by ${\mathcal M'}$. Then ${\mathcal M'}$ is the list $[ 4,8, 6,12,24, 24, 9,18,36, 5,10,20,40, 15,30,120, 45 ]$, where $[\cdots ]$ is used to emphasize that ${\mathcal M'}$ is a list. Note that $24$ appears twice in the last list meaning that we have two congruences modulo $24$ in ${\mathcal C'}$. This concludes the first part of the proof. Next, we reduce the covering ${\mathcal C'}$ modulo $3$ and obtain three coverings, say ${\mathcal C_0'}$, ${\mathcal C_1'}$, ${\mathcal C_2'}$, whose moduli are ${\mathcal M_0'}$, ${\mathcal M_1'}$, ${\mathcal M_2'}$, respectively. By Lemma \ref{lem:2} the moduli in ${\mathcal M_0} = \{ 4,8,5,10,20,40 \}$ can be used in all three coverings, and each modulus in $${\mathcal M_1} = [ 2*,4*,8*,8*,3*,6*,12*,5*,10*,40*,15* ]$$ can be used in just one of the coverings. We use * to the right of a modulus to indicate that it can be used in at most one of the three coverings ${\mathcal C_0'}$, ${\mathcal C_1'}$, ${\mathcal C_2'}$. After relabeling, we can assume that $2*$ is in ${\mathcal M_0'}$. Moreover, by Corollary \ref{cor:1}, we can take congruences with moduli $3*,6*,12*,15*$ in only one of ${\mathcal C_0'}$, ${\mathcal C_1'}$, ${\mathcal C_2'}$. (If we have just one or two congruences with moduli divisible by $3$ in a covering, we can discard them.) It is relatively easy to see that we can take $3*,6*,12*,15*$ to be in ${\mathcal M_1'}$ or ${\mathcal M_2'}$. Indeed, $2*,4,8$ are already in ${\mathcal M_0'}$. If $8*$ is also in ${\mathcal M_0'}$, no additional congruences are needed to construct the covering ${\mathcal C_0'}$. Note that the congruences with moduli $3*,6*,12*$ can cover a congruence class modulo $4$ (and some integers outside of it). If $8*$ is in ${\mathcal M_1'}$ or ${\mathcal M_2'}$ and $3*,6*,12*,15*$ are in ${\mathcal M_0'}$, we can swap the congruences with moduli $3*,6*,12*,15*$ and the congruence modulo $8*$ and we still have three coverings. Thus, after relabeling we can assume that $3*,6*,12*,15*$ are in ${\mathcal M_1'}$. Next, we analyze how to allocate the moduli $5*,10*,40*$. We claim that ${\mathcal M_2'}$ contains at least one of the moduli $5*,10*,40*$. Otherwise, $${\mathcal M_2'} \subseteq [ 4,8,5,10,20,40, 4*,8*,8* ].$$ Using Corollary \ref{cor:1} we discard any of $5,10,20,40$ from ${\mathcal M_2'}$, leaving us with the impossible task of constructing a covering using only congruences with moduli $4,8,4*,8*,8*$. So, we allocated $5$ out of $11$ moduli in ${\mathcal M_1}$ and have partial information about three of the remaining six moduli. It is possible to allocate the moduli $4*,8*,8*,5*,10*,40*$ and to construct the three coverings. However, we will show that it cannot be done without violating condition (ii) of Lemma \ref{lem:2}. To this end, we consider two cases. \bigskip \noindent{\it Case I:} The congruences with moduli $20$ and $40$ in ${\mathcal C'}$ are not in the same class modulo $5$. \bigskip By Lemma \ref{lem:2} the congruences with moduli $20$ and $40$ are not in the same class modulo $5$ in ${\mathcal C_0'}$, ${\mathcal C_1'}$, and ${\mathcal C_2'}$, as well. First, note that currently ${\mathcal M_0'}$ contains $\{2*, 4,8,5,10,20,40 \}$ which is not sufficient to construct a covering. (Discard $5,10,20,40$ using Corollary \ref{cor:1} and we are left only with moduli $2*,4,8$.) We assign $40*$ to ${\mathcal M_0'}$, so that covering is possible with moduli from ${\mathcal M_0'}$. All the remaining five moduli $4*,8*,8*,5*,10*$ are divisors of $40*$, so the remaining two coverings cannot benefit from us assigning a different modulus instead of $40$ to ${\mathcal M_0'}$. Next, we concentrate on allocating $5*,10*$. We proved above that ${\mathcal M_2'}$ contains at least one of the moduli $5*,10*$ (since $40*$ is already allocated to ${\mathcal M_0'}$). There are two subcases. \bigskip \noindent{\it Subcase A:} Exactly one of the moduli $5*,10*$ is in ${\mathcal M_2'}$. \bigskip In this subcase, $${\mathcal M_2'} \subseteq [ 4,8,5,10,20,40, 4*,8*,8*, 5*],$$ or $${\mathcal M_2'} \subseteq [ 4,8,5,10,20,40, 4*,8*,8*, 10*].$$ By Corollary \ref{cor:3} we can replace $5,10,20,40,$ and one of $5*$ and $10*$ by a congruence modulo $8$. Now, we need to construct a covering using moduli $4,8,8$ and some of $4*,8*,8*$. The covering ${\mathcal C_2'}$ can be completed only if all three moduli $4*,8*,8*$ are in ${\mathcal M_2'}$. Without loss of generality, we may assume we are left with $${\mathcal M_1' } = [ 4,8,5,10,20,40, 3*,6*,12*,15*, 5* ].$$ We finish this subcase by proving the following lemma. \begin{lemma} \label{lem:3} There is no covering with congruences whose moduli form the list $$ [ 4,8, 3,6,12,5,5, 10,20,40,15 ],$$ such that the congruence with modulus $20$ and the congruence with modulus $40$ are not in the same class modulo $5$. \end{lemma} \begin{proof} We assume that there is such a covering and use Lemma \ref{lem:2} to reduce the covering modulo $3$. We get three coverings where each has moduli in the list $ [ 4,8, 5,5, 10,20,40 ]$ and each of the moduli in $[ 1*, 2*, 4*, 5* ]$ is used in exactly one covering. Consider one of the three coverings, say ${\mathcal C'' }$, which does not use the moduli $1*$ and $2*$. The moduli of ${\mathcal C'' }$ are in the list $ [ 4,4*, 8, 5,5, 5*, 10,20,40 ]$. Next, we apply Lemma \ref{lem:1} with $p=5$ to the congruences with moduli $5,5,5*,10,20,40$. After reduction modulo $5$ we need to place the reduced congruences with moduli $1,1,1,2,4,8$ in five bins and take the intersection of the congruences in the five bins. Consider a bin which does not contain a congruence with modulus one of $1,1,1,2$. This bin contains the congruence modulo $4$ or the congruence modulo $8$ but not both, since $4$ and $8$ are not in the same bin (the congruences with moduli $20$ and $40$ are not in the same class modulo $5$). So, $D$ is inside a residue class modulo $4$. Thus, we can replace the congruences with moduli $5,5,5*,10,20,40$ by a single congruence modulo $4$. This is a contradiction because it requires building a covering with congruences with moduli $4,4,4,8$. \end{proof} \bigskip \noindent{\it Subcase B:} Both moduli $5*,10*$ are in ${\mathcal M_2'}$. \bigskip Here $$ {\mathcal M_2'}\subseteq [ 4,8,5,10,20,40, 4*,8*,8*, 5*,10*].$$ We apply Lemma \ref{lem:1} with $p=5$ to the congruences with moduli $5,10,20,40,5*,10*$. Proceeding word for word as in the proof of Lemma \ref{lem:3} we get that we can replace the congruences with moduli $5,10,20,40,5*,10*$ by a single congruence modulo $4$. Now, we need to construct a covering using moduli $4,8,4$ and some of the moduli $4*,8*,8*$. This can be done only if $4*$ and at least one $8*$ are in ${\mathcal M_2'}$. This leaves $${\mathcal M_1' } = [ 4,8,5,10,20,40, 8*, 3*,6*,12*,15*].$$ Next, by using Corollary \ref{cor:3} we replace the congruences with moduli $5,10,20,40, 15*$ by a single congruence modulo $24$. So, we need to construct a covering using moduli $4,8, 8, 3,6,12,24$. Assume there is such a covering and reduce it modulo $3$. We get three new coverings with common moduli $4,8,8$ and moduli to be used by just one of the three coverings: $1*,2*,4*,8*$. Consider the covering which does not contain $1*,2*$. The moduli of the congruences of this covering are at most $4,8,8,4*,8*$, which is a contradiction. This completes Case I. \bigskip \noindent{\it Case II:} The congruences with moduli $20$ and $40$ in ${\mathcal C'}$ are in the same class modulo $5$. \bigskip First, we claim that ${\mathcal M_2'}$ contains at least two of the moduli $5*,10*,40*$. Assume otherwise. Then ${\mathcal M_2'}$ contains at most the moduli $4,8,5,10,20,40, 4*,8*, 8*$ and one of the moduli $5*,10*, 40*$. By Corollary \ref{cor:3} we can discard from ${\mathcal C_2'}$ the congruences with moduli $5,10,20,40$ and the congruence with modulus $5*$, $10*$, or $40*$ since two of these congruences are in the same class modulo $5$. This leaves us at most with moduli $4,8,4*,8*,8*$ which are not sufficient to construct a covering, proving the claim. Thus, ${\mathcal M_0'}$ contains at most one of the moduli $5*,10*,40*$. Allocating just one of $5*, 10*, 40*$ to ${\mathcal M_0'}$ does not help construct ${\mathcal C_0'}$. For example, if $5*$ is in ${\mathcal M_0'}$ and $10*,40*$ are not in ${\mathcal M_0'}$, again, using Corollary \ref{cor:3} we can discard from ${\mathcal C_0'}$ the congruences with moduli $5,10,20,40,5*$. To complete ${\mathcal C_0'}$ we need to assign to ${\mathcal M_0'}$ one of the moduli $4*,8*,8*$. It is possible to construct the covering ${\mathcal C_0'}$ by assigning to it one congruence modulo $8*$ and it is the efficient way to do it. (The coverings ${\mathcal C_1'}$ and ${\mathcal C_2'}$ cannot benefit from swapping with ${\mathcal C_0'}$ a congruence modulo $4$ with a congruence modulo $8$.) So, one modulus $8*$ is allocated to ${\mathcal M_0'}$. Next, we analyze how to split the moduli $5*,10*,40*$ among ${\mathcal M_1'}$ and ${\mathcal M_2'}$. We claim that both moduli $5*$ and $10*$ are in ${\mathcal M_2'}$. Assume otherwise. Then ${\mathcal M_2'}$ contains at most the moduli $4,8,5,10,20,40, 4*,8*, 40*$ and one of the moduli $5*,10*$. Apply Lemma \ref{lem:1} to the congruences in the above list which are multiples of $5$. Note that in Case II, $4$ and $8$ are in the same bin and we have either $1$ or $2$ in a bin depending on whether $5*$ or $10*$ is in ${\mathcal M_2'}$. The only way that $D$ is a nonempty set is when we have $$|1 | 2 | 4,8 | 1 | 8| \quad \mbox { or } \quad |1 | 2 | 4,8 | 2 | 8| $$ in the five bins (here we use $|$ as a separator between the bins and for brevity, instead of writing `the congruence modulo $2$ is in a certain bin' we just write `$2$ is in the bin'). Thus, we can replace the congruences with moduli which are multiples of $5$ by a single congruence modulo $8$. Now, we need to construct a covering with moduli from the list $4,8,4*,8*,8$ which is impossible. Thus, we need to allocate both $5*$ and $10*$ to ${\mathcal M_2'}$. There are two subcases depending on how we allocate $40*$. \bigskip \noindent {\it Subcase A:} The modulus $40*$ is in ${\mathcal M_1'}$. \bigskip In this subcase ${\mathcal M_2'}$ contains the moduli $4,8,5,10,20,40, 5*,10*$ and some of the moduli $4*,8*$. We apply Lemma \ref{lem:1} again to the congruences with moduli $5,10,20,40, 5*,10*$. In this case $D$ is nonempty only if we have $|1 | 2 | 4, 8 | 1 | 2|$ in the bins (again, $4$ and $8$ must be in the same bin). Therefore, we can replace the congruences with moduli $5,10,20,40, 5*,10*$ by two congruences with moduli $4,8$ respectively. Now, we are left with moduli $4,8,4,8$ and some of $4*,8*$. So, we need to allocate $4*$ to ${\mathcal M_2'}$. This leaves for ${\mathcal M_1'}$ the moduli $4,8,5,10,20,40,3*,6*,12*,15*,8*,40*$. We apply Lemma \ref{lem:1} again, this time to the congruences with moduli $5,10,20,40,15*,40*$. Again, $D$ is nonempty only if the content of the bins is $|1 | 2 | 4, 8| 3 | 8|$ (in some order). Thus, we can replace the congruences with moduli $5,10,20,40,15*,40*$ by a single congruence modulo $24$. We are left with moduli $4,8,3*,6*,12*,24*,8*$. We already proved in Case 1, Subcase B that it is not possible to construct a covering with moduli from the last list. The same proof works word for word here, too, so we are done with this subcase. \bigskip \noindent{\it Subcase B:} The modulus $40*$ is in ${\mathcal M_2'}$. \bigskip In this subcase, ${\mathcal M_2'}$ contains the moduli $4,8,5,10,20,40, 5*,10*,40*$ and some of the moduli $4*,8*$. We apply Lemma \ref{lem:1} again to the congruences with moduli divisible by $5$. The congruences reduced modulo $5$ have moduli $1,2,4,8,1,2,8$ respectively. We need to place seven congruences in five bins, so at least one bin contains only one congruence modulo $2, 4$, or $8$. Thus, one can replace the congruences with moduli $5,10,20,40, 5*,10*,40*$ by a single congruence modulo $2$. This leaves ${\mathcal M_2'}$ with moduli $2,4,8$ and some of $4*,8*$, so we allocate $8*$ to ${\mathcal M_2'}$. Now, ${\mathcal M_1'} = [4,8,5,10,20,40,3*,6*,12*,15*,4*]$. We apply Corollary \ref{cor:3} to the congruences with moduli $5,10,20,40,15*$. Since the congruences with moduli $20$ and $40$ in ${\mathcal C'}$ are in the same class modulo $5$, we can discard the congruences with moduli $5,10,20,40,15*$. This leaves us the moduli $4,8,3,6,12,4$. We apply Corollary \ref{cor:3} again to replace the congruences with moduli $3,6,12$ by a single congruence modulo $4$. Finally, we are left with moduli $4,8,4,4$ which is not sufficient to construct a covering. Having exhausted all cases, we obtain the proof of the theorem. \end{proof} \section{Covering systems with minimum least common multiple of the moduli} \label{sec:4} In this section we solve the problem of minimizing the least common multiple of the moduli of a distinct covering system with a fixed minimum modulus $m$ in the cases $m=3$ and $m=4$. First, we need a lemma which will help us reduce the number of cases we need to consider. This lemma is Theorem 1 of Simpson and Zeilberger \cite{SimZeil} with the extra condition that the minimum modulus does not change. \begin{lemma} \label{lem:4} Let ${\mathcal C}$ be a distinct covering with minimum modulus $m$, and least common multiple of the moduli $L = L_1 q^{\alpha}$, where $q$ is a prime, $\alpha \geq 1$, and $q \nmid L_1 m$. Suppose $p$ is a prime which does not divide $L_1$, and $m \leq p < q$. Then one can construct a distinct covering ${\mathcal C_1}$ with the same minimum modulus $m$, and such that the least common multiple of the moduli divides $L_1 p^\alpha$. \end{lemma} \begin{proof} This proof relies on the coordinate notation for congruences we introduced. Note, that the $j$th coordinate corresponds to the $j$th prime in the prime factorization of $L$. Up to this point, everywhere the coordinate notation was used, the $j$th prime divisor of $L$ was simply the $j$th prime. The proof of this lemma and the example immediately after the lemma are the only places where coordinate notation is used and the $j$th prime divisor of $L$ may be different from the $j$th prime. Let $q$ be the $j$th prime in the prime factorization of $L$. Write all congruences in ${\mathcal C}$ in coordinate notation. We keep in ${\mathcal C_1}$ the congruences in which all base $q$ digits in the $j$th coordinate are $\leq p-1$ with no change and discard the remaining congruences. In the congruences which survived, we interpret the $j$th component modulo $p$. We claim that the congruences in ${\mathcal C_1}$ form a covering with least common multiple of the moduli $L_1 p^\alpha$. First, consider a residue class $r_1$ modulo $L_1 p^\alpha$. Write $r_1$ in coordinate notation. Note that all digits in the $j$th position do not exceed $p-1$. The residue class $r_1$ corresponds to a residue class $r$ modulo $L_1 q^\alpha$ where we have kept all digits in all positions the same (but interpreted the digits in $j$th position modulo $q$). Since ${\mathcal C}$ is a covering, there is a congruence $c$ in ${\mathcal C}$ which covers the residue class $r$. The congruence $c$ corresponds to a congruence $c_1$ in ${\mathcal C_1}$, where both congruences have the same digits in all positions in coordinate notation. Clearly, $c_1$ covers $r_1$, so ${\mathcal C_1}$ is a covering. Moreover, ${\mathcal C}$ is a distinct covering, so by construction ${\mathcal C_1}$ is a distinct covering (all we do is replace $q$ by $p$ in the prime factorization of the moduli and discard some congruences). Now, we need to show that the minimum modulus of ${\mathcal C_1}$ is still $m$. First, since $q \nmid m$ the congruence modulo $m$ is not discarded. Next, since the new congruences we created all have moduli which are multiples of $p$ and $p \geq m$ we did not include in ${\mathcal C_1}$ any congruences with moduli less than $m$. \end{proof} For example, consider the covering ${\mathcal C}$ with $L = 80 = 2^4 5$, $(1)$, $(01)$, $(001)$, $(0001)$, $(*|\ 4)$, $(0|\ 3)$, $(00|\ 2)$, $(000|\ 1)$, $(0000|\ 0)$. Proceeding as in the proof of Lemma \ref{lem:4} with $q=5$ and $p=3$, we get the covering ${\mathcal C_1}$ with $L = 48 = 2^4 3$, $(1)$, $(01)$, $(001)$, $(0001)$, $(00|\ 2)$, $(000|\ 1)$, $(0000|\ 0)$. Now, we turn to Theorem \ref{lcm}. Erd\H{o}s constructed a covering ${\mathcal C}$ with least modulus $m = 3$. Krukenberg \cite{Krukenberg} also constructed a covering ${\mathcal C}$ with least modulus $m = 3$, $L({\mathcal C}) = 120$, without using the moduli $40$ and $120$. Here is a covering with the above properties $(11)$, $(101)$, $(*|\ 2 )$, $(0|\ 1)$, $(100 |\ 1)$, $(10|\ 0)$, $(*|\ *|\ 4)$, $(0|\ *|\ 3)$, $(*|\ 0|\ 2)$, $(0|\ 0 |\ 1)$, $(01|\ *|\ 0)$, $(00|\ 0|\ 0)$. Next, we prove Theorem \ref{lcm}. \begin{proof} [Proof of Theorem \ref{lcm}] First, we deal with part (i), the case $m=3$. We need to show that if $n$ is less than $120$ there is no distinct covering having as moduli only divisors of $n$ which are at least $3$. Since the only $n$ less than $120$ for which ${\displaystyle \sum_{d|n, d \geq 3} \frac{1}{ d} \geq 1}$ are $24,36,48,60,72,84,90, 96$, and $108$, we only need to examine the numbers in this list. We can eliminate some cases using the work of Krukenberg on coverings with least common multiple of the moduli of the form $2^a 3^b$, see Theorem \ref{Krukenberg2}. As proved by Krukenberg, there is no covering with $m=3$, and $L= 24$, or $36$, or $48$, or $72$, or $96$, or $108$. What is left is to consider the cases when $m = 3$ and $L =60$, or $84$, or $90$. By Lemma \ref{lem:4}, if there is a covering with $m = 3$ and $L = 84$, then there is a covering with $m = 3$ and $L = 60$. To finish the proof, we need to show that there is no covering with $m = 3$ and $L=60$ or $L=90$. First, assume that there is a covering ${\mathcal C}$ with $m = 3$ and $L=60$. Then the moduli of the congruences are $4$; $3$, $6$, $12$; $5$, $10$, $20$; $15$, $30$, and $60$. Reduce the covering modulo $3$. We have to construct three coverings with shared moduli: $4$, $5$, $10$, $20$, and moduli used by just one covering: $1*$, $2*$, $4*$, $5*$, $10*$, $20*$. Consider the covering, say ${\mathcal C_1}$ which does not include $1*$, and includes at most one of $5*$, $10*$, and $20*$. Its moduli are $4$, $5$, $10$, $20$, some of $2*, 4*$, and at most one of $5*, 10*, 20*$. We can discard from ${\mathcal C_1}$ all congruences with moduli which are multiples of $5$ (there at most four of them). Thus, ${\mathcal C_1}$ includes both congruences with moduli $2*$ and $4*$. Let ${\mathcal C_2}$ be the covering including the congruence modulo $1*$. Then the moduli of the congruences in ${\mathcal C_3}$ are at most $4$, $5$, $10$, $20$, $5*$, $10*$, $20*$. The sum of the reciprocals of these moduli is at most $.95 < 1$, so a covering with $m = 3$ and $L=60$ does not exist. Finally, assume that there is a covering ${\mathcal C}$ with $m = 3$ and $L=90$. Then the moduli of the congruences are $3$, $6$; $9$, $18$; $5$, $10$; $15$, $30$; $45$, and $90$. Reduce the covering modulo $5$. We obtain five coverings with shared moduli: $3$, $6$, $9$, $18$, and moduli used by just one covering: $1*$, $2*$, $3*$, $6*$, $9*$, $18*$. Consider the two coverings that do not contain any congruences modulo $1*$, $2*$, or $3*$. Since the sum of the reciprocals of $3, 6, 9, 18$ is $2/3$, both coverings need all three moduli $6*$, $9*$, $18*$. Thus, a covering with $m = 3$ and $L=90$ does not exist completing the proof of part (i) of the theorem. Now, we turn to part (ii), the case $m=4$. Krukenberg \cite{Krukenberg} constructed a covering ${\mathcal C}$ with $m = 4$ and $L({\mathcal C}) = 360$. Here is a covering which uses as moduli all divisors of $360$ which are at least $4$, except $360$. It is $(11)$, $(101)$, $(0|\ 2)$, $(100|\ 2)$, $(01|\ 1)$, $(*|\ 02)$, $(0|\ 01)$, $(100|\ 01)$, $(10|\ 00)$, $(*|\ *|\ 4)$, $(0|\ *|\ 3)$, $(100|\ *|\ 3)$, $(00|\ *|\ 2)$, $(*|\ 1|\ 0)$, $(0|\ 1|\ 1)$, $(10|\ 1|\ 1)$, $(100|\ 1|\ 2)$, $(*|\ 00|\ 0)$, $(0|\ 00|\ 1)$, $(01|\ 00|\ 2)$. We need to show that if $n$ is less than $360$ there is no covering using only distinct divisors of $n$ which are at least $4$. Since the only positive integers $n$ less than $360$ for which ${\displaystyle \sum_{d|n, d \geq 4} \frac{1}{ d} \geq 1}$ are $120 ,168 , 180 ,240 ,252, 280 ,288, 300$, and $336$ we only need to examine these values of $n$. Since, $120 | 240$, it is sufficient to show that $240$ does not work. Using Lemma \ref{lem:4} we can reduce the cases $n=168= 2^3 \cdot 3 \cdot 7$, $n=252=2^2 \cdot 3^2 \cdot 7$, and $n = 336 = 2^4 \cdot 3 \cdot 7$ to $n=120$, $n=180$, and $n=240$ respectively. As proved by Krukenberg, Theorem \ref{Krukenberg2}, $n = 288 = 2^5 3^2$ does not work either. Let us consider the case $n=280$. Here and below, we assume as we may, all divisors of $n$ which are at least $4$, appear as a modulus of some congruence. The sum of the reciprocals of the divisors of $280$ which are at least $4$ is $1.0714\ldots$. However, the congruences modulo $4$, $5$, and $7$ cover a portion of the integers with density $1 - \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{6}{7}= \frac{17}{35}$. Since $ \frac{1}{4} + \frac{1}{5} + \frac{1}{7} - \frac{17}{35} = .1071\ldots$, there is no covering with $m = 4$ and $L = 280$. Next, let $n =300$. The sum of the reciprocals of the divisors of $300$ which are at least $4$ is $1.06$. However, the intersection of the congruences modulo $4$, $5$, and $15$ is at least $ \frac{1}{15} = .0666\ldots $. Therefore, there is no covering with $m = 4$ and $L = 280$. We are left with two remaining cases: $n=180$ and $n=240$. We consider each case in a separate lemma. \begin{lemma} \label{lem:5} There is no distinct covering with $m = 4$ and $L=180$. \end{lemma} \begin{proof} Assume that ${\mathcal C}$ is a covering with moduli $4$; $6$, $12$; $9$, $18$, $36$; $5$, $10$, $20$; $15$, $30$, $60$; $45$, $90$, and $180$. Let $S$ be the set of congruences with moduli $4$, $6$, $12$, $9$, $18$, $36$. Note that the density of the integers covered by congruences in $S$ is at most $\frac{2}{3}$. Indeed, the sum of the reciprocals of the moduli of congruences in $S$ is $\frac{25}{36}$ and the set of integers covered by the congruences modulo $4$ and modulo $9$ intersect. Next, reduce ${\mathcal C}$ modulo $5$. We need to construct five coverings with common moduli $4$, $6$, $12$, $9$, $18$, $36$ and moduli used by just one covering: $1*$, $2*$, $4*$, $3*$, $6*$, $12*$, $9*$, $18*$, $36*$. Consider the three coverings containing the congruences with moduli $1*$, $2*$, $3*$. One can see that either the congruence modulo $2$ and $S$ do not form a covering or the congruence modulo $3$ and $S$ do not form a covering. Otherwise, the set uncovered by $S$ is inside a residue class modulo $2$ and inside a residue class modulo $3$, that is, inside a residue class modulo $6$. This is not possible since the set uncovered by $S$ has density at least $1/3$. Thus, the three coverings containing the congruences with moduli $1*$, $2*$, $3*$ contain at least one more congruence. We can assume it has modulus $36*$ (all other $*$ moduli are divisors of $36$). So, the fourth covering and the fifth covering need to split the moduli $4*$, $6*$, $12*$, $9*$, $18*$. Now, $\frac{1}{4} +\frac{1}{6} + \frac{1}{12} + \frac{1}{9} + \frac{1}{18} = \frac{2}{3} $. Recall that the set uncovered by $S$ has density at least $1/3$. Thus, the only possible way to construct the remaining two coverings is if one covering uses moduli $4*$ and $12*$ and the other covering uses $6*$, $9*$, $18*$. Therefore, we need to be able to construct a covering using the moduli in the list $[4, 4,6, 12, 12,9, 18, 36]$. By Corollary \ref{cor:3} we can replace the congruences with moduli $9$, $18$, $36$ by a single congruence modulo $12$. The moduli of the congruences of the resulting covering are in the list $[4, 4,6, 12, 12,12]$. Since the sum of the reciprocals of the elements of the list $[4, 4,6, 12, 12,12]$ is less that one, it is not possible to construct at least one of the five coverings we needed to construct. \end{proof} The proof of the next lemma is somewhat complicated. However, we expect that the methods used in the proof of the lemma will be useful when analyzing coverings with least common multiple of the moduli of the form $2^a 3^b 5^c$. \begin{lemma} \label{lem:6} There is no distinct covering with $m = 4$ and $L=240$. \end{lemma} \begin{proof} In the proof of this lemma we use the notation $(n,r_n)$ to denote the congruence $x \equiv \modd{r_n} {n}$. Assume that there is a covering $${\mathcal C} = \{ (n, r_n)\ |\ n \in \{ 4,8,16, 6,12,24,48,5,10,20,40,80,15,30,60,120,240\} \}.$$ We introduce notation for some of the parts of ${\mathcal C}$. Let $${\mathcal C_1} = \{ (n, r_n)\ |\ n \in \{ 4,8,16 \} \}, \quad {\mathcal C_3} = \{ (n, r_n)\ |\ n \in \{ 6,12,24,48 \} \},$$ $${\mathcal C_5} = \{ (n, r_n)\ |\ n \in \{ 10,20,40,80 \} \} \mbox{, and } \quad {\mathcal C_{15}} = \{ (n, r_n)\ |\ n \in \{ 15,30,60,120,240 \} \}.$$ Also, let $R$ be the set of the $9$ integers in $[0,15]$ representing the $9$ residue classes modulo $16$ which are not covered by the congruences in ${\mathcal C_1}$. Note that by Lemma \ref{obvious}, without loss of generality, we can assume that the congruences modulo $4$, $8$, and $16$ do not intersect. Let $R_0 = R \cap \{ x \equiv \modd{0} {2} \}$ and $R_1 = R \cap \{ x \equiv \modd{1} {2} \}$. For each $r \in R$ denote by $a_3(r)$ the number of residue classes modulo $48$ of the form $ x \equiv \modd{r} {16} , \ x \equiv \modd{a} {3}$, which are covered by ${\mathcal C_3}$. One way to visualize this is that the residue class $(\modd{r} {16})$ splits into three fibers modulo $48$. The quantity $a_3(r)$ counts how many of these fibers are covered by ${\mathcal C_3}$. Similarly, for each $r \in R$ denote by $a_5(r)$ the number of residue classes modulo $80$ of the form $ x \equiv \modd{r} {16} , \ x \equiv \modd{b} {5}$, which are covered by ${\mathcal C_5}$. Then the number of residue classes modulo $240$ which are not covered by any of the congruences in ${\mathcal C_1}$, ${\mathcal C_3}$, ${\mathcal C_5}$, nor by the congruence $(5, r_5)$ is at least \begin{equation} \label{left mod 15} A:= \sum_{r \in R} (3 - a_3(r))(4-a_5(r)). \end{equation} Note that the congruences in ${\mathcal C_{15}}$ can cover at most $5$ residue classes modulo $240$ which are in the residue class $(\modd{r} {16})$. Thus, for each $r \in R$ we have $(3-a_3(r))(4 - a_5(r)) \leq 5$. Clearly, $(3-a_3(r))(4 - a_5(r)) \neq 5$. So, for each $r \in R$ we have \begin{equation} \label{at most 4} (3 - a_3(r))(4-a_5(r)) \leq 4. \end{equation} Furthermore, for each $r \in R$, \begin{equation} \label{2*3>5} \mbox{ if } \quad a_3(r)a_5(r) \neq 0, \mbox{ then } a_3(r)a_5(r) \geq 2. \end{equation} We give one more observation. Suppose $r_1 \in R$, $r_2 \in R$, and $r_1 \not\equiv \modd{r_2} {2}$. Then the number of residue classes modulo $240$ which are either $\equiv \modd{r_1} {16}$ or $\equiv \modd{r_2} {16}$ and can be covered by ${\mathcal C_{15}}$ is at most $6$. Indeed, the congruence modulo $15$ can cover at most two such classes, and each of $(30,r_{30})$, $(60, r_{60})$, $(120, r_{120})$, and $(240, r_{240})$ can cover at most one. So, in this case \begin{equation} \label{odd even} (3 - a_3(r_1))(4-a_5(r_1))+ (3 - a_3(r_2))(4-a_5(r_2)) \leq 6. \end{equation} We can rewrite \eqref{left mod 15} as \begin{equation} \label{S3 S5} A = \sum_{r \in R} (12 - 4a_3(r) - 3a_5(r)+a_3(r)a_5(r)) = 108 - 4S_3 - 3S_5 + O, \end{equation} where $$S_3 = \sum_{r \in R} a_3(r) \mbox{, }\quad S_5 = \sum_{r \in R} a_5(r) \quad \mbox{, and }\quad O = \sum_{r \in R} a_3(r)a_5(r).$$ The quantity $O$ measures the amount of overlap between ${\mathcal C_3}$ and ${\mathcal C_5}$. Ideally, we want $O$ to be small. If possible, cover one set of $r$'s by ${\mathcal C_3}$ and a different set of $r$'s by ${\mathcal C_5}$, while $S_3$ and $S_5$ are large, that is, cover a lot without much overlap. At least in the case of this lemma, this proves impossible. Next, we get bounds for $S_3$ and $S_5$. For $n \in \{ 2,4,6,8,16 \}$ define $$ M_n = \max_{0 \leq j < n} | R \cap \{ x \equiv \modd{j} {n} \} |.$$ Here $M_n$ is the size of the largest portion of $R$ in a residue class modulo $n$. Then the congruence $(6, r_6)$ can contribute at most $M_2$ to $S_3$, the congruence $(12, r_{12})$ can contribute at most $M_4$, etc. Thus, $S_3 \leq M_2 + M_4 + M_8 + M_{16}$. Similarly, $S_5 \leq M_2 + M_4 + M_8 + M_{16}$. Define $$D_3 = (M_2 + M_4 + M_8 + M_{16}) - S_3 \mbox{ and }\quad D_5 = (M_2 + M_4 + M_8 + M_{16}) - S_5.$$ In a certain sense, $D_3$ and $D_5$ measure the difference between the largest amount we could possibly cover, and what we cover in reality with ${\mathcal C_3}$ and ${\mathcal C_5}$, respectively. For example, if $R$ consists of $1$ class $r$ such that $ r \equiv \modd{0} {2}$, and $8$ classes $r_1$ such that $ r_1 \equiv \modd{1} {2}$, and if we have a congruence $(6, r_6)$ with $r_6 \equiv \modd{0} {2}$, then $D_3 \geq 7$ (we could have covered $8$ residue classes and covered just $1$ instead). Also, the number of residue classes modulo $240$ which can be covered by ${\mathcal C_{15}}$ and are not covered by ${\mathcal C_{2}}$ does not exceed $9 +M_2 + M_4 + M_8 + M_{16}$. Therefore, if ${\mathcal C}$ is a covering, then $A \leq 9 +M_2 + M_4 + M_8 + M_{16}$. Recall that $A$ is the number of residue classes modulo $240$ which are not covered by any of the congruences in ${\mathcal C_1}$, ${\mathcal C_3}$, ${\mathcal C_5}$, nor by the congruence $(5, r_5)$. These classes need to be covered by ${\mathcal C_{15}}$. Define $$D_{15} = 9+ (M_2 + M_4 + M_8 + M_{16})- A.$$ Since by assumption ${\mathcal C}$ is a covering, $D_{15} \geq 0$. Using \eqref{S3 S5}, we get $$9 + 8(M_2 + M_4 + M_8 + M_{16}) \geq 108 + 4D_3 + 3D_5 + D_{15} + O.$$ Since $M_4 \leq 4$, $M_8 \leq 2$, and $M_{16} \leq 1$, we obtain \begin{equation} \label{M2} 8M_2 \geq 43 + 4D_3 + 3D_5 + D_{15} + O. \end{equation} Next, we consider several cases, depending on the structure of ${\mathcal C_1}$. Without loss of generality, we can assume $r_4 = 0$. Since the set of all integers is invariant to translation by an integer, if $\{ (n, r_n) | n \in {\mathcal L} \}$, where ${\mathcal L}$ is a list of moduli, is a covering, then for any integer $a$, $\{ (n, r_n+a) | n \in {\mathcal L}\}$ is also a covering. \bigskip \noindent{\it Case I.} $r_8 \equiv r_{16} \equiv \modd{1} {2}$. \bigskip In this case, $|R_0|=4$, $|R_1|=5$, and $M_2 = 5$. From \eqref{M2} we get $0 \geq 3 + 4D_3 + 3D_5 + D_{15} + O$. Since $D_3 \geq 0$, $D_5 \geq 0$, $D_{15} \geq 0$, and $O \geq 0$, we get a contradiction. There is no covering in Case I. \bigskip \noindent{\it Case II.} $r_8 \equiv \modd{1} {2}$ and $ r_{16} \equiv \modd{0} {2}$. \bigskip In this case, $|R_0|=3$, $|R_1|=6$, and $M_2 = 6$. From \eqref{M2} we get $5 \geq 4D_3 + 3D_5 + D_{15} + O$. Thus, $D_3 \leq 1$ and $D_5 \leq 1$. Hence, $r_6 \equiv \modd{1} {2}$ and $r_{10} \equiv \modd{1} {2}$. We obtain that $a_3(r) \geq 1$ and $a_5(r) \geq 1$ for all $r \in R_1$, so $O \geq 6$, a contradiction in this case, too. \bigskip \noindent {\it Case III.} $r_8 \equiv \modd{0} {2}$ and $ r_{16} \equiv \modd{1} {2}$. \bigskip In this case, $|R_0|=2$, $|R_1|=7$, and $M_2 =7$. From \eqref{M2} we get $13 \geq 4D_3 + 3D_5 + D_{15} + O$. Therefore $D_3 \leq 3$ and $D_5 \leq 4$. Again, $r_6 \equiv r_{10} \equiv \modd{1} {2}$. So, $a_3(r) \geq 1$ and $a_5(r) \geq 1$ for all $r \in R_1$. By \eqref{2*3>5}, $a_3(r)a_5(r) \geq 2$ for all $r \in R_1$. Therefore, $O \geq 14$, so a covering does not exist in this case, too. \bigskip \noindent{\it Case IV.} $r_8 \equiv r_{16} \equiv \modd{0} {2}$. \bigskip Here, $|R_0|=1$, $|R_1|=8$, and $M_2 =8$. So, $R_0 = \{ r_0 \}$ where $r_0$ is an even integer in $[0,15]$. In this case, we can cover a lot with ${\mathcal C_3}$ and ${\mathcal C_5}$ but the overlap between them is too big and again we fall short of constructing a covering. First, note that ${\displaystyle \sum_{r \in R_1} a_3(r) \leq 8 + 4 + 2 + 1 = 15}$. Therefore, there exists $r_1 \in R_1$ such that $a_3(r_1) \leq 1$. By \eqref{at most 4}, we get $a_5(r_1) \geq 2$. Thus, $r_{10} \equiv r_{20} \equiv \modd{1} {2}$ (if any of the congruences in ${\mathcal C_5}$ are used to cover $R_0$, they should be the ones with the largest moduli since $|R_0| = 1$). Since, $r_{10} \equiv r_{20} \equiv \modd{1} {2}$, we have $a_5 (r_0) \leq 2$, and by \eqref{at most 4} we get $a_3(r_0) \geq 1$. Therefore, $r_{48} \equiv \modd{0} {2}$. Similarly, as above, ${\displaystyle \sum_{r \in R_1} a_5(r) \leq 15}$. Therefore, there exists $r_1' \in R_1$ such that $a_5(r_1') \leq 1$. By \eqref{at most 4}, we get $a_3(r_1') \geq 2$. Thus, $r_{6} \equiv r_{12} \equiv \modd{1} {2}$. So, $a_3 (r_0) \leq 2$. We proved above that $a_3(r_0) \geq 1$, so $a_3 (r_0)$ is either $1$ or $2$. Assume that $a_3 (r_0)=1$. Then \eqref{at most 4} implies $a_5(r_0) \geq 2$, so $r_{40} \equiv r_{80} \equiv \modd{0} {5}$. Hence, $a_5 (r) \leq 2$ for all $r \in R_1$. This implies $(3 - a_3(r_1))(4 - a_5(r_1)) \geq 4$. Also, $(3 - a_3(r_0))(4 - a_5(r_0)) \geq 4$. Thus, $$(3 - a_3(r_1))(4 - a_5(r_1)) + (3 - a_3(r_0))(4 - a_5(r_0)) \geq 8,$$ which contradicts \eqref{odd even}. So, $a_3(r_0)=2$, and $r_{24} \equiv r_{48} \equiv \modd{0} {2}$. Next, let $R_1' = \{ r \in R_1 \ | \ r \not\equiv \modd{r_{12}} {4} \}$. For all $r \in R_1'$ we have $a_3(r) = 1$. Also, $$ \sum_{r \in R_1'} a_5(r) \leq 4 + 4 + 2 + 1 = 11,$$ so there exists $r_1^* \in R_1'$ with $a_5(r_1^*) \leq 2$. Then $(3 - a_3(r_1^*))(4 - a_5(r_1^*)) \geq 4$. By \eqref{odd even} we get $( (3 - a_3(r_0))(4 - a_5(r_0)) \leq 2$. Thus, $a_5 (r_0) = 2$, and $r_{40} \equiv r_{80} \equiv \modd{0} {2}$. We have allocated all congruences in ${\mathcal C_3}$ and ${\mathcal C_5}$ to $R_0$ and $R_1$ (both $R_0$ and $R_1$ get two congruences from ${\mathcal C_3}$ and two from ${\mathcal C_5}$). Since $M_2 = 8$, \eqref{M2} becomes \begin{equation} \label{21} 21 \geq 4D_3 + 3D_5 + D_{15} + O. \end{equation} However, $D_3 \geq 1$, since $(24, r_{24})$ covers just one class modulo $48$, and $D_5 \geq 1$ since we did not use $(40, r_{40})$ in the most efficient way either. Also, $a_3(r) a_5(r) \neq 0$ for all $r \in R$, so by \eqref{2*3>5} $a_3(r) a_5(r) \geq 2$ for all $r \in R$, and $O \geq 18$. Substituting in \eqref{21} we get $21 \geq 4 + 3 + 18$, a contradiction. \end{proof} We just considered the last remaining case, and this concludes the proof of Theorem~\ref{lcm}. \end{proof} \section{Open problems and further work} \label{open pr} Recall that Krukenberg constructed a distinct covering system with least modulus $5$ and largest modulus $108$. He also conjectured that one cannot replace $108$ by a smaller constant. \bigskip \noindent {\bf Problem 1.} Prove or disprove that if the least modulus of a distinct covering system is $5$, then its largest modulus is at least $108$. We can show that if the least modulus of a distinct covering system is $5$, then its largest modulus is at least $84$. However, the result is too weak and the proof too long, to be included in this paper. Krukenberg also provided a description of the covering systems with least common multiple of the moduli of the form $2^a 3^b$, see Theorem \ref{Krukenberg2}. \bigskip \noindent{\bf Problem 2.} Describe the distinct covering systems with least common multiple of the moduli of the form $2^a 3^b 5^c$ where $a$, $b$, and $c$ are positive integers. Krukenberg \cite{Krukenberg} already provided such description in the case when $L = 2^a 3^b 5^c$ and one of the exponents $a$, $b$, and $c$ is zero. Using Krukenberg's results and the results of this paper one can find such a description when $a \geq b \geq c \geq 1$ and the minimum modulus $m=2,3,4$ with one exception. Extra work is needed to show that there is no distinct covering system with $m=4$ and $L=900$ (our proof of this is too long and technical to be included here). The more interesting case is when $m \geq 5$. Furthermore, Krukenberg constructed a distinct covering system with $m=5$ and $L=1440$. \bigskip \noindent {\bf Problem 3.} Prove or disprove that if the least modulus of a distinct covering system is $5$, then the least common multiple of its moduli is at least $1440$. Krukenberg also constructed a distinct covering system not using the modulus $3$, with all moduli squarefree integers. It is not known whether there exists a distinct covering system with squarefree moduli and least modulus $3$. \bigskip \noindent {\bf Problem 4.} Prove or disprove that the least modulus of any distinct covering system with squarefree moduli is $2$. Showing that the least modulus of any distinct covering system with squarefree moduli is $2$ will lead to a complete solution of the {\it minimum modulus problem} in the squarefree case. \bigskip \noindent{\bf Problem 5.} Find the largest integer $c$ such that there exists a finite set of congruences with distinct moduli with the property that every integer satisfies at least $c$ of the congruences. In other words, what is the largest number of times we can cover the integers by a finite system of congruences with distinct positive moduli? For a positive integer $n$ let $c(n)$ be the largest number of times we can cover all integers using congruences with moduli $1,2,\ldots,n$ respectively. Clearly, $c(1)=1$. Also, $c(n) \leq c(n+1)$ for all positive integers $n$ (having more congruences allows us to cover more). Furthermore, $c(n+1) \leq c(n) + 1$ for all $n$. Indeed, if a certain integer is covered $c(n)$ times by certain congruences with moduli $1,\ldots,n$, it can be covered at most once more by a congruence with modulus $n+1$. Recall that by Theorem \ref{Krukenberg1} there is no distinct covering with moduli in the interval $[2,11]$, and there is a distinct covering with moduli $2,3,4,6,12$. Therefore, $c(2) = \cdots = c(11) = 1$, and $c(12) = 2$. Moreover, Krukenberg constructed a distinct covering with least modulus $13$ and largest modulus $52562109600$. Therefore, $c(52562109600) \geq 3$. Moreover, the sequence $\left ( c(n)\right ) _{n=1}^\infty $ is bounded. Recall that Balister et al.\ \cite{Bal0} showed that the least modulus of any distinct covering system does not exceed $616000$. If we consider a system of congruences with moduli $1,2, \ldots , n$ respectively, where $n > 616000$, there is an integer $m$ which is not covered by any of the congruences with moduli $616001, 616002, \ldots, n$. Even if $m$ is covered by each of the congruences with moduli $1,\ldots, 616000$, then $m$ is covered $616000$ times. Thus, $c(n) \leq 616000$ for all $n$. Thus, ${\displaystyle c = \lim_{n \to \infty} c(n)}$ exists and Problem 5 is to find $c$. Problem 5 was considered by Harrington \cite{Har} who constructed three distinct covering systems with nonintersecting sets of moduli, thus establishing $c \geq 4$. We have a heuristic based on several assumptions showing that $c$ is either $4$ or $5$. \section{Acknowledgments} The authors greatly appreciate the help of Michael Filaseta and the work of the anonymous referee whose close read of the paper and numerous suggestions helped improve the paper. \appendix \section{Some details on the computations in the proof of Theorem~\ref{8}} \label{Appendix 1} Here we provide some details on how we showed that $T_m < 1$ for all $m$ in $[51,616000]$. Using the methods below, we only needed to calculate $19$ values of $T_m$ in the interval $[67, 616000]$. Recall that ${\displaystyle T_m = \sum_{\substack{ m \leq n \leq 8m \\ P(n) < \sqrt{7m+1}}} \frac{1}{n}}$, where $P(n)$ denotes the largest prime divisor of $n$. First, we computed and stored $P(n)$ for all $n$ from $2$ to $8\cdot 616000=4928000$. Next note that $T_{m-1} \leq T_m + a_{m-1}$, where we define $a_{m-1}$ to be $\frac{1}{m-1}$ when $P(m-1) < \sqrt{7m-6}$, and we define $a_{m-1}$ to be $0$ when $P(m-1) \geq \sqrt{7m-6}$. Indeed, \begin{align*} T_{m-1} &= \sum_{\substack{ m-1 \leq n \leq 8m-8 \\ P(n) < \sqrt{7m-6}}} \frac{1}{n} \\ & = a_{m-1} + \sum_{\substack{ m \leq n \leq 8m-8 \\ P(n) < \sqrt{7m-6}}} \frac{1}{n}\\ & \leq a_{m-1} + \sum_{\substack{ m \leq n \leq 8m \\ P(n) < \sqrt{7m+1}}} \frac{1}{n}. \end{align*} So, we computed $T_{616000}=0.6886632306756396\ldots$, and then using the inequality $T_{m-1} \leq T_m + a_{m-1}$ we get that $T_{615999} \leq T_{616000} +\frac {1}{615999}$. Iterating this method, we backtracked down to the last value of $m$ where the sum is less than $1$, which got us to $286068$. Thus, because $$\sum_{m=286068}^{615999} a_m + T_{616000}=0.9999973928615169\ldots<1,$$ we get that $T_m < 1$ for all $m \in [286068, 616000]$. Next, we computed $$T_{286067}=0.6897176227760186\ldots,$$ and backtracked again; we got $$\sum_{m=134747}^{286066} a_m + T_{286067}=0.9999955266833742\ldots<1.$$ Through these jumps, we confirmed that $T_m <1$ for each $m$ in the intervals \begin{align*} [286068,616000],[134747, 286067],[65512,134746],[3&3049,65511],[17044,33048], \\ [8837,17043],[4807,8836],[2739,4806],[1597,2738]&,[976,1596],[610,975], \\ [385,609],[254,384],[176,253],[126,175],[96,125]&,[77,95],[67,76]. \end{align*} We had to compute $19$ values of $T_m$ to get to $m=67$. We then computed directly all values of $T_m$ for $m \in [3,66]$. \section{Construction of a distinct covering} \label{Appendix 2} Here we provide an example of a distinct covering system with a congruence modulo $180$ and the remaining moduli in $[4,56]$. The moduli we use are $$4,8,16;6,12,24,48;9,18,36;5,10,20,40; 15,30;45,180;7,14,21,28,35,42,56,$$ where the semicolons are used to separate the moduli involved in different stages of our argument below. The congruences modulo $4,8,16$ which we use are $(11)$, $(101)$, and $(1001)$. The uncovered set after the first stage consists of a residue class modulo $2$, $(0)$, and a residue class modulo $16$, $(1000)$. Splitting modulo $3$, the uncovered set is $(0 |\ 0,1,2)$ and $(1000 |\ 0,1,2)$. Next, we use the congruences modulo $12,24,48$ to cover $(1000)$. The congruences modulo $12,24,48$ given by $(10 | \ 0)$, $(100 |\ 1)$, and $(1000 |\ 2)$ accomplish this. We use the congruence modulo $6$ given by $(0 |\ 2)$. After the second stage, the uncovered set is $( 0 |\ 0,1)$. We use the congruences modulo $9,18,36$ to attack the residue class $(0 |\ 1)$, which is the same as $(0 |\ 10,11,12)$. We take the congruences modulo $9,18,36$ to be $( * | 12)$, $(0 |\ 11)$, and $(01 |\ 10)$. The uncovered set after the third stage is $(0 |\ 0)$ and $(00 |\ 10)$. We split the uncovered set modulo $5$, to get $(0 |\ 0 |\ 0,1,2,3,4 )$ and $(00 |\ 10 |\ 0,1,2,3,4 )$. The congruences modulo $5$, $10$, and $20$ are $(* |\ * |\ 4)$, $(0 |\ * |\ 3)$, and $(00 |\ * |\ 2)$. Now, the uncovered set is $(0 |\ 0 |\ 0,1)$, $(01 |\ 0 |\ 2 )$, and $(00 |\ 10 |\ 0,1 )$. The congruence modulo $40$ is $(011 |\ * |\ 2 )$. The congruences modulo $15$ and $30$ are $(* |\ 0 |\ 1)$ and $(0 |\ 0 |\ 0)$, and they cover $(0 |\ 0 |\ 0,1)$. We are left with the uncovered set $(010 |\ 0 |\ 2 )$ and $(00 |\ 10 |\ 0,1 )$. We use the congruences modulo $45$ and $180$, $(* |\ 10 |\ 1 )$ and $(00 |\ 10 |\ 0 )$ to cover $(00 |\ 10 |\ 0,1 )$. 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Published in {\it Journal of Integer Sequences}, October 27 2022. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}