\hsize=12.6cm
\def\mod#1{\quad({\rm mod}\ #1)}
%205
{\bf Lemma 7.11} {\it The six congruences
$$\matrix{0\mod{2}\cr
0\mod{3}\cr
1\mod{4}\cr
3\mod{8}\cr
7\mod{12}\cr
23\mod{24}}$$
form a set of covering congruences.}
\medskip

{\bf Proof.}  First, we show that each of the 24 integers $0, 1, \ldots, 23$ satisfies at least
one of these six congruences.  Every even integer $k$ satisfies $k\equiv 0\mod{2}$.  For
odd integers, we have
$$\matrix{1\equiv 1\mod{4}\cr
3\equiv 0\mod{3}\cr
5\equiv 1\mod{4}\cr
7\equiv 7\mod{12}\cr
9\equiv 0\mod{3}\cr
11\equiv 3\mod{8}\cr
13\equiv 1\mod{4}\cr
15\equiv 0\mod{3}\cr
17\equiv 1\mod{4}\cr
19\equiv 7\mod{12}\cr
21\equiv 0\mod{3}\cr
23\equiv 23\mod{24}}$$
%206
For every integer $k$, there is a unique integer $r\in \{0, 1, \dots, 23\}$ such that
$$k\equiv r\mod{24}.$$
Choose $i$ so that
$$r\equiv a_i\mod{m_i},$$
where $a_i\mod{m_i}$ is one of our six congruences.  Each of the six moduli 2, 3,
4, 6, 12, and 24 divides 24, so $m_i$ divides 24 and
$$k\equiv r\mod{m_i}.$$
Therefore,
$$k\equiv a_i\mod{m_i}.$$
This completes the proof.
\bye