我想创建一个这样的表格，末尾带有花括号

Question

这是我目前得到的结果。如果有人能帮我在现有列之间插入 2^2x5 和 2x11，那就太好了。

\setlength{\tabcolsep}{12pt}
\paragraph{}
\begin{tabular}{c c c c}
$n\mod19=3$ & $n\equiv3\pmod{19}$ & $n\equiv3\pmod{19}$ & $19$\\[0.25cm]
$n\mod20=1$ & $n\equiv1\pmod{20}$ & $n\equiv2\pmod{21}$ & $3\cdot7$\\[0.25cm]   
$n\mod21=2$ & $n\equiv2\pmod{21}$ & $n\equiv1\pmod{220}$ & $2^2\cdot5\cdot11$      \\[0.25cm] 
$n\mod22=1$ & $n\equiv1\pmod{22}$ & $n\equiv0\pmod{23}$ & $23$\\[0.25cm] 
$n\mod23=0$ & $n\equiv0\pmod{23}$ & 
\end{tabular}
\paragraph{}

Answer 1

\documentclass{report}
\usepackage{amsmath}
\usepackage[a4paper]{geometry}
\begin{document}

{\setlength{\tabcolsep}{12pt}
$\left.
\begin{tabular}{c c c c}
    $n\mod19=3$ & $n\equiv3\pmod{19}$ & $n\equiv3\pmod{19}$ & $19$\\[0.25cm]
    $n\mod20=1$ & $n\equiv1\pmod{20}$ & $n\equiv2\pmod{21}$ & $3\cdot7$\\[0.25cm]   
    $n\mod21=2$ & $n\equiv2\pmod{21}$ & $n\equiv1\pmod{220}$ & $2^2\cdot5\cdot11$      \\[0.25cm] 
    $n\mod22=1$ & $n\equiv1\pmod{22}$ & $n\equiv0\pmod{23}$ & $23$\\[0.25cm] 
    $n\mod23=0$ & $n\equiv0\pmod{23}$ & 
\end{tabular}
\right\}$ 
text
}
    

\bigskip
{\setlength{\arraycolsep}{12pt}
    $\left.
    \begin{array}{c c c c}
        n\mod19=3 & n\equiv3\pmod{19} & n\equiv3\pmod{19} & 19\\[0.25cm]
        n\mod20=1 & n\equiv1\pmod{20} & n\equiv2\pmod{21} & 3\cdot7\\[0.25cm]   
        n\mod21=2 & n\equiv2\pmod{21} & n\equiv1\pmod{220} & 2^2\cdot5\cdot11      \\[0.25cm] 
        n\mod22=1 & n\equiv1\pmod{22} & n\equiv0\pmod{23} & 23\\[0.25cm] 
        n\mod23=0 & n\equiv0\pmod{23} & 
    \end{array}
    \right\}$ 
    text
}
    
\end{document}

Answer 2

与。{NiceTabular}​nicematrix

\documentclass{article}
\usepackage{nicematrix}

\begin{document}

\setlength{\tabcolsep}{12pt}

\begin{NiceTabular}{c c c c}
  $n\mod19=3$ & $n\equiv3\pmod{19}$ & $n\equiv3\pmod{19}$ & $19$\\[0.25cm]
  $n\mod20=1$ & $n\equiv1\pmod{20}$ & $n\equiv2\pmod{21}$ & $3\cdot7$\\[0.25cm]   
  $n\mod21=2$ & $n\equiv2\pmod{21}$ & $n\equiv1\pmod{220}$ & $2^2\cdot5\cdot11$      \\[0.25cm] 
  $n\mod22=1$ & $n\equiv1\pmod{22}$ & $n\equiv0\pmod{23}$ & $23$\\[0.25cm] 
  $n\mod23=0$ & $n\equiv0\pmod{23}$ & 
\CodeAfter
  \SubMatrix{.}{1-1}{4-4}{\}}
\end{NiceTabular}

\end{document}

Answer 3

如果您不关心构造组件的垂直对齐，则简单的嵌套align将aligned提供：

\documentclass{article}

\usepackage{amsmath,eqparbox}

\usepackage[margin=1in]{geometry}% Just for this example

% https://tex.stackexchange.com/a/34412/5764
\makeatletter
\NewDocumentCommand{\eqmathbox}{o O{c} m}{%
  \IfValueTF{#1}
    {\def\eqmathbox@##1##2{\eqmakebox[#1][#2]{$##1##2$}}}
    {\def\eqmathbox@##1##2{\eqmakebox{$##1##2$}}}
  \mathpalette\eqmathbox@{#3}
}
\makeatother

\begin{document}

  \begin{align*}
    & \begin{aligned}
      n \mod 19 &= 3 \\
      n \mod 20 &= 1 \\
      n \mod 21 &= 2 \\
      n \mod 22 &= 1 \\
      n \mod 23 &= 0
    \end{aligned} &
    & \begin{aligned}
      n &\equiv 3 \pmod{19} \\
      n &\equiv 1 \pmod{20} & \eqmathbox[prp1]{2^2 \cdot 5} \\
      n &\equiv 2 \pmod{21} \\
      n &\equiv 1 \pmod{22} & \eqmathbox[prp1]{2 \cdot 11}  \\
      n &\equiv 0 \pmod{23}
    \end{aligned} &
    &\left.\kern-\nulldelimiterspace\begin{aligned}
      n &\equiv 3 \pmod{19}  & \eqmathbox[prp2]{19} \\
      n &\equiv 2 \pmod{21}  & \eqmathbox[prp2]{3 \cdot 7} \\
      n &\equiv 1 \pmod{220} & \eqmathbox[prp2]{2^2 \cdot 5 \cdot 11} \\
      n &\equiv 0 \pmod{23}  & \eqmathbox[prp2]{23}
    \end{aligned} \right\}
  \end{align*}

\end{document}