# delete a column in the table [duplicate]

I have the following table where I would want to delete the column $2 \neq \Box$ but had no luck as it keeps giving error when I compile it.

\documentclass{article}
\setlength{\textwidth}{16cm}
\setlength{\oddsidemargin}{0pt}
\setlength{\evensidemargin}{0pt}
\setlength{\parskip}{3mm}
\setlength{\parindent}{0mm}
\usepackage{amsmath}
\usepackage{amsthm}
\DeclareMathOperator{\lcm}{lcm}
\DeclareMathOperator{\ord}{ord}
\usepackage{tabularx,ragged2e,booktabs,caption}
\usepackage[figuresleft]{rotating}
\usepackage{rotating}
\newcolumntype{C}{>{\Centering\arraybackslash}X}
\newcommand\swb{{\scriptstyle\Box}} % "small white box"
\usepackage{amssymb,bm}
\usepackage{upgreek} %use uptau greek letter

\begin{document}
\begin{sidewaystable}
\setlength\tabcolsep{6pt} % default value: 6pt
\begin{tabularx}{\textwidth}{@{}*{14}{C}@{}}
\toprule
$\delta_\mu$ & $2 \neq \swb$ & \multicolumn{4}{c}{$5 \cdot 29 \neq \swb$} & \multicolumn{4}{c}{$13 \cdot 1789 \neq \swb$} & \multicolumn{4}{c}{$5333 \cdot 97324757 \neq \swb$}\\
\cmidrule(lr){3-6} \cmidrule(l){7-10} \cmidrule(l){11-14}
&   & $5=\swb$ & $29\neq\swb$ & $5\neq\swb$ & $29=\swb$ & $13=\swb$ & $1789\neq\swb$ & $13\neq\swb$ & $1789=\swb$ & $5333=\swb$ & $97324757\neq\swb$ & $5333\neq\swb$ & $97324757=\swb$ \\
\midrule
Primes that satisfy the \footnote{Refer Appendix A for the PARI code to find the complete list of $G_1, G_2, G_3, G_4, G_5, G_6$} condition $\delta_\mu \neq \swb$  &
$q\equiv\pm3\pmod{8}$ &
$q\equiv\pm1\pmod{5}$ &
$q\equiv 2,3,8,10,\allowbreak11,12,14,\allowbreak15,17,18,\allowbreak 19,21,26, \allowbreak 27\pmod{29}$ &
$q\equiv\pm2\pmod{5}$ &
$q\equiv 1,4,5,6,\allowbreak7,9,13,\allowbreak 16,20,\allowbreak 22,23,\allowbreak 24,25,28\pmod{29}$ &
$q\equiv 1,3,4,9,\allowbreak10,12\pmod{13}$ &
$q \equiv$ $G_1$ \text{(mod} \allowbreak $1789$) &
$q\equiv2,5,6,\allowbreak 7,8,11\pmod{13}$ &
$q \equiv$ $G_2$ \text{(mod} \allowbreak $1789$) &
$q \equiv$ $G_3$ \text{(mod} \allowbreak $5333$) &
$q \equiv$ $G_4$ \text{(mod} \allowbreak $97324757$) &
$q \equiv$ $G_5$ \text{(mod} \allowbreak $5333$) &
$q \equiv$ $G_6$ \text{(mod} \allowbreak $97324757$) \\
\midrule
Period of $w_k$  & $24$ & $30$ & $102$ & $30$ & $102$ & $30$ & $2670$ & $30$ & $2670$ & $750$ & $97306362$ & $750$ & $97306362$ \\
\midrule
$\ord(\widetilde{P})$  &  & $5$ & $17$ & $5$ & $17$ & $10$ & $890$ & $10$ & $890$ & $125$ & $48653181$ & $125$ & $48653181$ \\
\midrule
$k$ which satisfies $w_k$ \footnote{Refer Appendix B for the lists of sets $A_{2}$, $A_5$, $A_{29^*}$, $A_{5}^*$, $A_{29}$, $A_{13}$, $A_{1789}^*$, $A_{13}^*$, $A_{1789}$, $A_{5333}$, $A_{97324757}^*$, $A_{5333}^*$, $A_{97324757}$} &  $A_{2}$ : 10 elements & $A_5$ : 21 elements & $A_{29^*}$ : 37 elements & $A_{5}^*$ : 9 elements & $A_{29}$ : 65 elements & $A_{13}$ : 22 elements & $A_{1789}^*$ : 1304 elements & $A_{13}^*$ : 8 elements & $A_{1789}$ : 1362 elements & $A_{5333}$ : 421 elements & $A_{97324757}^*$ : 48584207 elements & $A_{5333}^*$ : 329 elements & $A_{97324757}$ : 48722155 elements\\
\bottomrule
\end{tabularx}\captionof{table}{Summary of the congruence conditions for $\mu \in Y = \{-4, -3, -2, -1, 0, 3\}$} \label{summary}
\end{sidewaystable}
\end{document}


\documentclass{article}
\setlength{\textwidth}{16cm}
\setlength{\oddsidemargin}{0pt}
\setlength{\evensidemargin}{0pt}
\setlength{\parskip}{3mm}
\setlength{\parindent}{0mm}
\usepackage{amsmath}
\usepackage{amsthm}
\DeclareMathOperator{\lcm}{lcm}
\DeclareMathOperator{\ord}{ord}
\usepackage{tabularx,ragged2e,booktabs,caption}
\usepackage[figuresleft]{rotating}
\usepackage{rotating}
\newcolumntype{C}{>{\Centering\arraybackslash}X}
\newcommand\swb{{\scriptstyle\Box}} % "small white box"
\usepackage{amssymb,bm}
\usepackage{upgreek} %use uptau greek letter

\begin{document}

\begin{sidewaystable}
\setlength\tabcolsep{6pt} % default value: 6pt
\begin{tabularx}{\textwidth}{@{}*{13}{C}@{}}
\toprule
$\delta_\mu$ & \multicolumn{4}{c}{$5 \cdot 29 \neq \swb$} & \multicolumn{4}{c}{$13 \cdot 1789 \neq \swb$} & \multicolumn{4}{c}{$5333 \cdot 97324757 \neq \swb$}\\
\cmidrule(lr){3-6} \cmidrule(l){7-10} \cmidrule(l){11-13}
&    $5=\swb$ & $29\neq\swb$ & $5\neq\swb$ & $29=\swb$ & $13=\swb$ & $1789\neq\swb$ & $13\neq\swb$ & $1789=\swb$ & $5333=\swb$ & $97324757\neq\swb$ & $5333\neq\swb$ & $97324757=\swb$ \\
\midrule
Primes that satisfy the \footnote{Refer Appendix A for the PARI code to find the complete list of $G_1, G_2, G_3, G_4, G_5, G_6$} condition $\delta_\mu \neq \swb$  &
$q\equiv\pm1\pmod{5}$ &
$q\equiv 2,3,8,10,\allowbreak11,12,14,\allowbreak15,17,18,\allowbreak 19,21,26, \allowbreak 27\pmod{29}$ &
$q\equiv\pm2\pmod{5}$ &
$q\equiv 1,4,5,6,\allowbreak7,9,13,\allowbreak 16,20,\allowbreak 22,23,\allowbreak 24,25,28\pmod{29}$ &
$q\equiv 1,3,4,9,\allowbreak10,12\pmod{13}$ &
$q \equiv$ $G_1$ \text{(mod} \allowbreak $1789$) &
$q\equiv2,5,6,\allowbreak 7,8,11\pmod{13}$ &
$q \equiv$ $G_2$ \text{(mod} \allowbreak $1789$) &
$q \equiv$ $G_3$ \text{(mod} \allowbreak $5333$) &
$q \equiv$ $G_4$ \text{(mod} \allowbreak $97324757$) &
$q \equiv$ $G_5$ \text{(mod} \allowbreak $5333$) &
$q \equiv$ $G_6$ \text{(mod} \allowbreak $97324757$) \\
\midrule
Period of $w_k$  &  $30$ & $102$ & $30$ & $102$ & $30$ & $2670$ & $30$ & $2670$ & $750$ & $97306362$ & $750$ & $97306362$ \\
\midrule
$\ord(\widetilde{P})$    & $5$ & $17$ & $5$ & $17$ & $10$ & $890$ & $10$ & $890$ & $125$ & $48653181$ & $125$ & $48653181$ \\
\midrule
$k$ which satisfies $w_k$ \footnote{Refer Appendix B for the lists of sets $A_{2}$, $A_5$, $A_{29^*}$, $A_{5}^*$, $A_{29}$, $A_{13}$, $A_{1789}^*$, $A_{13}^*$, $A_{1789}$, $A_{5333}$, $A_{97324757}^*$, $A_{5333}^*$, $A_{97324757}$} & $A_5$ : 21 elements & $A_{29^*}$ : 37 elements & $A_{5}^*$ : 9 elements & $A_{29}$ : 65 elements & $A_{13}$ : 22 elements & $A_{1789}^*$ : 1304 elements & $A_{13}^*$ : 8 elements & $A_{1789}$ : 1362 elements & $A_{5333}$ : 421 elements & $A_{97324757}^*$ : 48584207 elements & $A_{5333}^*$ : 329 elements & $A_{97324757}$ : 48722155 elements\\
\bottomrule
\end{tabularx}\captionof{table}{Summary of the congruence conditions for $\mu \in Y = \{-4, -3, -2, -1, 0, 3\}$} \label{summary}
\end{sidewaystable}

\end{document}

• This is great but when I compile I notice that the horizontal line on $$5=\Box$$ is short. Is there a way you can fix it please. – shahrina ismail Nov 9 '16 at 6:12
• please change \cmidrule(lr){3-6} to \cmidrule(lr){2-6}. This draws midrule for cols 2 to 6 instead of from 3 to 6. – Rama Krishna Majety Nov 9 '16 at 14:10