=== multienum 패키지 ===
KTUGOperate:12267 (by [Progress])

연습문제 해답 등을 작성하기에 좋은 툴로 multienum 패키지가 있습니다. (CTAN에서 찾아보세요)
다음 소스를 컴파일해보시고요.

{{{
%This is a 2-page sample illustrating how to use the
%multienum package

\documentclass{article}
\setlength{\textwidth}{6in}
\setlength{\textheight}{8.5in}
\setlength{\topmargin}{-0.5in}
\setlength{\oddsidemargin}{0.25in}
\usepackage{multicol,multienum}


\begin{document}
\begin{center}
{\Large\bf Sample formating using {\tt multienumerate}}
\end{center}

\bigskip
Sometimes we want to typeset the solutions to exercises. This
is easy to do using the {\tt multienumerate} environment.
\subsection*{Answers to All Exercises}
\begin{multienumerate}
\mitemxxxx{Not}{Linear}{Not}{Quadratic}
\mitemxxxo{Not}{Linear}{No; if $x=3$, then $y=-2$.}
\mitemxx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or
$(s,3s-6)$}{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-
u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$}
\mitemxxxx{$(2,-1,3)$}{None}{$(2,1,0,1)$}{$(0,0,0,0)$}
\end{multienumerate}


\bigskip
\hrule

\bigskip

We can also enumerate the items using an even-only or odd only
counter.
\subsection*{Answers to Even-Numbered Exercises}
\begin{multienumerate}[evenlist]
\mitemxxxx{Not}{Linear}{Not}{Quadratic}
\mitemxxxo{Not}{Linear}{No; if $x=3$, then $y=-2$.}
\mitemxx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or
$(s,3s-6)$}{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-
u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$}
\mitemxxxx{$(2,-1,3)$}{None}{$(2,1,0,1)$}{$(0,0,0,0)$}
\end{multienumerate}

\hrule

\subsection*{Answers to Odd-Numbered Exercises}
\begin{multienumerate}[oddlist]
\mitemxxxx{Not}{Linear}{Not}{Quadratic}
\mitemxxxo{Not}{Linear}{No; if $x=3$, then $y=-2$.}
\mitemxx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or
$(s,3s-6)$}{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-
u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$}
\mitemxxxx{$(2,-1,3)$}{None}{$(2,1,0,1)$}{$(0,0,0,0)$}
\end{multienumerate}

\bigskip
\hrule

\bigskip

Sometimes we want to create sublists which are
enumerated using an alpha counter.

\begin{multienumerate}
\mitemx{Which of the following numbers is the solution of the
equation
$x+3=7$:}
\begin{multienumerate}
\mitemxxxx{1}{2}{3}{4}
\end{multienumerate}
\mitemx{The value of $\log_28$ is:}
\begin{multienumerate}
\mitemxxxx{1}{$-1$}{3}{$-3$}
\end{multienumerate}
\end{multienumerate}
\pagebreak

\begin{multicols}{2}
\subsection*{Answers to All Exercises}
\begin{multienumerate}
\mitemxx{Not}{Linear}
\mitemxx{Not}{Quadratic}
\mitemxx{Not}{Linear}
\mitemx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or
$(s,3s-6)$}
\mitemx{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-
u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$}
\mitemxx{$(2,-1,3)$}{None}
\mitemxx{$(2,1,0,1)$}{$(0,0,0,0)$}
\mitemxx{Not}{Linear}
\mitemxx{Not}{Quadratic}
\mitemxx{Not}{Linear}
\mitemx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or
$(s,3s-6)$}
\mitemx{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-
u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$}
\mitemxx{$(2,-1,3)$}{None}
\mitemxx{$(2,1,0,1)$}{$(0,0,0,0)$}
\mitemxx{Not}{Linear}
\mitemxx{Not}{Quadratic}
\mitemxx{Not}{Linear}
\mitemx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or
$(s,3s-6)$}
\mitemx{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-
u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$}
\mitemxx{$(2,-1,3)$}{None}
\mitemxx{$(2,1,0,1)$}{$(0,0,0,0)$}
\end{multienumerate}

\subsection*{Multiple Choice}
\begin{multienumerate}
\mitemx{Which of the following numbers is the solution of the
equation
$x+3=7$:}
\begin{multienumerate}
\mitemxxxx{1}{2}{3}{4}
\end{multienumerate}
\mitemx{The value of $\log_28$ is:}
\begin{multienumerate}
\mitemxxxx{1}{$-1$}{3}{$-3$}
\end{multienumerate}
\mitemx{Which of the following numbers is the solution of the
equation
$x+3=7$:}
\begin{multienumerate}
\mitemxxxx{1}{2}{3}{4}
\end{multienumerate}
\mitemx{The value of $\log_28$ is:}
\begin{multienumerate}
\mitemxxxx{1}{$-1$}{3}{$-3$}
\end{multienumerate}
\mitemx{Which of the following numbers is the solution of the
equation
$x+3=7$:}
\begin{multienumerate}
\mitemxxxx{1}{2}{3}{4}
\end{multienumerate}
\mitemx{The value of $\log_28$ is:}
\begin{multienumerate}
\mitemxxxx{1}{$-1$}{3}{$-3$}
\end{multienumerate}
\end{multienumerate}
\end{multicols}

\end{document}
}}}
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