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I am trying to make a large list of exercises with multicols and enumerate the thing is that when I just use enumerate and vfill there is no problem vfill does the work but when I use multicols vfill doesn't add the space it`s suppose to add.

\begin{multicols}{2}
\begin{enumerate}
\begin{large}
\item  $\dfrac{2}{8}=\dfrac{1}{4} $ \vfill
\item  $\dfrac{2}{6}=\dfrac{1}{3} $ \vfill
\item  $\dfrac{4}{6}=\dfrac{2}{3} $ \vfill
\item  $\dfrac{2}{8}=\dfrac{1}{4} $ \vfill
\item  $\dfrac{6}{8}=\dfrac{3}{4} $ \vfill
\item  $\dfrac{2}{10}=\dfrac{1}{5} $ \vfill
\item  $\dfrac{4}{10}=\dfrac{2}{5} $ \vfill
\item  $\dfrac{4}{14}=\dfrac{2}{7} $ \vfill
\item  $\dfrac{2}{14}=\dfrac{1}{7} $ \vfill
\item  $\dfrac{8}{22}=\dfrac{4}{11} $ \vfill
\item  $\dfrac{6}{26}=\dfrac{3}{13} $ \vfill
\item  $\dfrac{12}{14}=\dfrac{6}{7} $ \vfill
\item  $\dfrac{8}{10}=\dfrac{4}{5} $ \vfill
\item  $\dfrac{8}{14}=\dfrac{4}{7} $ \vfill
\item  $\dfrac{16}{18}=\dfrac{8}{9} $ \vfill
\columnbreak 
\item  $\dfrac{14}{16}=\dfrac{7}{8} $ \vfill
\item  $\dfrac{20}{22}=\dfrac{10}{11} $ \vfill
\item  $\dfrac{22}{24}=\dfrac{11}{12} $ \vfill
\item  $\dfrac{2}{12}=\dfrac{1}{6} $ \vfill
\item  $\dfrac{2}{14}=\dfrac{1}{7} $ \vfill
\item  $\dfrac{2}{16}=\dfrac{1}{8} $ \vfill
\item  $\dfrac{10}{22}=\dfrac{5}{11} $ \vfill
\item  $\dfrac{14}{32}=\dfrac{7}{16} $ \vfill
\item  $\dfrac{10}{14}=\dfrac{5}{7} $ \vfill
\item  $\dfrac{6}{10}=\dfrac{3}{5} $ \vfill
\item  $\dfrac{16}{22}=\dfrac{8}{11} $ \vfill
\item  $\dfrac{22}{26}=\dfrac{11}{13} $ \vfill
\item  $\dfrac{2}{20}=\dfrac{1}{10} $ \vfill
\item  $\dfrac{14}{24}=\dfrac{7}{12} $ \vfill
\item  $\dfrac{26}{28}=\dfrac{13}{14} $ \vfill
\end{large}
\end{enumerate} 
\end{multicols}

The output is the as follows

enter image description here

How can I solve this, I think columnbreak does something that doesn't allow \vfill to work properly

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  • 2
    \vfill is not supposed to add any space, unless TeX needs to fill up some area. In this case it isn't requested to.
    – egreg
    May 15, 2015 at 20:37
  • 2
    as egreg says, also the whole point of a list environment is that it sets up the desired spacing, adding \vfill (or any space command) after every item is just wrong, you should set the list parameter for inter-item space (either directly or via enumitem package) May 15, 2015 at 20:48

1 Answer 1

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Please always post complete documents. The inter-item separation is a parameter of a list:

\documentclass{article}
\usepackage{enumitem,multicol,amsmath}
\begin{document}
\begin{multicols}{2}\large
\begin{enumerate}[itemsep=9pt]

\item  $\dfrac{2}{8}=\dfrac{1}{4} $ 
\item  $\dfrac{2}{6}=\dfrac{1}{3} $ 
\item  $\dfrac{4}{6}=\dfrac{2}{3} $ 
\item  $\dfrac{2}{8}=\dfrac{1}{4} $ 
\item  $\dfrac{6}{8}=\dfrac{3}{4} $ 
\item  $\dfrac{2}{10}=\dfrac{1}{5} $ 
\item  $\dfrac{4}{10}=\dfrac{2}{5} $ 
\item  $\dfrac{4}{14}=\dfrac{2}{7} $ 
\item  $\dfrac{2}{14}=\dfrac{1}{7} $ 
\item  $\dfrac{8}{22}=\dfrac{4}{11} $ 
\item  $\dfrac{6}{26}=\dfrac{3}{13} $ 
\item  $\dfrac{12}{14}=\dfrac{6}{7} $ 
\item  $\dfrac{8}{10}=\dfrac{4}{5} $ 
\item  $\dfrac{8}{14}=\dfrac{4}{7} $ 
\item  $\dfrac{16}{18}=\dfrac{8}{9} $ 

\item  $\dfrac{14}{16}=\dfrac{7}{8} $ 
\item  $\dfrac{20}{22}=\dfrac{10}{11} $ 
\item  $\dfrac{22}{24}=\dfrac{11}{12} $ 
\item  $\dfrac{2}{12}=\dfrac{1}{6} $ 
\item  $\dfrac{2}{14}=\dfrac{1}{7} $ 
\item  $\dfrac{2}{16}=\dfrac{1}{8} $ 
\item  $\dfrac{10}{22}=\dfrac{5}{11} $ 
\item  $\dfrac{14}{32}=\dfrac{7}{16} $ 
\item  $\dfrac{10}{14}=\dfrac{5}{7} $ 
\item  $\dfrac{6}{10}=\dfrac{3}{5} $ 
\item  $\dfrac{16}{22}=\dfrac{8}{11} $ 
\item  $\dfrac{22}{26}=\dfrac{11}{13} $ 
\item  $\dfrac{2}{20}=\dfrac{1}{10} $ 
\item  $\dfrac{14}{24}=\dfrac{7}{12} $ 
\item  $\dfrac{26}{28}=\dfrac{13}{14} $ 

\end{enumerate} 
\end{multicols}

\end{document}

or automatically padding, but not balancing automatically

\documentclass[twocolumn]{article}
\usepackage{enumitem,amsmath}
\begin{document}

\begin{enumerate}[itemsep=\fill]

\item  $\dfrac{2}{8}=\dfrac{1}{4} $ 
\item  $\dfrac{2}{6}=\dfrac{1}{3} $ 
\item  $\dfrac{4}{6}=\dfrac{2}{3} $ 
\item  $\dfrac{2}{8}=\dfrac{1}{4} $ 
\item  $\dfrac{6}{8}=\dfrac{3}{4} $ 
\item  $\dfrac{2}{10}=\dfrac{1}{5} $ 
\item  $\dfrac{4}{10}=\dfrac{2}{5} $ 
\item  $\dfrac{4}{14}=\dfrac{2}{7} $ 
\item  $\dfrac{2}{14}=\dfrac{1}{7} $ 
\item  $\dfrac{8}{22}=\dfrac{4}{11} $ 
\item  $\dfrac{6}{26}=\dfrac{3}{13} $ 
\item  $\dfrac{12}{14}=\dfrac{6}{7} $ 
\item  $\dfrac{8}{10}=\dfrac{4}{5} $ 
\item  $\dfrac{8}{14}=\dfrac{4}{7} $ 
\item  $\dfrac{16}{18}=\dfrac{8}{9} $ 

\newpage

\item  $\dfrac{14}{16}=\dfrac{7}{8} $ 
\item  $\dfrac{20}{22}=\dfrac{10}{11} $ 
\item  $\dfrac{22}{24}=\dfrac{11}{12} $ 
\item  $\dfrac{2}{12}=\dfrac{1}{6} $ 
\item  $\dfrac{2}{14}=\dfrac{1}{7} $ 
\item  $\dfrac{2}{16}=\dfrac{1}{8} $ 
\item  $\dfrac{10}{22}=\dfrac{5}{11} $ 
\item  $\dfrac{14}{32}=\dfrac{7}{16} $ 
\item  $\dfrac{10}{14}=\dfrac{5}{7} $ 
\item  $\dfrac{6}{10}=\dfrac{3}{5} $ 
\item  $\dfrac{16}{22}=\dfrac{8}{11} $ 
\item  $\dfrac{22}{26}=\dfrac{11}{13} $ 
\item  $\dfrac{2}{20}=\dfrac{1}{10} $ 
\item  $\dfrac{14}{24}=\dfrac{7}{12} $ 
\item  $\dfrac{26}{28}=\dfrac{13}{14} $ 

\end{enumerate} 


\end{document}
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  • sorry for the imcomplite code and thanks that does what I want but what happens if the list has more o less items Do I have to play with the itemsep? or Can I put the distance as vfill or something similar.
    – Very23
    May 15, 2015 at 22:01
  • 1
    I@Very23 if you don't balance the columns you can make it stretch but the balancing routine packs the columns, I'll add an example not balancing: May 15, 2015 at 22:40

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