# Problem with \vfill within enumerate and multicols

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 its 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

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

• \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 '15 at 20:37
• 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) – David Carlisle May 15 '15 at 20:48

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}
`
• 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 '15 at 22:01
• 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: – David Carlisle May 15 '15 at 22:40