How do I move the \fboxsep in the middle

Question

I am trying to create these number bonds with squares for the numbers but I am not sure how to move the box towards the middle.

\documentclass{article}
\usepackage{enumitem,multicol,xcolor}
\fboxsep=4mm \fboxrule=0.5mm % set these parameters globally

\begin{document}
\begin{multicols}{3}

\begin{enumerate}[label=(\alph*)]
\item $2 + \fcolorbox{black}{white}{\null} = 5$
\item $\fcolorbox{black}{white}{\null} + 2 = 5$

\item $1 + \fcolorbox{black}{white}{\null} = 5$
\item $\fcolorbox{black}{white}{\null} + 1 = 5$

\item $3 + \fcolorbox{black}{white}{\null} = 5$
\item $\fcolorbox{black}{white}{\null} + 3 = 5$
\end{enumerate}

\end{multicols}
\end{document}

Answer 1

Don't use \null as it doesn't have any height. Instead, set a \phantom number; it'll allow you to have a box that roughly matches the size of the contents. Also, define a command that can do all of the adjustments for you; this will make global changes much more easy.

\documentclass{article}

\usepackage{multicol,enumitem}

\newcommand{\boxnum}[1]{{%
  \setlength{\fboxsep}{4mm}%
  \setlength{\fboxrule}{0.5mm}%
  \fbox{\phantom{#1}}%
}}

\begin{document}

\begin{multicols}{3}
  \begin{enumerate}[label=(\alph*)]
    \item $2 + \boxnum{3} = 5$
    \item $\boxnum{3} + 2 = 5$

    \item $1 + \boxnum{4} = 5$
    \item $\boxnum{4} + 1 = 5$

    \item $3 + \boxnum{2} = 5$
    \item $\boxnum{2} + 3 = 5$
  \end{enumerate}
\end{multicols}

\end{document}

Answer 2

I will assume that by "move the box towards the middle", you want the boxes to be vertically centered on the math axis. (Aside: The math axis is an invisible horizontal line that bisects symbols such as -, =, and +.) Moreover, I will assume that you'd like to achieve this goal without enlarging the frameboxes.

If these assumptions are correct, I suggest you encase the frameboxes in \vcenter{\hbox{...}} wrappers; \vcenter is a command that centers its argument vertically on the math axis.

By the way, I think it's easier to write \boxed{} than \fcolorbox{black}{white}{\null}.

\documentclass{article}
\usepackage{enumitem,multicol,xcolor,amsmath}
\fboxsep=4mm \fboxrule=0.5mm
\newcommand\mybox{%
   \vcenter{\hbox{$\boxed{}$}}} % feel free to come up with a snazzier name
\begin{document}

\noindent \textcolor{red}{Before}
\begin{multicols}{3}
\begin{enumerate}[label=(\alph*)]
\item $2 + \boxed{} = 5$
\item $\boxed{} + 2 = 5$

\item $1 + \boxed{} = 5$
\item $\boxed{} + 1 = 5$

\item $3 + \boxed{} = 5$
\item $\boxed{} + 3 = 5$
\end{enumerate}
\end{multicols}

\bigskip
\noindent \textcolor{red}{After}
\begin{multicols}{3}
\begin{enumerate}[label=(\alph*)]
\item $2 + \mybox = 5$
\item $\mybox + 2 = 5$

\item $1 + \mybox = 5$
\item $\mybox + 1 = 5$

\item $3 + \mybox = 5$
\item $\mybox + 3 = 5$
\end{enumerate}
\end{multicols}

\end{document}

Answer 3

Use a phantom of a symbol that's centered along the math axis, such as \sum and, to get a square, measure its height plus depth. Finally, decide about the value of \fboxsep to get the desired width of the square.

\documentclass{article}
\usepackage{enumitem,multicol}

\fboxsep=4mm \fboxrule=0.5mm % set these parameters globally

\newcommand{\fillnumber}{%
  \begingroup
  \setlength{\fboxrule}{0.5mm}%
  \setlength{\fboxsep}{1mm}%
  \sbox0{$\sum$}%
  \fbox{$\vphantom{\sum}$\hspace{\dimeval{\ht0+\dp0}}}%
  \endgroup
}

\begin{document}
\begin{multicols}{3}

\begin{enumerate}[label=(\alph*)]
\item $2 + \fillnumber = 5$
\item $\fillnumber + 2 = 5$

\item $1 + \fillnumber = 5$
\item $\fillnumber + 1 = 5$

\item $3 + \fillnumber = 5$
\item $\fillnumber + 3 = 5$
\end{enumerate}

\end{multicols}
\end{document}

Using tasks is, in my opinion, better than multicols.

\documentclass{article}
\usepackage{tasks}

\fboxsep=4mm \fboxrule=0.5mm % set these parameters globally

\newcommand{\fillnumber}{%
  \begingroup
  \setlength{\fboxrule}{0.5mm}%
  \setlength{\fboxsep}{1mm}%
  \sbox0{$\sum$}%
  \fbox{$\vphantom{\sum}$\hspace{\dimeval{\ht0+\dp0}}}%
  \endgroup
}

\begin{document}

\begin{tasks}(3)
\task $2 + \fillnumber = 5$
\task $\fillnumber + 2 = 5$
\task $1 + \fillnumber = 5$
\task $\fillnumber + 1 = 5$
\task $3 + \fillnumber = 5$
\task $\fillnumber + 3 = 5$
\end{tasks}

\end{document}