Right now I have the following for a vertical addition problem but I would also like it to be numbered without messing everything else up.

\begin{array}{c@{\,}c}
& 1 \\
+ & 1 \\
\hline
& 2 \\
\end{array}
-

You could also just use an enumerate type of environment. With that I would recommend you use the aligned environment which will allow specification of alignment of the numbers, but array can be used as well:

## Note:

• The addition problem used a macro only to improve readability in this example.

## Code:

\documentclass{article}
\usepackage{amsmath}

\begin{array}{c@{\,}c}
& 1 \\
+ & 1 \\
\hline
& 2 \\
\end{array}
}%

\begin{aligned}[t]% Choose alignement: t, b, or c
& 1 \\
+ & 1 \\
\hline
& 2
\end{aligned}
}%

\begin{document}\noindent
Using \texttt{array} environemnt
\begin{enumerate}
\item $\ProblemInArray$
\item $\ProblemInArray$
\end{enumerate}
%
Using \texttt{aligned} environemnt
\begin{enumerate}
\item $\ProblemInAligned$
\item $\ProblemInAligned$
\end{enumerate}
\end{document}
-

You can try,

$\begin{array}{@{}cr} & 1 \\ + & 10 \\ \cline{2-2} & 11 \end{array}$

You may want also to adjust the space between the columns by using

\setlength\arraycolsep{0.1em}

There are many ways to number the equations, but for this type of problem, I would either use the multienum package or incorporate the numbering in the array manually, or create a counter to auto increment. Here is a rather longer example. Of course it will look better if the + sign was on the right like we do in my part of the world.

\documentclass[fleqn]{article}
\usepackage{amsmath,multienum,booktabs}
\renewcommand{\regularlisti}{\setcounter{multienumi}{0}%
\renewcommand{\labelenumi}%
$\begin{array}{lrr} \text{(i)}& & 1 \\ &+ & 10 \\ \cmidrule(lr){3-3} & & 11 \\ \end{array}$
$\begin{array}{lrr} \text{(ii)}& & 1 \\ &+ & 10 \\ \cmidrule(lr){3-3} & & 11 \\ \end{array}$
$\begin{array}{crr@{}c@{}r@{}c@{}l} \text{(iii)} &237 &100a &{}+{}&10b &{}+{}&c\\ \addlinespace \text{(iv)} &732 &100c &{}+{}&10b &{}+{}&a\\ \cmidrule(lr){2-2}\cmidrule(lr){3-7} \text{(v)} &495 &\multicolumn{1}{l@{}}{100(a-c-1)} &+&90 &+&(10+c-a)\\ \addlinespace \text{(vi)} &594 &\multicolumn{1}{l@{}}{100(10+c-a)} &+&90 &+&(a-c-1)\\ \cmidrule(lr){2-2}\cmidrule(lr){3-7} \text{(vii)} &1089 &\multicolumn{1}{c}{900} &+&180 &+&9\\ \cmidrule(lr){2-2}\cmidrule(lr){3-7} \specialrule{0pt}{-1pt}{-1pt} \cmidrule(lr){2-2}\cmidrule(lr){3-7} \end{array}$
the op requested a numbered problem, so $$...$$ would be more appropriate. –  barbara beeton Mar 17 '12 at 14:08