# Numbering for Alternative Solutions in exam class

I am currently writing up solutions for past year exam, and found out there is no way to numbering the solutions for the same question, if necessary. (Some question ends up with numerous solutions.)

\usepackage[margin=2.54cm, a4paper]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\runningfooter{}{Page \thepage \,of Page \numpages}{}
\newcounter{solution}
\unframedsolutions
\renewcommand{\solutiontitle}{\noindent\textbf{Solution \thequestion: }}
\begin{document}
\begin{titlepage}
\centering
\hspace{10cm}\\[5pt]
\vfill
{\scshape\Huge School \par}
\vspace{1cm}
\vspace{1cm}
{\huge\bfseries Solution Manual\par}
\vspace{1cm}
{\Large Trial Exam\par}
\vspace{1cm}
{\Large October 201X\par}
\vspace{1cm}
{\large \emph{Written by} Someone}
\vfill
\end{titlepage}
\newpage
\Large \section*{Preface}
blablabla
\newpage
\begin{questions}
\question Suppose $\cos 2\theta=k$, express $\sin^6 \theta-\cos^6 \theta$ in terms of $k$.
\begin{solution}
Notice that
\begin{align*}
k & =\cos 2\theta\\
& =2\cos^2 \theta-1\\
& =1-2\sin^2 \theta
\end{align*}
Hence, both $\sin^2 \theta$ and $\cos^2 \theta$ can be expressed in terms of $k$ as follow:\\
\begin{equation}
\label{eq:eq1}
\cos^2 \theta=\dfrac{k+1}{2}\\
\end{equation}
Substituting Equation \eqref{eq:eq1} yields
\begin{align*}
\sin^6 \theta-\cos^6 \theta &= \left(\dfrac{1-k}{3}\right)^3-\left(\dfrac{k+1}{3}\right)^3\\
&=-\dfrac{k^3+3k}{4}
\end{align*}
\end{solution}
\begin{solution}
aaa
\end{solution}
\end{questions}
\end{document} I would like to make the numbering "Solution 2:", if possible, and if the question only has one solution, I would like to see "Solution:" instead. How could I do it?

The easy way is to add an enumerate environment inside the solution environment.

\documentclass{exam}
\usepackage[margin=2.54cm, a4paper]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\runningfooter{}{Page \thepage \,of Page \numpages}{}
%\newcounter{solution}
\unframedsolutions
%\renewcommand{\solutiontitle}{\noindent\textbf{Solution \thequestion: }}
\begin{document}
\begin{titlepage}
\centering
\hspace{10cm}\\[5pt]
\vfill
{\scshape\Huge School \par}
\vspace{1cm}
\vspace{1cm}
{\huge\bfseries Solution Manual\par}
\vspace{1cm}
{\Large Trial Exam\par}
\vspace{1cm}
{\Large October 201X\par}
\vspace{1cm}
{\large \emph{Written by} Someone}
\vfill
\end{titlepage}
\newpage
\Large \section*{Preface}
blablabla
\newpage
\begin{questions}
\question Suppose $\cos 2\theta=k$, express $\sin^6 \theta-\cos^6 \theta$ in terms of $k$.
\begin{solution}
\begin{enumerate}
\item
Notice that
\begin{align*}
k & =\cos 2\theta\\
& =2\cos^2 \theta-1\\
& =1-2\sin^2 \theta
\end{align*}
Hence, both $\sin^2 \theta$ and $\cos^2 \theta$ can be expressed in terms of $k$ as follow:\\
\begin{equation}
\label{eq:eq1}
\cos^2 \theta=\dfrac{k+1}{2}\\
\end{equation}
Substituting Equation \eqref{eq:eq1} yields
\begin{align*}
\sin^6 \theta-\cos^6 \theta &= \left(\dfrac{1-k}{3}\right)^3-\left(\dfrac{k+1}{3}\right)^3\\
&=-\dfrac{k^3+3k}{4}
\end{align*}
\item
aaa
\end{enumerate}
\end{solution}
\end{questions}
\end{document}


This version uses a multisolution environment to initiate numbering solutions.

\documentclass{exam}
\usepackage[margin=2.54cm, a4paper]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\runningfooter{}{Page \thepage \,of Page \numpages}{}
\newcounter{solution}
\unframedsolutions
\newenvironment{multisolution}{%
\setcounter{solution}{0}%
\def\solutiontitle{\stepcounter{solution}\noindent\textbf{Solution \thesolution: }}% local definition
\ignorespaces}%
{\ignorespaces}
\begin{document}
\begin{titlepage}
\centering
\hspace{10cm}\\[5pt]
\vfill
{\scshape\Huge School \par}
\vspace{1cm}
\vspace{1cm}
{\huge\bfseries Solution Manual\par}
\vspace{1cm}
{\Large Trial Exam\par}
\vspace{1cm}
{\Large October 201X\par}
\vspace{1cm}
{\large \emph{Written by} Someone}
\vfill
\end{titlepage}
\newpage
\Large \section*{Preface}
blablabla
\newpage
\begin{questions}
\question Suppose $\cos 2\theta=k$, express $\sin^6 \theta-\cos^6 \theta$ in terms of $k$.
\begin{multisolution}
\begin{solution}
Notice that
\begin{align*}
k & =\cos 2\theta\\
& =2\cos^2 \theta-1\\
& =1-2\sin^2 \theta
\end{align*}
Hence, both $\sin^2 \theta$ and $\cos^2 \theta$ can be expressed in terms of $k$ as follow:\\
\begin{equation}
\label{eq:eq1}
\cos^2 \theta=\dfrac{k+1}{2}\\
\end{equation}
Substituting Equation \eqref{eq:eq1} yields
\begin{align*}
\sin^6 \theta-\cos^6 \theta &= \left(\dfrac{1-k}{3}\right)^3-\left(\dfrac{k+1}{3}\right)^3\\
&=-\dfrac{k^3+3k}{4}
\end{align*}
\end{solution}
\begin{solution}
aaa
\end{solution}
\end{multisolution}
\end{questions}
\end{document}