# How can I make an odd-numbered list of questions/solutions?

I'm trying to make two odd-numbered lists, one with excercises and the other one with solutions, but the latter should have only 4k+1 odd-numbered items. Example: List of questions: 1) 3) 5) 7) etc List of solutions 1) 5) 9) 13) Which commands should I use??

• Take a look at the enumitem for a package which gives you control over the appearance of lists such as enumerate. – Skillmon May 28 '17 at 17:15
• What is k here? And how does (say) having q questions translate into 4k + 1 items? – Werner May 28 '17 at 17:16
• @Werner I guess he means $k \in \mathbb{N}^+$ (or he wants 4001 odd-numbered items, who knows) – Skillmon May 28 '17 at 17:23
• Sorry, I'm going to edit my quesiton. – Juani May 28 '17 at 17:44

I hope I got your question right. Here is a rather primitive way of accomplishing what I hope you want:

\documentclass[]{article}

\let\itembak\item
\makeatletter
\newcommand{\odditems}[1][2]{%
\setcounter{\@listctr}{\numexpr1-#1}%
\renewcommand{\item}[1][]{%
\ifx\relax##1\relax%
\itembak%
\else%
\itembak[##1]%
\fi%
}%
}
\makeatother

\begin{document}
\noindent
first:
\begin{enumerate}
\odditems
\item first
\item second
\item third
\end{enumerate}
second:
\begin{enumerate}
\odditems[4]
\item first
\item second
\item third
\end{enumerate}
\end{document}


EDIT: The enumerate is nestable (but in a counter intuitive way):

\begin{enumerate}
\odditems
\item first
\item second
\item third
\begin{enumerate}
\odditems[1]% this second call of \odditems allows the nested use
\item third.first
\item third.second
\item third.third
\end{enumerate}
\item fourth
\end{enumerate}

• @Juani though I am glad that it works and is what you wanted, you should not accept this answer just now. It is (almost) always better to wait a few hours so you might get better answers. – Skillmon May 28 '17 at 17:49

The following example provides an automated way of setting questions and answers. (Expandable) calculations are provided by xfp and can be generalised as needed:

\documentclass{article}

\usepackage{xfp}

\newcounter{question}\newcounter{question@}
\newenvironment{questions}
{\setcounter{question@}{0}%
\begin{enumerate}
\let\olditem\item
\renewcommand{\item}{%
\setcounter{question}{\fpeval{2*\value{question@}}}\refstepcounter{question}%
\olditem[\thequestion)]\leavevmode\stepcounter{question@}%
}}
{\end{enumerate}}

\begin{enumerate}
\let\olditem\item
\renewcommand{\item}{%
}}
{\end{enumerate}}

\begin{document}

Questions:
\begin{questions}
\item first
\item second
\item third
\item fourth
\end{questions}


The modification of \item doesn't allow for nesting of other lists within questions or answers.
• I don't get why you need xfp for that. You could use \addtocounter{answer@}{4} instead of the \fpeval{4*\value{anwer@} (same is true for question). What is the advantage of using the package? – Skillmon May 28 '17 at 18:36
• @skillmon True; I reference the fact that with xfp one can generalize the calculation to anything arbitrary. – Werner May 28 '17 at 18:40