Enumerate Horizontally with multicols

I would like to have a maths exercise in a document so that the questions are numbered thus:

1. Question 1
a) subquestion 1      b) subquestion 2    c)subquestion 3
d) subquestion 4      e) subquestion 5    f)subquestion 6
etc.

2. Question 2


I've managed to achieve something similar using the multicols package, but this renders the letters in vertical order, rather than horizontal. The code of my attempt is detailed below:

\documentclass{report}
\usepackage{multicol,amsmath,amsfonts}
\begin{document}
\subsubsection*{Exercise 18.1}
\begin{enumerate}
\item Prove the following by induction on $n$, where $n\in\mathbb Z^+$.
\begin{enumerate}
\begin{multicols}{2}
\item $\displaystyle\sum_{r=0}^n 2^r=2^{n+1}-1$
\item $5+5^2+\cdots+5^n=\displaystyle\frac{5(5^n-1)}{4}$
\item $\displaystyle\sum_{r=0}^n a^r=\frac{a(a^n-1)}{a-1};~a\in\mathbb R$\\
\item The sum of the first $n$ odd numbers is $n^2$.
\end{multicols}
\item $\displaystyle(1+a)+2(2+a)+3(3+a)+\cdots+n(n+a)=\frac{n}{6}(n+1)(3a+2n+1);~a\in\mathbb R$
\item $5(4)+6(5)+7(6)+\cdots+n(n-1)=\displaystyle\frac{1}{3}(n^3-n-60);~\forall n\geq 5$
\item $1^4+2^4+3^4+\cdots+n^4=\displaystyle\frac{n}{30}(n+1)(2n+1)(3n^2+3n-1)$
\end{enumerate}
\item Prove the following statements by induction.
\begin{enumerate}
\begin{multicols}{2}
\item $\displaystyle\sum_{r=1}^n r^5=\frac{n^2}{12}(n+1)^2(2n^2+2n-1)$
\item $\displaystyle\sum_{r=1}^n \frac{1}{r(r+1)}=\frac{r}{r+1}$
\item $\displaystyle\prod_{r=1}^n r^{12}=(r!)^{12}$
\end{multicols}
\end{enumerate}
\end{enumerate}
\end{document}


Which renders the following output: Anyone know how I can achieve the horizontal ordered numbering?

The tasks package is made for that. I added enumitem, to easily customise the enumerate environment (wide option). Also, I simplifies your your code with everymath{\displaystyle}at the beginning of the enumerate environment. The labels can be easily customised with the counter-format key. Last, not least, if an item is longer than one column, it it is typed as a \parbox. Alternatively, you can let it spread for more than one column with the star version of \tasks:

\documentclass{report}
\usepackage{multicol,amsmath,amsfonts}
\usepackage{enumitem}

\begin{document}

\subsubsection*{Exercise 18.1}
\begin{enumerate}[wide = 0pt]\everymath{\displaystyle}
\item Prove the following by induction on $n$, where $n\in\mathbb Z^+$.
\task $\sum_{r=0}^n 2^r=2^{n+1}-1$
\task $5+5^2+\cdots+5^n=\frac{5(5^n-1)}{4}$
\task $\sum_{r=0}^n a^r=\frac{a(a^n-1)}{a-1};~a\in\mathbb R$\\
\task The sum of the first $n$ odd numbers is $n^2$.
\task* $(1+a)+2(2+a)+3(3+a)+\cdots+n(n+a)=\frac{n}{6}(n+1)(3a+2n+1);~a\in\mathbb R$
\task* $5(4)+6(5)+7(6)+\cdots+n(n-1)=\frac{1}{3}(n^3-n-60);~\forall n\geq 5$
\task* $1^4+2^4+3^4+\cdots+n^4=\frac{n}{30}(n+1)(2n+1)(3n^2+3n-1)$
\item Prove the following statements by induction.
\task $\sum_{r=1}^n r^5=\frac{n^2}{12}(n+1)^2(2n^2+2n-1)$
\task $\sum_{r=1}^n \frac{1}{r(r+1)}=\frac{r}{r+1}$
\task $\prod_{r=1}^n r^{12}=(r!)^{12}$


• @user93370: To have a roman counter, the syntax is [counter-format={tsk[r].}]. For the second problem, add something like [label-width=1.5em]. – Bernard Jul 5 '17 at 19:12