How do I encode problems 3.)
and 5.)
that is consistent with the remaining code? The display is what I want.
\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}
\newlength\labelwd
\settowidth\labelwd{\bfseries viii.)}
\usepackage{tasks}
\settasks{counter-format =tsk[a].), label-format=\bfseries, label-offset=1em, label-align=right, label-width
=\labelwd, before-skip =\smallskipamount, after-item-skip=0pt}
\usepackage{enumitem}
\setlist[enumerate,1]{% (
leftmargin=*, itemsep=12pt, label={\textbf{\arabic*.)}}}
\begin{document}
\begin{center}\Large{\textbf{Elementary Number Theory}}\end{center}\vskip0.25in
\begin{enumerate}[itemsep=\baselineskip]
\item How many positive integers less than 100 have a remainder of 3 upon division by 7?
\begin{tasks}(3)
\task 10
\task 11
\task 12
\task 13
\task 14
\end{tasks}
\end{enumerate}
\begin{enumerate}[start=2, itemsep=\baselineskip]
\item For every natural number $n$, $\tau(n)$ is the number of positive divisors of $n$. Evaluate $\tau^{3}(12)$.
\begin{tasks}(3)
\task 1
\task 2
\task 3
\task 4
\task 6
\end{tasks}
\end{enumerate}
\noindent {\textbf{3.) }}$p$ and $q$ are prime numbers greater than 2. which of the following statements must be true? \\
\hspace*{3em} \hphantom{3.)\ }
\begin{tabular}{r l}
{\bf I} & \hspace*{-0.5em}$p + q$ is even. \\
{\bf II} & \hspace*{-0.5em}$pq$ is odd. \\
{\bf III} & \hspace*{-0.5em}$p^{2} - q^{2}$ is even
\end{tabular}
\begin{tabbing}
\hspace*{2em} \= \hspace{2.5in} \= \kill
\> {\textbf{a.) }}I only \> {\textbf{b.) }}II only \\
\> {\textbf{c.) }}I and II only \> {\textbf{d.) }}I and III only \\
\> {\textbf{e.) }}I, II, and III
\end{tabbing}
\vskip0.25in
\begin{enumerate}[start=4, itemsep=\baselineskip]
\item How many integers less than 1000 are such that the remainder upon division by each of 2, 3, 4, 5, 6, and 7 is 1?
\begin{tasks}(3)
\task 0
\task 1
\task 2
\task 3
\task 4
\end{tasks}
\end{enumerate}
\noindent {\textbf{5.) }}$n$ is a positive integer. Which of the following quantities is divisible by 3? \\
\hspace*{3em} \hphantom{3.)\ }
\begin{tabular}{r l}
{\bf I} & \hspace*{-0.5em}$n^{3} - 1$ \\
{\bf II} & \hspace*{-0.5em}$n^{3} + 1$ \\
{\bf III} & \hspace*{-0.5em}$n^{3} + 2n$
\end{tabular}
\begin{tabbing}
\hspace*{2em} \= \hspace{2.5in} \= \kill
\> {\textbf{a.) }}I only \> {\textbf{b.) }}II only \\
\> {\textbf{c.) }}I and II only \> {\textbf{d.) }}II and III only \\
\> {\textbf{e.) }}I, II, and III
\end{tabbing}
\end{document}
Best Answer
I would do it this way:
A variant: