49

I don't want to display the modulo symbol, I want to programmatically calculate n modulo 3 and display the result.

8 Answers 8

37

You can also use \intcalcMod from the intcalc package:

\documentclass{article}
\usepackage{amsmath}
\usepackage{ifthen}
\usepackage{intcalc}

\newcounter{mycount}
\newcommand\Nmodiii[1]{%
\setcounter{mycount}{0}\whiledo{\value{mycount}<#1}
  {$\themycount\pmod 3=\intcalcMod{\value{mycount}}{3}$\\\stepcounter{mycount}}
}
\begin{document}

\noindent A little example: $8\pmod 3=\intcalcMod{8}{3}$

\noindent And a little loop:\\
\Nmodiii{20}

\end{document}

enter image description here

The code that appears in the link posted in a comment, there are some spurious blank spaces producing an undesired indentation of the first line; here's a corrected version:

\documentclass{article}
\usepackage{ifthen}
\usepackage{forloop}
\usepackage{fmtcount}
\usepackage{intcalc}
\usepackage{multicol}

\begin{document}
\begin{multicols}{2}
\newcounter{i}
\noindent\forloop{i}{1}{\value{i} < 101}{%
    \ifthenelse{\equal{\intcalcMod{\value{i}}{15}}{0}}{
        FizzBuzz
    }{%
        \ifthenelse{\equal{\intcalcMod{\value{i}}{3}}{0}}{
            Fizz
        }{%
            \ifthenelse{\equal{\intcalcMod{\value{i}}{5}}{0}}{
                Buzz
            }{%
                \thei
            }
        }
    }\\
}
\end{multicols}
\end{document}
13
  • ! Missing number, treated as zero. <to be read again> v l.32 }
    – mcandre
    Nov 11, 2011 at 1:13
  • Thanks, that helps. I'm writing FizzBuzz for LaTeX, so I need to use the mod result in further calculations, not just display it.
    – mcandre
    Nov 11, 2011 at 1:14
  • @mcandre: nothing prevents you from using the result in your calculations! Nov 11, 2011 at 1:15
  • Hmm. When I do this, I get ! Missing number, treated as zero. github.com/mcandre/mcandre/blob/master/latex/fizzy.tex
    – mcandre
    Nov 11, 2011 at 1:30
  • You are using "value" in your code and you should use \value (with a backslash). Also, you could have included that code in an edit to your original question. Nov 11, 2011 at 1:34
41

There are several nice answers using different packages. I'd like to note that TeX uses integer arithmetics, so it is easy to program the standard formula a-(a/b)*b, where / means integer division.

Plain TeX solution:

\newcount\tmpcnta
\def\modulo#1#2{\tmpcnta=#1
        \divide\tmpcnta by #2
        \multiply\tmpcnta by #2
        \multiply\tmpcnta by -1
        \advance\tmpcnta by #1\relax
        \the\tmpcnta}
\modulo{17}{3}
\modulo{19}{3}
\bye

LaTeX solution:

\documentclass{article} 
\makeatletter
\newcommand\modulo[2]{\@tempcnta=#1
        \divide\@tempcnta by #2
        \multiply\@tempcnta by #2
        \multiply\@tempcnta by -1
        \advance\@tempcnta by #1\relax
        \the\@tempcnta}
\makeatother
\begin{document} 
\modulo{17}{3}
\modulo{19}{3}
\end{document}
3
  • I used a similar solution in a calendar.
    – starblue
    Nov 11, 2011 at 14:21
  • 1
    Why can't you write \divide#1 by #2 or \divide\number\numexpr#1\relax by #2?
    – A.Ellett
    Dec 19, 2013 at 7:52
  • 1
    @A.Ellett \divide mutates (changes) the thing being divided (as do \multiply and \advance), and so we want to change our own counter \tmpcnta rather than #1. Nov 2, 2017 at 20:30
24

The "expandable" version, using e-TeX's \numexpr:

\def\truncdiv#1#2{((#1-(#2-1)/2)/#2)}
\def\moduloop#1#2{(#1-\truncdiv{#1}{#2}*#2)}
\def\modulo#1#2{\number\numexpr\moduloop{#1}{#2}\relax}

\truncdiv and \moduloop can be plugged into other expressions. It's necessary to do like this because \numexpr performs rounded integer division.

3
  • the same idea was used in the calendarweek TeX package.
    – ogerard
    Jun 5, 2013 at 9:27
  • 2
    Hmm, from \truncdiv{0}{64} I get -1 and so \modulo{0}{64} gives 64. Nov 2, 2017 at 20:26
  • 1
    I'm using the following for now, which works for positive #2: \def\moduloop#1#2{\ifnum \numexpr(#1 - (#1/#2)*(#2))\relax < 0 (#1 - (#1/#2)*(#2) + #2) \else (#1 - (#1/#2)*(#2)) \fi} and \def\truncdiv#1#2{((#1 - \moduloop{#1}{#2})/(#2))} Nov 2, 2017 at 20:44
22

The fp package is small and provides the functionality to do quite complex arithmetic. In the minimal example below the macro \modulo{<a>}{<b>} stores the result of <a> mod <b> in the macro \result, which is then directly printed:

enter image description here

\documentclass{article}
\usepackage[nomessages]{fp}% http://ctan.org/pkg/fp
\newcommand{\modulo}[2]{%
  \FPeval{\result}{trunc(#1-(#2*trunc(#1/#2,0)),0)}\result%
}
\begin{document}
Some modular arithmetic:
\begin{itemize}
  \item $512 \pmod{7}=\modulo{512}{7}$
  \item $6 \pmod{4}=\modulo{6}{4}$
  \item $15 \pmod{4}=\modulo{15}{4}$
  \item $1234567 \pmod{3}=\modulo{1234567}{3}$
\end{itemize}
\end{document}

Since the result is stored in \result, it can be used later in the text as well, until another execution of \modulo will overwrite \result.

Similar functionality in terms of mathematical functions is provided with pgf as well.

7
  • ! Undefined control sequence. \FP@@upn ...on \string "#2\string "}\edef \FP@tmp {[#2]}\expandafter \FP@upn... l.38 }
    – mcandre
    Nov 11, 2011 at 0:42
  • Do you receive this error when compiling my MWE?
    – Werner
    Nov 11, 2011 at 0:47
  • Yes. Specs: TeXworks on Mac OS X 10.7.2.
    – mcandre
    Nov 11, 2011 at 0:53
  • What distribution of TeX do you have installed? I have TeX Live 2011 with fp verion 1995/04/02.
    – Werner
    Nov 11, 2011 at 0:58
  • 2
    I like that a LaTeX solution to doing modular arithmetic is to load a package designed to do floating point calculations. WHAT!?
    – Seamus
    Feb 2, 2012 at 17:08
15

Another solution is to use pgfmath

\documentclass{article} 
\input{pgfutil-common.tex}
\usepackage{pgfkeys,pgfmath}
\begin{document}

\pgfmathparse{mod(20,6)} \pgfmathresult %displays 2.0
\pgfmathtruncatemacro{\myint}{ \pgfmathresult}  

\myint %displays 2 
\end{document}  
1
  • It seems that only the package pgfmath is necessary. \input{pgfutil-common.tex} and \usepackage{pgfkeys} could be removed.
    – dexteritas
    May 24, 2020 at 20:35
15

LaTeX3 (the expl3 package) also has a facility for computing modulus (moduli?), namely \int_mod:nn.

\documentclass{article}
\usepackage{expl3}
\ExplSyntaxOn
\newcommand{\mymod}[2]{\int_mod:nn{#1}{#2}}
\ExplSyntaxOff
\begin{document}
  The residue of $45$ modulo $19$ is $\mymod{45}{19}$.
\end{document}
8

You can use the calculator package then, type this code:

\MODULO{14}{3}{\sol}

$14\pmod{3}=\sol$
2
  • Nice, a new package! But you might want to make it clearer that you are the package author...
    – cgnieder
    Jun 12, 2012 at 10:35
  • How nice! :) Feel free to add your package to our list. And by the way, welcome to TeX.sx! :) Jun 12, 2012 at 10:54
2

You may use calc package as long as the absolute values of the numbers are not exceeding 2^31-1=2147483647. Otherwise you may use bigintcalc package.

\documentclass{article}
\usepackage{amsmath}
\usepackage{calc}
\newcounter{modulo}
\newcommand\modulo[2]{%
  \setcounter{modulo}{#1-(#1/#2)*#2}%
  \arabic{modulo}%
}
\begin{document}
\begin{align*}
131 \equiv \modulo{131}{3} &\pmod{3} \\
131 \equiv \modulo{131}{5} &\pmod{5} \\
131 \equiv \modulo{131}{7} &\pmod{7} \\
131 \equiv \modulo{131}{8} &\pmod{8} \\
-97 \equiv \modulo{-97}{3} &\pmod{3} \\
-97 \equiv \modulo{-97}{5} &\pmod{5} \\
-97 \equiv \modulo{-97}{7} &\pmod{7} \\
-97 \equiv \modulo{-97}{8} &\pmod{8} \\
\end{align*}
\end{document}

result of \modulo commands

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