# How do I calculate n modulo 3 in LaTeX?

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

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}


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}

• ! Missing number, treated as zero. <to be read again> v l.32 } – mcandre Nov 11 '11 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 '11 at 1:14
• @mcandre: nothing prevents you from using the result in your calculations! – Gonzalo Medina Nov 11 '11 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 '11 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. – Gonzalo Medina Nov 11 '11 at 1:34

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
\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
\the\@tempcnta}
\makeatother
\begin{document}
\modulo{17}{3}
\modulo{19}{3}
\end{document}

• I used a similar solution in a calendar. – starblue Nov 11 '11 at 14:21
• Why can't you write \divide#1 by #2 or \divide\number\numexpr#1\relax by #2? – A.Ellett Dec 19 '13 at 7:52
• @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. – ShreevatsaR Nov 2 '17 at 20:30

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:

\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.

• ! Undefined control sequence. \FP@@upn ...on \string "#2\string "}\edef \FP@tmp {[#2]}\expandafter \FP@upn... l.38 } – mcandre Nov 11 '11 at 0:42
• Do you receive this error when compiling my MWE? – Werner Nov 11 '11 at 0:47
• Yes. Specs: TeXworks on Mac OS X 10.7.2. – mcandre Nov 11 '11 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 '11 at 0:58
• 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 '12 at 17:08

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.

• the same idea was used in the calendarweek TeX package. – ogerard Jun 5 '13 at 9:27
• Hmm, from \truncdiv{0}{64} I get -1 and so \modulo{0}{64} gives 64. – ShreevatsaR Nov 2 '17 at 20:26
• 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))} – ShreevatsaR Nov 2 '17 at 20:44

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}


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}


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

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

$14\pmod{3}=\sol$

• Nice, a new package! But you might want to make it clearer that you are the package author... – cgnieder Jun 12 '12 at 10:35
• How nice! :) Feel free to add your package to our list. And by the way, welcome to TeX.sx! :) – Paulo Cereda Jun 12 '12 at 10:54

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}