Least common multiple

Compute the least common multiple of two integers.

Given m and n, the least common multiple is the smallest positive integer that has both m and n as factors. For example, the least common multiple of 12 and 18 is 36, because 12 is a factor (12 × 3 = 36), and 18 is a factor (18 × 2 = 36), and there is no positive integer less than 36 that has both factors. As a special case, if either m or n is zero, then the least common multiple is zero.

One way to calculate the least common multiple is to iterate all the multiples of m, until you find one that is also a multiple of n.

If you already have gcd for greatest common divisor, then this formula calculates lcm.

$\operatorname {lcm} (m,n)={\frac {|m\times n|}{\operatorname {gcd} (m,n)}}$

One can also find lcm by merging the prime decompositions of both m and n.

References: MathWorld, Wikipedia.

Ada

lcm_test.adb: <lang Ada>with Ada.Text_IO; use Ada.Text_IO;

procedure Lcm_Test is

  function Gcd (A, B : Integer) return Integer is
     M : Integer := A;
     N : Integer := B;
     T : Integer;
  begin
     while N /= 0 loop
        T := M;
        M := N;
        N := T mod N;
     end loop;
     return M;
  end Gcd;

  function Lcm (A, B : Integer) return Integer is
  begin
     if A = 0 or B = 0 then
        return 0;
     end if;
     return abs (A * B) / Gcd (A, B);
  end Lcm;

begin

  Put_Line ("LCM of 12, 18 is" & Integer'Image (Lcm (12, 18)));
  Put_Line ("LCM of -6, 14 is" & Integer'Image (Lcm (-6, 14)));
  Put_Line ("LCM of 35, 0 is" & Integer'Image (Lcm (35, 0)));

end Lcm_Test;</lang>

Output:

LCM of 12, 18 is 36
LCM of -6, 14 is 42
LCM of 35, 0 is 0

AutoHotkey

<lang autohotkey>LCM(Number1,Number2) {

If (Number1 = 0 || Number2 = 0)
 Return
Var := Number1 * Number2
While, Number2
 Num := Number2, Number2 := Mod(Number1,Number2), Number1 := Num
Return, Var // Number1

}

Num1 = 12 Num2 = 18 MsgBox % LCM(Num1,Num2)</lang>

AWK

<lang awk># greatest common divisor function gcd(m, n, t) { # Euclid's method while (n != 0) { t = m m = n n = t % n } return m }

least common multiple

function lcm(m, n, r) { if (m == 0 || n == 0) return 0 r = m * n / gcd(m, n) return r < 0 ? -r : r }

Read two integers from each line of input.
Print their least common multiple.

{ print lcm($1, $2) }</lang>

Example input and output:

$ awk -f lcd.awk
12 18
36
-6 14
42
35 0
0

BBC BASIC

Works with: BBC BASIC for Windows

     DEF FN_LCM(M%,N%)
     IF M%=0 OR N%=0 THEN =0 ELSE =ABS(M%*N%)/FN_GCD_Iterative_Euclid(M%, N%)
     
     DEF FN_GCD_Iterative_Euclid(A%, B%)
     LOCAL C%
     WHILE B%
       C% = A%
       A% = B%
       B% = C% MOD B%
     ENDWHILE
     = ABS(A%)

</lang>

bc

Translation of: AWK

<lang bc>/* greatest common divisor */ define g(m, n) { auto t

/* Euclid's method */ while (n != 0) { t = m m = n n = t % n } return (m) }

/* least common multiple */ define l(m, n) { auto r

if (m == 0 || n == 0) return (0) r = m * n / g(m, n) if (r < 0) return (-r) return (r) }</lang>

C

<lang C>#include <stdio.h>

int gcd(int m, int n) {

       int tmp;
       while(m) { tmp = m; m = n % m; n = tmp; }       
       return n;

}

int lcm(int m, int n) {

       return m / gcd(m, n) * n;

}

int main() {

       printf("lcm(35, 21) = %d\n", lcm(21,35));
       return 0;

}</lang>

C#

<lang csharp>public static int Lcm(int m, int n)

   {
     int r = 0;
     Func<int, int, int> gcd = delegate(int m2, int n2)
                                 {
                                   while (n2!=0)
                                   {
                                     var t2 = m2;
                                     m2 = n2;
                                     n2 = t2%n2;
                                   }
                                   return m2;
                                 };
     
     try
     {
       if (m == 0 || n == 0)
         throw new ArgumentException();
       r = Math.Abs(m*n)/gcd(m, n);
     }
     catch(Exception exception)
     {
       Console.WriteLine(exception.Message);
     }
     return (r<0) ? -r : r;
   }</lang>

Clojure

<lang Clojure> (defn gcd

     [a b]
     (if (zero? b)
     a
     (recur b, (mod a b))))

(defn lcm

     [a b]
     (/ (* a b) (gcd a b)))

</lang>

Common Lisp

Common Lisp provides the lcm function. It can accept two or more (or less) parameters.

Here is one way to reimplement it.

<lang lisp>CL-USER> (defun my-lcm (&rest args) (reduce (lambda (m n) (cond ((or (= m 0) (= n 0)) 0) (t (abs (/ (* m n) (gcd m n)))))) args :initial-value 1)) MY-LCM CL-USER> (my-lcm 12 18) 36 CL-USER> (my-lcm 12 18 22) 396</lang>

In this code, the lambda finds the least common multiple of two integers, and the reduce transforms it to accept any number of parameters. The reduce operation exploits how lcm is associative, (lcm a b c) == (lcm (lcm a b) c); and how 1 is an identity, (lcm 1 a) == a.

D

<lang d>import std.stdio, std.bigint;

T lcm(T)(T m, T n) {

   if (m == 0) return m;
   if (n == 0) return n;
   T r = (m * n) / gcd(m, n);
   return (r < 0) ? -r : r;

}

T gcd(T)(T a, T b) {

   while (b != 0) {
       auto t = b;
       b = a % b;
       a = t;
   }
   return a;

}

void main() {

   writeln(lcm(12, 18));
   writeln(lcm(BigInt("2562047788015215500854906332309589561"),
               BigInt("6795454494268282920431565661684282819")));

}</lang> Output:

36
15669251240038298262232125175172002594731206081193527869

DWScript

<lang delphi>PrintLn(Lcm(12, 18));</lang> Output:

Euphoria

<lang euphoria>function gcd(integer m, integer n)

   integer tmp
   while m do
       tmp = m
       m = remainder(n,m)
       n = tmp
   end while
   return n

end function

function lcm(integer m, integer n)

   return m / gcd(m, n) * n

end function</lang>

Factor

The vocabulary math.functions already provides lcm.

<lang factor>USING: math.functions prettyprint ; 26 28 lcm .</lang>

This program outputs 364.

One can also reimplement lcm.

<lang factor>USING: kernel math prettyprint ; IN: script

gcd ( a b -- c )

   [ abs ] [
       [ nip ] [ mod ] 2bi gcd
   ] if-zero ;

lcm ( a b -- c )

   [ * abs ] [ gcd ] 2bi / ;

26 28 lcm .</lang>

Forth

<lang forth>: gcd ( a b -- n )

 begin dup while tuck mod repeat drop ;

lcm ( a b -- n )

 over 0= over 0= or if 2drop 0 exit then
 2dup gcd abs */ ;</lang>

Frink

Frink has a built-in LCM function that handles arbitrarily-large integers. <lang frink> println[lcm[2562047788015215500854906332309589561, 6795454494268282920431565661684282819]] </lang>

GAP

<lang gap># Built-in LcmInt(12, 18);

36</lang>

Go

<lang go>package main

import (

   "fmt"
   "math/big"

)

var m, n big.Int

func init() {

   m.SetString("2562047788015215500854906332309589561", 10)
   n.SetString("6795454494268282920431565661684282819", 10)

}

func main() {

   var d big.Int
   big.GcdInt(&d, nil, nil, &m, &n)
   fmt.Println(d.Mul(d.Div(&m, &d), &n))

}</lang> Output:

15669251240038298262232125175172002594731206081193527869

Haskell

That is already available as the function lcm in the Prelude. Here's the implementation:

<lang haskell>lcm :: (Integral a) => a -> a -> a lcm _ 0 = 0 lcm 0 _ = 0 lcm x y = abs ((x `quot` (gcd x y)) * y)</lang>

Icon and Unicon

The lcm routine from the Icon Programming Library uses gcd. The routine is

<lang Icon>link numbers procedure main() write("lcm of 18, 36 = ",lcm(18,36)) write("lcm of 0, 9 36 = ",lcm(0,9)) end</lang>

Library: Icon Programming Library

numbers provides lcm and gcd and looks like this: <lang Icon>procedure lcm(i, j) #: least common multiple

  if (i =  0) | (j = 0) then return 0	
  return abs(i * j) / gcd(i, j)

end</lang>

J

J provides the dyadic verb *. which returns the least common multiple of its left and right arguments.

<lang j> 12 *. 18 36

  12 *. 18 22

36 132

  *./ 12 18 22

396

  0 1 0 1 *. 0 0 1 1  NB. for boolean arguments (0 and 1) it is equivalent to "and"

0 0 0 1

  *./~ 0 1

0 0 0 1</lang>

Java

<lang java>import java.util.Scanner;

public class LCM{

  public static void main(String[] args){
     Scanner aScanner = new Scanner(System.in);
  
     //prompts user for values to find the LCM for, then saves them to m and n
     System.out.print("Enter the value of m:");
     int m = aScanner.nextInt();
     System.out.print("Enter the value of n:");
     int n = aScanner.nextInt();
     int lcm = (n == m || n == 1) ? m :(m == 1 ? n : 0);
     /* this section increases the value of mm until it is greater  
     / than or equal to nn, then does it again when the lesser 
     / becomes the greater--if they aren't equal. If either value is 1,
     / no need to calculate*/
     if (lcm == 0) {
        int mm = m, nn = n;
        while (mm != nn) {
            while (mm < nn) { mm += m; }
            while (nn < mm) { nn += n; }
        }  
        lcm = mm;
     }
     System.out.println("lcm(" + m + ", " + n + ") = " + lcm);
  }

}</lang>

K

<lang K> gcd:{:[~x;y;_f[y;x!y]]}

  lcm:{_abs _ x*y%gcd[x;y]}

  lcm .'(12 18; -6 14; 35 0)

36 42 0

  lcm/1+!20

232792560</lang>

Liberty BASIC

<lang lb>print "Least Common Multiple of 12 and 18 is ";LCM(12,18) end

function LCM(m,n)

   LCM=abs(m*n)/GCD(m,n)
   end function

function GCD(a,b)

   while b
       c = a
       a = b
       b = c mod b
   wend
   GCD = abs(a)
   end function
</lang>

Logo

<lang logo>to abs :n

 output sqrt product :n :n

end

to gcd :m :n

 output ifelse :n = 0 [ :m ] [ gcd :n modulo :m :n ]

end

to lcm :m :n

 output quotient (abs product :m :n) gcd :m :n

end</lang>

Demo code:

<lang logo>print lcm 38 46</lang>

Output:

Lua

<lang lua>function gcd( m, n )

   while n ~= 0 do
       local q = m
       m = n
       n = q % n
   end
   return m

end

function lcm( m, n )

   return ( m ~= 0 and n ~= 0 ) and m * n / gcd( m, n ) or 0

end

print( lcm(12,18) )</lang>

OCaml

<lang ocaml>let rec gcd u v =

 if v <> 0 then (gcd v (u mod v))
 else (abs u)

let lcm m n =

 match m, n with
 | 0, _ | _, 0 -> 0
 | m, n -> abs (m * n) / (gcd m n)

let () =

 Printf.printf "lcm(35, 21) = %d\n" (lcm 21 35)</lang>

PARI/GP

Built-in function: <lang parigp>lcm</lang>

Perl

Using GCD: <lang Perl>sub gcd { my ($a, $b) = @_; while ($a) { ($a, $b) = ($b % $a, $a) } $b }

sub lcm { my ($a, $b) = @_; ($a && $b) and $a / gcd($a, $b) * $b or 0 }

print lcm(1001, 221);</lang> Or by repeatedly increasing the smaller of the two until LCM is reached:<lang perl>sub lcm { use integer; my ($x, $y) = @_; my ($a, $b) = @_; while ($a != $b) { ($a, $b, $x, $y) = ($b, $a, $y, $x) if $a > $b; $a = $b / $x * $x; $a += $x if $a < $b; } $a }

print lcm(1001, 221);</lang>

Perl 6

This function is provided as an infix so that it can be used productively with various metaoperators. <lang>say 3 lcm 4; # infix say [lcm] 1..20; # reduction say ~(1..10 Xlcm 1..10) # cross</lang> Output:

12
232792560
1 2 3 4 5 6 7 8 9 10 2 2 6 4 10 6 14 8 18 10 3 6 3 12 15 6 21 24 9 30 4 4 12 4 20 12 28 8 36 20 5 10 15 20 5 30 35 40 45 10 6 6 6 12 30 6 42 24 18 30 7 14 21 28 35 42 7 56 63 70 8 8 24 8 40 24 56 8 72 40 9 18 9 36 45 18 63 72 9 90 10 10 30 20 10 30 70 40 90 10

PHP

Translation of: D

<lang php>echo lcm(12, 18) == 36;

function lcm($m, $n) {

   if ($m == 0 || $n == 0) return 0;
   $r = ($m * $n) / gcd($m, $n);
   return abs($r);

}

function gcd($a, $b) {

   while ($b != 0) {
       $t = $b;
       $b = $a % $b;
       $a = $t;
   }
   return $a;

}</lang>

PicoLisp

Using 'gcd' from Greatest common divisor#PicoLisp: <lang PicoLisp>(de lcm (A B)

  (abs (*/ A B (gcd A B))) )</lang>

Prolog

SWI-Prolog knows gcd. <lang Prolog>lcm(X, Y, Z) :- Z is abs(X * Y) / gcd(X,Y). </lang> Example :

 ?- lcm(18,12, Z).
Z = 36.

PureBasic

<lang PureBasic>Procedure GCDiv(a, b); Euclidean algorithm

 Protected r
 While b
   r = b
   b = a%b
   a = r
 Wend
 ProcedureReturn a

EndProcedure

Procedure LCM(m,n)

 Protected t
 If m And n
   t=m*n/GCDiv(m,n)
 EndIf
 ProcedureReturn t*Sign(t)

EndProcedure</lang>

Python

gcd

Using the fractions libraries gcd function: <lang python>>>> import fractions >>> def lcm(a,b): return abs(a * b) / fractions.gcd(a,b) if a and b else 0

>>> lcm(12, 18) 36 >>> lcm(-6, 14) 42 >>> assert lcm(0, 2) == lcm(2, 0) == 0 >>> </lang>

Prime decomposition

This imports Prime decomposition#Python <lang python> import operator from prime_decomposition import decompose

def lcm(a, b):

   if a and b:
       da = list(decompose(abs(a)))
       db = list(decompose(abs(b)))
       merge= da
       for d in da:
           if d in db: db.remove(d)
       merge += db
       return reduce(operator.mul, merge, 1)
   return 0

if __name__ == '__main__':

   print( lcm(12, 18) )    # 36
   print( lcm(-6, 14) )    # 42
   assert lcm(0, 2) == lcm(2, 0) == 0</lang>

Iteration over multiples

<lang python>>>> def lcm(*values): values = set([abs(int(v)) for v in values]) if values and 0 not in values: n = n0 = max(values) values.remove(n) while any( n % m for m in values ): n += n0 return n return 0

>>> lcm(-6, 14) 42 >>> lcm(2, 0) 0 >>> lcm(12, 18) 36 >>> lcm(12, 18, 22) 396 >>> </lang>

Repeated modulo

Translation of: Tcl

<lang python>>>> def lcm(p,q): p, q = abs(p), abs(q) m = p * q if not m: return 0 while True: p %= q if not p: return m // q q %= p if not q: return m // p

>>> lcm(-6, 14) 42 >>> lcm(12, 18) 36 >>> lcm(2, 0) 0 >>> </lang>

Qi

<lang qi> (define gcd

 A 0 -> A
 A B -> (gcd B (MOD A B)))

(define lcm A B -> (/ (* A B) (gcd A B))) </lang>

Retro

This is from the math extensions library included with Retro.

<lang Retro>: gcd ( ab-n ) [ tuck mod dup ] while drop ;

lcm ( ab-n ) 2over gcd [ * ] dip / ;</lang>

Ruby

Ruby has an Integer#lcm method, which finds the least common multiple of two integers.

<lang ruby>irb(main):001:0> require 'rational' => true irb(main):002:0> 12.lcm 18 => 36</lang>

I can also write my own lcm method. This one takes any number of arguments, and works by iterating the multiples of m until it finds a multiple of n.

<lang ruby>def lcm(*args)

 args.inject(1) do |m, n|
   next 0 if m == 0 or n == 0
   i = m
   loop do
     break i if i % n == 0
     i += m
   end
 end

end</lang>

Scala

Scheme

Seed7

<lang seed7>$ include "seed7_05.s7i";

const func integer: gcd (in var integer: a, in var integer: b) is func

 result
   var integer: result is 0;
 local
   var integer: help is 0;
 begin
   while a <> 0 do
     help := b rem a;
     b := a;
     a := help;
   end while;
   result := b;
 end func;

const func integer: lcm (in integer: a, in integer: b) is

 return a div gcd(a, b) * b;

const proc: main is func

 begin
   writeln("lcm(35, 21) = " <& lcm(21, 35));
 end func;</lang>

Original source: [1]

Tcl

<lang tcl>proc lcm {p q} {

   set m [expr {$p * $q}]
   if {!$m} {return 0}
   while 1 {

set p [expr {$p % $q}] if {!$p} {return [expr {$m / $q}]} set q [expr {$q % $p}] if {!$q} {return [expr {$m / $p}]}

}</lang> Demonstration <lang tcl>puts [lcm 12 18]</lang> Output:

TI-83 BASIC

<lang ti83b>lcm(12, 18)

              36</lang>

UNIX Shell

$\operatorname {lcm} (m,n)=\left|{\frac {m\times n}{\operatorname {gcd} (m,n)}}\right|$

Works with: Bourne Shell

<lang bash>gcd() { # Calculate $1 % $2 until $2 becomes zero. until test 0 -eq "$2"; do # Parallel assignment: set -- 1 2 set -- "$2" "`expr "$1" % "$2"`" done

# Echo absolute value of $1. test 0 -gt "$1" && set -- "`expr 0 - "$1"`" echo "$1" }

lcm() { set -- "$1" "$2" "`gcd "$1" "$2"`" set -- "`expr "$1" \* "$2" / "$3"`" test 0 -gt "$1" && set -- "`expr 0 - "$1"`" echo "$1" }

lcm 30 -42

=> 210</lang>

C Shell

<lang csh>alias gcd eval \set gcd_args=( \!*:q ) \\ @ gcd_u=$gcd_args[2] \\ @ gcd_v=$gcd_args[3] \\ while ( $gcd_v != 0 ) \\ @ gcd_t = $gcd_u % $gcd_v \\ @ gcd_u = $gcd_v \\ @ gcd_v = $gcd_t \\ end \\ if ( $gcd_u < 0 ) @ gcd_u = - $gcd_u \\ @ $gcd_args[1]=$gcd_u \\ '\'

alias lcm eval \set lcm_args=( \!*:q ) \\ @ lcm_m = $lcm_args[2] \\ @ lcm_n = $lcm_args[3] \\ gcd lcm_d $lcm_m $lcm_n \\ @ lcm_r = ( $lcm_m * $lcm_n ) / $lcm_d \\ if ( $lcm_r < 0 ) @ lcm_r = - $lcm_r \\ @ $lcm_args[1] = $lcm_r \\ '\'

lcm result 30 -42 echo $result

=> 210</lang>

Vala

<lang vala> int lcm(int a, int b){

   /*Return least common multiple of two ints*/
   // check for 0's                                                            
   if (a == 0 || b == 0)

return 0;

   // Math.abs(x) only works for doubles, Math.absf(x) for floats              
   if (a < 0)
       a *= -1;
   if (b < 0)

b *= -1;

   int x = 1;
   while (true){
       if (a * x % b == 0)
           return a*x;
       x++;
   }

}

void main(){

   int	a = 12;
   int	b = 18;

   stdout.printf("lcm(%d, %d) = %d\n",	a, b, lcm(a, b));

} </lang>