# Least common multiple

Least common multiple
You are encouraged to solve this task according to the task description, using any language you may know.

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.

## Contents

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) * (abs (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;

Output:

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

## 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 = 12Num2 = 18MsgBox % LCM(Num1,Num2)

12 18
36
-6 14
42
35 0
0


## BASIC

### Applesoft BASIC

ported from BBC BASIC

10 DEF FN MOD(A) = INT((A / B - INT(A / B)) * B + .05) * SGN(A / B)20 INPUT"M=";M%30 INPUT"N=";N%40 GOSUB 10050 PRINT R60 END 100 REM LEAST COMMON MULTIPLE M% N%110 R = 0120 IF M% = 0 OR N% = 0 THEN RETURN130 A% = M% : B% = N% : GOSUB 200"GCD140 R = ABS(M%*N%)/R150 RETURN 200 REM GCD ITERATIVE EUCLID A% B%210 FOR B = B% TO 0 STEP 0220     C% = A%230     A% = B240     B = FN MOD(C%)250 NEXT B260 R = ABS(A%)270 RETURN

### BBC BASIC

       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%)

## bc

Translation of: AWK
/* 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)}

We utilize the fact that Bracmat simplifies fractions (using Euclid's algorithm). The function den$number returns the denominator of a number. (gcd= a b. !arg:(?a.?b) & den$(!a*!b^-1)    * (!a:<0&-1|1)    * !a);out$(gcd$(12.18) gcd$(-6.14) gcd$(35.0) gcd$(117.18)) Output: 36 42 35 234 ## 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;} ## C++ Library: Boost #include <boost/math/common_factor.hpp>#include <iostream> int main( ) { std::cout << "The least common multiple of 12 and 18 is " << boost::math::lcm( 12 , 18 ) << " ,\n" << "and the greatest common divisor " << boost::math::gcd( 12 , 18 ) << " !" << std::endl ; return 0 ;} Output: The least common multiple of 12 and 18 is 36 , and the greatest common divisor 6 !  ## C# 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; } ## Clojure (defn gcd [a b] (if (zero? b) a (recur b, (mod a b)))) (defn lcm [a b] (/ (* a b) (gcd a b)));; to calculate the lcm for a variable number of arguments(defn lcmv [& v] (reduce lcm v))  ## COBOL  IDENTIFICATION DIVISION. PROGRAM-ID. show-lcm. ENVIRONMENT DIVISION. CONFIGURATION SECTION. REPOSITORY. FUNCTION lcm . PROCEDURE DIVISION. DISPLAY "lcm(35, 21) = " FUNCTION lcm(35, 21) GOBACK . END PROGRAM show-lcm. IDENTIFICATION DIVISION. FUNCTION-ID. lcm. ENVIRONMENT DIVISION. CONFIGURATION SECTION. REPOSITORY. FUNCTION gcd . DATA DIVISION. LINKAGE SECTION. 01 m PIC S9(8). 01 n PIC S9(8). 01 ret PIC S9(8). PROCEDURE DIVISION USING VALUE m, n RETURNING ret. COMPUTE ret = FUNCTION ABS(m * n) / FUNCTION gcd(m, n) GOBACK . END FUNCTION lcm. IDENTIFICATION DIVISION. FUNCTION-ID. gcd. DATA DIVISION. LOCAL-STORAGE SECTION. 01 temp PIC S9(8). 01 x PIC S9(8). 01 y PIC S9(8). LINKAGE SECTION. 01 m PIC S9(8). 01 n PIC S9(8). 01 ret PIC S9(8). PROCEDURE DIVISION USING VALUE m, n RETURNING ret. MOVE m to x MOVE n to y PERFORM UNTIL y = 0 MOVE x TO temp MOVE y TO x MOVE FUNCTION MOD(temp, y) TO Y END-PERFORM MOVE FUNCTION ABS(x) TO ret GOBACK . END FUNCTION gcd. ## Common Lisp Common Lisp provides the lcm function. It can accept two or more (or less) parameters. CL-USER> (lcm 12 18)36CL-USER> (lcm 12 18 22)396 Here is one way to reimplement it. 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-LCMCL-USER> (my-lcm 12 18)36CL-USER> (my-lcm 12 18 22)396 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 import std.stdio, std.bigint, std.math; T gcd(T)(T a, T b) pure nothrow { while (b) { immutable t = b; b = a % b; a = t; } return a;} T lcm(T)(T m, T n) pure nothrow { if (m == 0) return m; if (n == 0) return n; return abs((m * n) / gcd(m, n));} void main() { lcm(12, 18).writeln; lcm("2562047788015215500854906332309589561".BigInt, "6795454494268282920431565661684282819".BigInt).writeln;} Output: 36 15669251240038298262232125175172002594731206081193527869 ## DWScript PrintLn(Lcm(12, 18)); Output: 36 ## Erlang % Implemented by Arjun Sunel-module(lcm).-export([main/0]). main() -> lcm(-3,4). gcd(A, 0) -> A; gcd(A, B) -> gcd(B, A rem B). lcm(A,B) -> abs(A*B div gcd(A,B)). Output: 12  ## Euphoria function gcd(integer m, integer n) integer tmp while m do tmp = m m = remainder(n,m) n = tmp end while return nend function function lcm(integer m, integer n) return m / gcd(m, n) * nend function ## Excel Excel's LCM can handle multiple values. type in a cell:  =LCM(A1:J1)  This will get the LCM on the first 10 cells in the first row. Thus :  12 3 5 23 13 67 15 9 4 2 3605940  ## Ezhil  ## இந்த நிரல் இரு எண்களுக்கு இடையிலான மீச்சிறு பொது மடங்கு (LCM), மீப்பெரு பொது வகுத்தி (GCD) என்ன என்று கணக்கிடும் நிரல்பாகம் மீபொம(எண்1, எண்2) @(எண்1 == எண்2) ஆனால் ## இரு எண்களும் சமம் என்பதால், மீபொம அந்த எண்ணேதான் பின்கொடு எண்1 @(எண்1 > எண்2) இல்லைஆனால் சிறியது = எண்2 பெரியது = எண்1 இல்லை சிறியது = எண்1 பெரியது = எண்2 முடி மீதம் = பெரியது % சிறியது @(மீதம் == 0) ஆனால் ## பெரிய எண்ணில் சிறிய எண் மீதமின்றி வகுபடுவதால், பெரிய எண்தான் மீபொம பின்கொடு பெரியது இல்லை தொடக்கம் = பெரியது + 1 நிறைவு = சிறியது * பெரியது @(எண் = தொடக்கம், எண் <= நிறைவு, எண் = எண் + 1) ஆக ## ஒவ்வோர் எண்ணாக எடுத்துக்கொண்டு தரப்பட்ட இரு எண்களாலும் வகுத்துப் பார்க்கின்றோம். முதலாவதாக இரண்டாலும் மீதமின்றி வகுபடும் எண்தான் மீபொம மீதம்1 = எண் % சிறியது மீதம்2 = எண் % பெரியது @((மீதம்1 == 0) && (மீதம்2 == 0)) ஆனால் பின்கொடு எண் முடி முடி முடி முடி அ = int(உள்ளீடு("ஓர் எண்ணைத் தாருங்கள் "))ஆ = int(உள்ளீடு("இன்னோர் எண்ணைத் தாருங்கள் ")) பதிப்பி "நீங்கள் தந்த இரு எண்களின் மீபொம (மீச்சிறு பொது மடங்கு, LCM) = ", மீபொம(அ, ஆ)  ## F# let rec gcd x y = if y = 0 then abs x else gcd y (x % y) let lcm x y = x * y / (gcd x y) ## Factor The vocabulary math.functions already provides lcm. USING: math.functions prettyprint ;26 28 lcm . This program outputs 364. One can also reimplement lcm. 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 . ## 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 */ ; ## Frink Frink has a built-in LCM function that handles arbitrarily-large integers.  println[lcm[2562047788015215500854906332309589561, 6795454494268282920431565661684282819]]  ## FunL FunL has function lcm in module integers with the following definition: def lcm( _, 0 ) = 0 lcm( 0, _ ) = 0 lcm( x, y ) = abs( (x\gcd(x, y)) y ) ## GAP # Built-inLcmInt(12, 18);# 36 ## Go package main import ( "fmt" "math/big") var m, n, z big.Int func init() { m.SetString("2562047788015215500854906332309589561", 10) n.SetString("6795454494268282920431565661684282819", 10)} func main() { fmt.Println(z.Mul(z.Div(&m, z.GCD(nil, nil, &m, &n)), &n))} Output: 15669251240038298262232125175172002594731206081193527869  ## Groovy def gcdgcd = { m, n -> m = m.abs(); n = n.abs(); n == 0 ? m : m%n == 0 ? n : gcd(n, m % n) } def lcd = { m, n -> Math.abs(m * n) / gcd(m, n) } [[m: 12, n: 18, l: 36], [m: -6, n: 14, l: 42], [m: 35, n: 0, l: 0]].each { t -> println "LCD of$t.m, $t.n is$t.l"    assert lcd(t.m, t.n) == t.l}
Output:
LCD of 12, 18 is 36
LCD of -6, 14 is 42
LCD of 35, 0 is 0

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

lcm :: (Integral a) => a -> a -> alcm _ 0 =  0lcm 0 _ =  0lcm x y =  abs ((x quot (gcd x y)) * y)

## Icon and Unicon

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

link numbers procedure main()write("lcm of 18, 36 = ",lcm(18,36))write("lcm of 0, 9 36 = ",lcm(0,9))end

numbers provides lcm and gcd and looks like this:

procedure lcm(i, j)		#: least common multiple   if (i =  0) | (j = 0) then return 0	   return abs(i * j) / gcd(i, j)end

## J

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

      12 *. 1836   12 *. 18 2236 132   *./ 12 18 22396   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 10 00 1

## 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);   }}

## JavaScript

Computing the least common multiple of an integer array, using the associative law:

$\operatorname{lcm}(a,b,c)=\operatorname{lcm}(\operatorname{lcm}(a,b),c),$

$\operatorname{lcm}(a_1,a_2,\ldots,a_n) = \operatorname{lcm}(\operatorname{lcm}(a_1,a_2,\ldots,a_{n-1}),a_n).$

function LCM(A)  // A is an integer array (e.g. [-50,25,-45,-18,90,447]){       var n = A.length, a = Math.abs(A[0]);    for (var i = 1; i < n; i++)     { var b = Math.abs(A[i]), c = a;       while (a && b){ a > b ? a %= b : b %= a; }        a = Math.abs(c*A[i])/(a+b);     }    return a;} /* For example:   LCM([-50,25,-45,-18,90,447]) -> 67050*/

## jq

Direct method

# Define the helper function to take advantage of jq's tail-recursion optimizationdef lcm(m; n):  def _lcm:    # state is [m, n, i]    if (.[2] % .[1]) == 0 then .[2] else (.[0:2] + [.[2] + m]) | _lcm end;  [m, n, m] | _lcm;

## Julia

Built-in function:

lcm(m,n)

## 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+!20232792560

## LabVIEW

Requires GCD. This image is a VI Snippet, an executable image of LabVIEW code. The LabVIEW version is shown on the top-right hand corner. You can download it, then drag-and-drop it onto the LabVIEW block diagram from a file browser, and it will appear as runnable, editable code.

## Lasso

define gcd(a,b) => {	while(#b != 0) => {		local(t = #b)		#b = #a % #b		#a = #t	}	return #a}define lcm(m,n) => {	 #m == 0 || #n == 0 ? return 0	 local(r = (#m * #n) / decimal(gcd(#m, #n)))	 return integer(#r)->abs} lcm(-6, 14)lcm(2, 0)lcm(12, 18)lcm(12, 22)lcm(7, 31)
Output:
42
0
36
132
217

## Liberty BASIC

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

## Logo

to abs :n  output sqrt product :n :nend 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 :nend

Demo code:

print lcm 38 46

Output:

874

## Lua

function gcd( m, n )    while n ~= 0 do        local q = m        m = n        n = q % n    end    return mend function lcm( m, n )    return ( m ~= 0 and n ~= 0 ) and m * n / gcd( m, n ) or 0end print( lcm(12,18) )

## Maple

The least common multiple of two integers is computed by the built-in procedure ilcm in Maple. This should not be confused with lcm, which computes the least common multiple of polynomials.

> ilcm( 12, 18 );                                   36

## Mathematica

LCM[18,12]-> 36

## MATLAB / Octave

 lcm(a,b)

## Maxima

lcm(a, b);   /* a and b may be integers or polynomials */ /* In Maxima the gcd of two integers is always positive, and a * b = gcd(a, b) * lcm(a, b),so the lcm may be negative. To get a positive lcm, simply do */ abs(lcm(a, b))

## МК-61/52

ИПA	ИПB	*	|x|	ПC	ИПA	ИПB	/	[x]	П9ИПA	ИПB	ПA	ИП9	*	-	ПB	x=0	05	ИПCИПA	/	С/П

## ML

### mLite

fun gcd (a, 0) = a      | (0, b) = b      | (a, b) where (a < b)               = gcd (a, b rem a)      | (a, b) = gcd (b, a rem b) fun lcm (a, b) = let val d = gcd (a, b)                 in a * b div d                 end

## NetRexx

/* NetRexx */options replace format comments java crossref symbols nobinary numeric digits 3000 runSample(arg)return -- ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~method lcm(m_, n_) public static  L_ = m_ * n_ % gcd(m_, n_)  return L_ -- ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~-- Euclid's algorithm - iterative implementationmethod gcd(m_, n_) public static  loop while n_ > 0    c_ = m_ // n_    m_ = n_    n_ = c_    end  return m_ -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~method runSample(arg) private static  parse arg samples  if samples = '' | samples = '.' then    samples = '-6 14 =    42 |' -               '3  4 =    12 |' -              '18 12 =    36 |' -               '2  0 =     0 |' -               '0 85 =     0 |' -              '12 18 =    36 |' -               '5 12 =    60 |' -              '12 22 =   132 |' -               '7 31 =   217 |' -             '117 18 =   234 |' -              '38 46 =   874 |' -           '18 12 -5 =   180 |' -           '-5 18 12 =   180 |' - -- confirm that other permutations work           '12 -5 18 =   180 |' -        '18 12 -5 97 = 17460 |' -              '30 42 =   210 |' -              '30 42 =     . |' - -- 210; no verification requested              '18 12'             -- 36   loop while samples \= ''    parse samples sample '|' samples    loop while sample \= ''      parse sample mnvals '=' chk sample      if chk = '' then chk = '.'      mv = mnvals.word(1)      loop w_ = 2 to mnvals.words mnvals        nv = mnvals.word(w_)        mv = mv.abs        nv = nv.abs        mv = lcm(mv, nv)        end w_      lv = mv      select case chk        when '.' then state = ''        when lv  then state = '(verified)'        otherwise     state = '(failed)'        end      mnvals = mnvals.space(1, ',').changestr(',', ', ')      say 'lcm of' mnvals.right(15.max(mnvals.length)) 'is' lv.right(5.max(lv.length)) state      end    end   return
Output:
lcm of          -6, 14 is    42 (verified)
lcm of            3, 4 is    12 (verified)
lcm of          18, 12 is    36 (verified)
lcm of            2, 0 is     0 (verified)
lcm of           0, 85 is     0 (verified)
lcm of          12, 18 is    36 (verified)
lcm of           5, 12 is    60 (verified)
lcm of          12, 22 is   132 (verified)
lcm of           7, 31 is   217 (verified)
lcm of         117, 18 is   234 (verified)
lcm of          38, 46 is   874 (verified)
lcm of      18, 12, -5 is   180 (verified)
lcm of      -5, 18, 12 is   180 (verified)
lcm of      12, -5, 18 is   180 (verified)
lcm of  18, 12, -5, 97 is 17460 (verified)
lcm of          30, 42 is   210 (verified)
lcm of          30, 42 is   210
lcm of          18, 12 is    36


## Nim

proc gcd(u, v): auto =  var    t = 0    u = u    v = v  while v != 0:    t = u    u = v    v = t %% v  abs(u) proc lcm(a, b): auto = abs(a * b) div gcd(a, b) echo lcm(12, 18)echo lcm(-6, 14)

## Objeck

Translation of: C
 class LCM {  function : Main(args : String[]) ~ Nil {    IO.Console->Print("lcm(35, 21) = ")->PrintLine(lcm(21,35));  }   function : lcm(m : Int, n : Int) ~ Int {    return m / gcd(m, n) * n;  }   function : gcd(m : Int, n : Int) ~ Int {    tmp : Int;    while(m <> 0) { tmp := m; m := n % m; n := tmp; };    return n;  }}

## 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)

## Oforth

lcm is already defined into Integer class :

12 18 lcm

## ooRexx

 say lcm(18, 12) -- calculate the greatest common denominator of a numerator/denominator pair::routine gcd private  use arg x, y   loop while y \= 0      -- check if they divide evenly      temp = x // y      x = y      y = temp  end  return x -- calculate the least common multiple of a numerator/denominator pair::routine lcm private  use arg x, y  return x / gcd(x, y) * y

## Order

Translation of: bc
#include <order/interpreter.h> #define ORDER_PP_DEF_8gcd ORDER_PP_FN( \8fn(8U, 8V,                            \    8if(8isnt_0(8V), 8gcd(8V, 8remainder(8U, 8V)), 8U))) #define ORDER_PP_DEF_8lcm ORDER_PP_FN( \8fn(8X, 8Y,                            \    8if(8or(8is_0(8X), 8is_0(8Y)),     \        0,                             \        8quotient(8times(8X, 8Y), 8gcd(8X, 8Y)))))// No support for negative numbers ORDER_PP( 8to_lit(8lcm(12, 18)) )   // 36

## PARI/GP

Built-in function:

lcm

## Pascal

Program LeastCommonMultiple(output); function lcm(a, b: longint): longint;  begin    lcm := a;    while (lcm mod b) <> 0 do      inc(lcm, a);  end; begin  writeln('The least common multiple of 12 and 18 is: ', lcm(12, 18));end.

Output:

The least common multiple of 12 and 18 is: 36


## Perl

Using GCD:

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) * $b or 0} print lcm(1001, 221); Or by repeatedly increasing the smaller of the two until LCM is reached: sub lcm { use integer; my ($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);

## Perl 6

This function is provided as an infix so that it can be used productively with various metaoperators.

say 3 lcm 4;            # infixsay [lcm] 1..20;        # reductionsay ~(1..10 Xlcm 1..10) # cross

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
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;} ## PicoLisp Using 'gcd' from Greatest common divisor#PicoLisp: (de lcm (A B) (abs (*/ A B (gcd A B))) ) ## PL/I  /* Calculate the Least Common Multiple of two integers. */ LCM: procedure options (main); /* 16 October 2013 */ declare (m, n) fixed binary (31); get (m, n); put edit ('The LCM of ', m, ' and ', n, ' is', LCM(m, n)) (a, x(1)); LCM: procedure (m, n) returns (fixed binary (31)); declare (m, n) fixed binary (31) nonassignable; if m = 0 | n = 0 then return (0); return (abs(m*n) / GCD(m, n));end LCM; GCD: procedure (a, b) returns (fixed binary (31)) recursive; declare (a, b) fixed binary (31); if b = 0 then return (a); return (GCD (b, mod(a, b)) ); end GCD;end LCM;  The LCM of 14 and 35 is 70  ## Prolog SWI-Prolog knows gcd. lcm(X, Y, Z) :- Z is abs(X * Y) / gcd(X,Y). Example:  ?- lcm(18,12, Z). Z = 36.  ## PureBasic Procedure GCDiv(a, b); Euclidean algorithm Protected r While b r = b b = a%b a = r Wend ProcedureReturn aEndProcedure Procedure LCM(m,n) Protected t If m And n t=m*n/GCDiv(m,n) EndIf ProcedureReturn t*Sign(t)EndProcedure ## Python ### gcd Using the fractions libraries gcd function: >>> 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>>>  ### Prime decomposition This imports Prime decomposition#Python from prime_decomposition import decomposetry: reduceexcept NameError: from functools import reduce def lcm(a, b): mul = int.__mul__ 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(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 ### Iteration over multiples >>> 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>>>  ### Repeated modulo Translation of: Tcl >>> 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>>>  ## Qi  (define gcd A 0 -> A A B -> (gcd B (MOD A B))) (define lcm A B -> (/ (* A B) (gcd A B)))  ## R  "%gcd%" <- function(u, v) {ifelse(u %% v != 0, v %gcd% (u%%v), v)} "%lcm%" <- function(u, v) { abs(u*v)/(u %gcd% v)} print (50 %lcm% 75)  ## Racket Racket already has defined both lcm and gcd funtions: #lang racket(lcm 3 4 5 6) ;returns 60(lcm 8 108) ;returns 216(gcd 8 108) ;returns 4(gcd 108 216 432) ;returns 108 ## Retro This is from the math extensions library included with Retro. : gcd ( ab-n ) [ tuck mod dup ] while drop ;: lcm ( ab-n ) 2over gcd [ * ] dip / ; ## REXX ### version 1 The lcm subroutine can handle any number of integers and/or arguments. The integers (negative/zero/positive) can be (as per the numeric digits) up to ten thousand digits. Usage note: note that the integers can be expressed as a list and/or specified as individual arguments. /*REXX pgm finds LCM (Least Common Multiple) of any number of integers.*/numeric digits 10000 /*can handle 10,000 digit numbers*/say 'the LCM of 19 & 0 is: ' lcm(19 0)say 'the LCM of 0 & 85 is: ' lcm( 0 85)say 'the LCM of 14 & -6 is: ' lcm(14, -6)say 'the LCM of 18 & 12 is: ' lcm(18 12)say 'the LCM of 18 & 12 & -5 is: ' lcm(18 12, -5)say 'the LCM of 18 & 12 & -5 & 97 is: ' lcm(18, 12, -5, 97)say 'the LCM of 2**19-1 & 2**521-1 is: ' lcm(2**19-1 2**521-1) /* [↑] 7th, 13th Mersenne primes*/exit /*stick a fork in it, we're done.*//*──────────────────────────────────LCM subroutine──────────────────────*/lcm: procedure; parse arg$,_; $=$ _;    do i=3  to arg(); $=$ arg(i); endparse var $x$                        /*obtain the first value in args.*/x=abs(x)                               /*use the absolute value of  X.  */          do  while $\=='' /*process the remainder of args. */ parse var$ ! $; !=abs(!) /*pick off the next arg (ABS val)*/ if !==0 then return 0 /*if zero, then LCM is also zero.*/ d=x*! /*calculate part of the LCM here */ do until !==0; parse value x//! ! with ! x end /*until*/ /* [↑] this is a short&fast GCD.*/ x=d%x /*divide the pre─calculated value*/ end /*while*/ /* [↑] process subsequent args. */return x /*return with the LCM of the args*/ output the LCM of 19 & 0 is: 0 the LCM of 0 & 85 is: 0 the LCM of 14 & -6 is: 42 the LCM of 18 & 12 is: 36 the LCM of 18 & 12 & -5 is: 180 the LCM of 18 & 12 & -5 & 97 is: 17460 the LCM of 2**19-1 & 2**521-1 is: 3599124170836896975638715824247986405702540425206233163175195063626010878994006898599180426323472024265381751210505324617708575722407440034562999570663839968526337  ### version 2 Translation of: REXX version 0 using different argument handling- Use as lcm(a,b,c,---) lcm2: procedurex=abs(arg(1))do k=2 to arg() While x<>0 y=abs(arg(k)) x=x*y/gcd2(x,y) endreturn x gcd2: procedurex=abs(arg(1))do j=2 to arg() y=abs(arg(j)) If y<>0 Then Do do until z==0 z=x//y x=y y=z end end endreturn x ### versions 1 and 2 compared The (performance) improvement of version 2 is due to the different argument handling at the cost of less freedom of argument specification (you must use lcm2(a,b,c) instead of the possible lcm1(a b,c). Consider, however, lcm1(3 -4, 5) which, of course, won't work as possibly intended. The performance improvement is more significant with ooRexx than with Regina. Note:$ in version 1 was replaced by d in order to adapt it for this test with ooRexx.

Parse Version vSay 'Version='vCall time 'R'Do a=0 To 100  Do b=0 To 100    Do c=0 To 100      x1.a.b.c=lcm1(a,b,c)      End    End  EndSay 'version 1' time('E')Call time 'R'Do a=0 To 100  Do b=0 To 100    Do c=0 To 100      x2.a.b.c=lcm2(a,b,c)      End    End  EndSay 'version 2' time('E')cnt.=0Do a=0 To 100  Do b=0 To 100    Do c=0 To 100      If x1.a.b.c=x2.a.b.c then cnt.0ok=cnt.0ok+1      End    End  EndSay cnt.0ok 'comparisons ok'Exit/*----------------------------------LCM subroutine----------------------*/lcm1: procedure; d=strip(arg(1) arg(2));  do i=3  to arg(); d=d arg(i); endparse var d x d                        /*obtain the first value in args.*/x=abs(x)                               /*use the absolute value of  X.  */          do  while d\==''             /*process the rest of the args.  */          parse var d ! d;   !=abs(!)  /*pick off the next arg (ABS val)*/          if !==0  then return 0       /*if zero, then LCM is also zero.*/          x=x*!/gcd1(x,!)               /*have  GCD do the heavy lifting.*/          end   /*while*/return x                               /*return with  LCM  of arguments.*//*----------------------------------GCD subroutine----------------------*/gcd1: procedure; d=strip(arg(1) arg(2));  do j=3  to arg(); d=d arg(j); endparse var d x d                        /*obtain the first value in args.*/x=abs(x)                               /*use the absolute value of  X.  */          do  while d\==''             /*process the rest of the args.  */          parse var d y d;   y=abs(y)  /*pick off the next arg (ABS val)*/          if y==0  then iterate        /*if zero, then ignore the value.*/               do  until y==0;  parse  value   x//y  y   with  y  x;   end          end   /*while*/return x                               /*return with  GCD  of arguments.*/ lcm2: procedurex=abs(arg(1))do k=2 to arg() While x<>0  y=abs(arg(k))  x=x*y/gcd2(x,y)  endreturn x gcd2: procedurex=abs(arg(1))do j=2 to arg()  y=abs(arg(j))  If y<>0 Then Do    do until z==0      z=x//y      x=y      y=z      end    end  endreturn x
Output:

Output of rexx lcmt and regina lcmt cut and pasted together:

Version=REXX-ooRexx_4.2.0(MT)_32-bit 6.04 22 Feb 2014
Version=REXX-Regina_3.9.0(MT) 5.00 16 Oct 2014
version 1 29.652000       version 1 23.821000
version 2 10.857000       version 2 21.209000
1030301 comparisons ok    1030301 comparisons ok

## Ruby

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

irb(main):001:0> require 'rational'=> trueirb(main):002:0> 12.lcm 18=> 36

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.

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  endend
irb(main):004:0> lcm 12, 18=> 36irb(main):005:0> lcm 12, 18, 22=> 396

## Run BASIC

print lcm(22,44) function lcm(m,n) while n   t = m   m = n   n = t mod n wendlcm = mend function

## Scala

def gcd(a: Int, b: Int):Int=if (b==0) a.abs else gcd(b, a%b)def lcm(a: Int, b: Int)=(a*b).abs/gcd(a,b)
lcm(12, 18)   // 36lcm( 2,  0)   // 0lcm(-6, 14)   // 42

## Scheme

> (lcm 108 8)216

$include "seed7_05.s7i"; const func integer: gcd (in var integer: a, in var integer: b) is func result var integer: gcd is 0; local var integer: help is 0; begin while a <> 0 do help := b rem a; b := a; a := help; end while; gcd := b; end func; const func integer: lcm (in integer: a, in integer: b) is return a div gcd(a, b) * b; const proc: main is func begin writeln("lcm(35, 21) = " <& lcm(21, 35)); end func; Original source: [1] ## Smalltalk Smalltalk has a built-in lcm method on SmallInteger: 12 lcm: 18 ## Sparkling function factors(n) { var f = {}; for var i = 2; n > 1; i++ { while n % i == 0 { n /= i; f[i] = f[i] != nil ? f[i] + 1 : 1; } } return f;} function GCD(n, k) { let f1 = factors(n); let f2 = factors(k); let fs = map(f1, function(factor, multiplicity) { let m = f2[factor]; return m == nil ? 0 : min(m, multiplicity); }); let rfs = {}; foreach(fs, function(k, v) { rfs[sizeof rfs] = pow(k, v); }); return reduce(rfs, 1, function(x, y) { return x * y; });} function LCM(n, k) { return n * k / GCD(n, k);} ## Swift Using the Swift GCD function. func lcm(a:Int, b:Int) -> Int { return abs(a * b) / gcd_rec(a, b)} ## Tcl proc lcm {p q} { set m [expr {$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}]}    }}

Demonstration

puts [lcm 12 18]

Output:

36


## TI-83 BASIC

lcm(12, 18)               36

## TSE SAL

// library: math: get: least: common: multiple <description></description> <version control></version control> <version>1.0.0.0.2</version> <version control></version control> (filenamemacro=getmacmu.s) [<Program>] [<Research>] [kn, ri, su, 20-01-2013 14:36:11]INTEGER PROC FNMathGetLeastCommonMultipleI( INTEGER x1I, INTEGER x2I ) // RETURN( x1I * x2I / FNMathGetGreatestCommonDivisorI( x1I, x2I ) ) //END // library: math: get: greatest: common: divisor <description>greatest common divisor whole numbers. Euclid's algorithm. Recursive version</description> <version control></version control> <version>1.0.0.0.3</version> <version control></version control> (filenamemacro=getmacdi.s) [<Program>] [<Research>] [kn, ri, su, 20-01-2013 14:22:41]INTEGER PROC FNMathGetGreatestCommonDivisorI( INTEGER x1I, INTEGER x2I ) // IF ( x2I == 0 )  //  RETURN( x1I )  // ENDIF // RETURN( FNMathGetGreatestCommonDivisorI( x2I, x1I MOD x2I ) ) //END PROC Main() // STRING s1[255] = "10" STRING s2[255] = "20" REPEAT  IF ( NOT ( Ask( "math: get: least: common: multiple: x1I = ", s1, _EDIT_HISTORY_ ) ) AND ( Length( s1 ) > 0 ) ) RETURN() ENDIF  IF ( NOT ( Ask( "math: get: least: common: multiple: x2I = ", s2, _EDIT_HISTORY_ ) ) AND ( Length( s2 ) > 0 ) ) RETURN() ENDIF  Warn( FNMathGetLeastCommonMultipleI( Val( s1 ), Val( s2 ) ) ) // gives e.g. 10 UNTIL FALSEEND

## UNIX Shell

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

Works with: Bourne Shell

## 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));}

## Wortel

Operator

@lcm a b

Number expression

!#~km a b

Function (using gcd)

&[a b] *b /a @gcd a b

## XPL0

include c:\cxpl\codes; func GCD(M,N);  \Return the greatest common divisor of M and Nint  M, N;int  T;[while N do     \Euclid's method    [T:= M;  M:= N;  N:= rem(T/N)];return M;]; func LCM(M,N);  \Return least common multipleint  M, N;return abs(M*N) / GCD(M,N); \Display the LCM of two integers entered on command lineIntOut(0, LCM(IntIn(8), IntIn(8)))

## zkl

fcn gcd(a,b){while(b){t:=a; a=b; b=t%b} a.abs()}fcn lcm(m,n){(m*n).abs()/gcd(m,n)}
Output:
zkl: lcm(12,18)
36
zkl: lcm(-6,14)
42
zkl: lcm(35,0)
0