31
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The least common multiple of a set of positive integers A is the smallest postive integer B such that, for each k in A, there exists a positive integer n such that k*n = B.

Given at least two positive integers as input, output their least common multiple.

Rules

  • Builtins are allowed, but if your solution uses one, you are encouraged to include an alternate solution that does not use GCD/LCM builtins. However, the alternate solution will not count towards your score at all, so it is entirely optional.
  • All inputs and outputs will be within the natively-representable range for your language. If your language is natively capable of arbitrarily-large integers, then your solution must work with arbitrarily large inputs and outputs.

Test cases

[7, 2] -> 14
[8, 1] -> 8
[6, 4, 8] -> 24
[8, 2, 1, 10] -> 40
[9, 6, 2, 1, 5] -> 90
[5, 5, 7, 1, 1] -> 35
[4, 13, 8, 8, 11, 1] -> 1144
[7, 2, 2, 11, 11, 8, 5] -> 3080
[1, 6, 10, 3, 4, 10, 7] -> 420
[5, 2, 9, 10, 3, 4, 4, 4, 7] -> 1260
[9, 7, 10, 9, 7, 8, 5, 10, 1] -> 2520
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  • 6
    \$\begingroup\$ Because it's a reasonably frequent misconception: the formula LCM(a,b) = ab/GCD(a,b) does not extend to more than two numbers (or, for that matter, to one number!). \$\endgroup\$ – Greg Martin Oct 1 '16 at 2:56

42 Answers 42

4
\$\begingroup\$

Actually, 12 1 byte

Golfing suggestions are still welcome, though I'm not sure how to improve on the raw LCM built-in. Try it online!

A 12-byte version without the built-in. Golfing suggestions welcome. Try it online!

╗2`╜@♀%ΣY`╓N

Ungolfing

          Implicit input array.
╗         Save array in register 0.
2`...`╓   Starting with f(0), find the first (two) x where f(x) returns a truthy value.
          These two values will be 0 and our LCM.
  ╜         Push array from register 0.
  @         Swap the top two values. Stack: x, array
  ♀%        Map % over x and array, returning (x % item) for each item in array.
  ΣY        If the sum of all the modulos equals 0, x is either 0 or our LCM.

N         Push the last (second) value of our results. This is our LCM.
          Implicit return.
\$\endgroup\$
  • \$\begingroup\$ You do realize you're allowed to use the builtin, right? \$\endgroup\$ – Mego Oct 1 '16 at 5:39
  • 1
    \$\begingroup\$ @Mego I'll add it in, but my understanding was that builtins were discouraged, so I didn't use it at first. \$\endgroup\$ – Sherlock9 Oct 1 '16 at 6:16
  • 1
    \$\begingroup\$ Builtins are allowed. They're not discouraged at all - I simply wanted to encourage non-builtin solutions to also be included because they are often much more interesting than the builtin. \$\endgroup\$ – Mego Oct 1 '16 at 6:30
  • 1
    \$\begingroup\$ I read that as actually, 1 byte. \$\endgroup\$ – programmer5000 Apr 17 '17 at 18:21
  • 2
    \$\begingroup\$ @programmer5000 I think that may be why the language is called actually... \$\endgroup\$ – Socratic Phoenix Sep 25 '17 at 20:22
17
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JavaScript (ES6), 36 bytes

f=(a,i=1)=>a.some(v=>i%v)?f(a,i+1):i

Starting from 1 it's the first number that can be divided by all.

f=(a,i=1)=>a.some(v=>i%v)?f(a,i+1):i
;

console.log(f([7, 2]));
console.log(f([8, 1]));
console.log(f([6, 4, 8]));
console.log(f([8, 2, 1, 10]));
console.log(f([9, 6, 2, 1, 5]));
console.log(f([5, 5, 7, 1, 1]));
console.log(f([4, 13, 8, 8, 11, 1]));
console.log(f([7, 2, 2, 11, 11, 8, 5]));
console.log(f([1, 6, 10, 3, 4, 10, 7]));
console.log(f([5, 2, 9, 10, 3, 4, 4, 4, 7]));
console.log(f([9, 7, 10, 9, 7, 8, 5, 10, 1]));

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  • \$\begingroup\$ Of course... I thought about doing a loop with this technique, but recursion is way shorter. \$\endgroup\$ – ETHproductions Sep 30 '16 at 19:26
  • 1
    \$\begingroup\$ This is genius... If I recall, some returns true if at least one element in the array satisfies the condition, right? \$\endgroup\$ – WallyWest Oct 4 '16 at 2:49
12
\$\begingroup\$

05AB1E/2sable, 2 bytes

.¿

Try it online! in 05AB1E
or 2sable

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11
\$\begingroup\$

Jelly, 3 bytes

æl/

Reduces by LCM. Try it online! or verify all test cases.

Alternate version, 6 bytes

ÆE»/ÆẸ

Try it online! or verify all test cases.

How it works

ÆE»/ÆẸ  Main link. Argument: A (array)

ÆE      Yield all prime exponents of each integer in A.
  »/    Reduce columns (exponents that correspond to the same prime) by maximum.
    ÆẸ  Turn the resulting array of prime exponents into the corresponding integer.
\$\endgroup\$
8
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Python, 69 65 52 50 bytes

A=lambda l,i=1:any(i%a for a in l)and A(l,i+1)or i

2 bytes saved thanks to Dennis!

Pretty straightforward recursive solution, you will need to make the recursion limit a bit higher for some of the test cases to work.

\$\endgroup\$
  • 1
    \$\begingroup\$ any takes a generator; you don't need the brackets. \$\endgroup\$ – Dennis Sep 30 '16 at 19:10
  • 3
    \$\begingroup\$ A=lambda l,i=1:all(i%a<1for a in l)or-~A(l,i+1) saves a few more bytes. \$\endgroup\$ – Dennis Sep 30 '16 at 19:58
8
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MATL, 7 bytes

&YFX>^p

No builtin.

Try it online!

Explanation

Let's take input [8, 2, 1, 10] as an example.

&YF    % Take array implicitly. Push vector of prime factors and matrix of exponents 
       % of factorization, where each row represents one of the input numbers
       %   STACK: [2 3 5], [3 0 0; 1 0 0; 0 0 0; 1 0 1]
X>     % Maximum of each column
       %   STACK: [2 3 5], [3 0 1]
^      % Element-wise power
       %   STACK: [8 1 5]
p      % Product of array
       %   STACK: 40
       % Implicitly display

EDIT (June 9, 2017): YF with two outputs has been modified in release 20.1.0: non-factor primes and their (zero) exponents are skipped. This doesn't affect the above code, which works without requiring any changes.

\$\endgroup\$
6
\$\begingroup\$

Julia (3 Bytes) [Working on Non-Built-in]

lcm     # Using LCM built-in (3 Bytes)

As Dennis pointed out, I keep forgetting that Julia automatically vectorizes inputs.

Example:

println(lcm(1,2,3,4,5,6,7,8,9)) #Prints 2520
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6
\$\begingroup\$

PowerShell v2+, 73 60 bytes

param($a)for($i=1;($a|?{!($i%$_)}).count-ne$a.count){$i++}$i

Takes input $a, loops upward from $i=1 with $i++, based on a conditional. The condition is ($a|?{!($i%$_)}).count being -notequal to $a.count. Meaning, the loop ends when the elements of $a that are divisors of $i is equal to the elements of $a. Then, a solitary $i is left on the pipeline, and output is implicit.

Test Cases

PS C:\Tools\Scripts\golfing> @(7,2),@(8,1),@(6,4,8),@(8,2,1,10),@(9,6,2,1,5),@(5,5,7,1,1),@(4,13,8,8,11,1)|%{($_-join',')+" -> "+(.\least-common-multiple.ps1 $_)}
7,2 -> 14
8,1 -> 8
6,4,8 -> 24
8,2,1,10 -> 40
9,6,2,1,5 -> 90
5,5,7,1,1 -> 35
4,13,8,8,11,1 -> 1144

PS C:\Tools\Scripts\golfing> @(7,2,2,11,11,8,5),@(1,6,10,3,4,10,7),@(5,2,9,10,3,4,4,4,7),@(9,7,10,9,7,8,5,10,1)|%{($_-join',')+" -> "+(.\least-common-multiple.ps1 $_)}
7,2,2,11,11,8,5 -> 3080
1,6,10,3,4,10,7 -> 420
5,2,9,10,3,4,4,4,7 -> 1260
9,7,10,9,7,8,5,10,1 -> 2520
\$\endgroup\$
4
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Mathematica, 3 bytes

LCM

Usage:

In[1]:= LCM[9, 7, 10, 9, 7, 8, 5, 10, 1]                                        

Out[1]= 2520
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  • 6
    \$\begingroup\$ The day that Mathematica matched Jelly is a day I never thought I'd see. \$\endgroup\$ – Steven H. Oct 1 '16 at 3:28
3
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Cheddar, 33 bytes

(n,i=1)f->n.any(i&(%))?f(n,i+1):i

Nothing super new.

Ungolfed

(n, i = 1) f ->
  n.any(j -> i % j) ?
    f(n, i + 1) :
    i

Basically this starts at one and keeps increasing until it finds an LCM

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3
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JavaScript (ES6), 63 59 bytes

f=([x,...a])=>a[0]?x*f(a)/(g=(m,n)=>n?g(n,m%n):m)(x,f(a)):x

Recursively finds the LCM of the last two elements.

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  • \$\begingroup\$ This is what my solution would have been: a=>a.reduce((l,n)=>l*n/(g=(m,n)=>n?g(n,m%n):m)(l,n)) \$\endgroup\$ – Neil Sep 30 '16 at 19:15
  • \$\begingroup\$ @Neil You can post that if you'd like. I doubt my technique can get that short... \$\endgroup\$ – ETHproductions Sep 30 '16 at 19:32
3
\$\begingroup\$

Dyalog APL, 2 bytes

∧/

Reduces by LCM. Test it on TryAPL.

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  • 4
    \$\begingroup\$ Congrats on 100k! \$\endgroup\$ – Copper Sep 30 '16 at 20:45
3
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JavaScript (ES6), 52 bytes

a=>a.reduce((l,n)=>l*n/(g=(m,n)=>n?g(n,m%n):m)(l,n))

I reduced this answer as much as I could but I'm obviously not going to get anywhere near the simplicity of @Hedi's answer.

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3
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Java 8, 75 59 121 89 bytes

Uses the Euclidean Algorithm and the fact that LCM(A, B)= A * B / GCD(A, B)

  • 16 bytes off. Thanks to @carusocomputing
  • Added Multi-Input +62 bytes
  • 32 bytes off. Thanks to @Olivier Grégoire

Code:

public static int lcm(int l, int c){
  for(int i=1;i<=l&&i<=c;++i) 
    if (i%l==0&&i%c==0)
      return l*c/i;
}
public static int lcm(int...x){
  int y=x[0];
  for(int j:x){
    y=lcm(j,y);
  }
  return y;
}

Remove line-breaks:

int g(int a,int b){return b<1?a:g(b,a%b);}

l->{int l=1;for(int n:a)l=l*n/g(l,n);return l;}
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  • \$\begingroup\$ Technically a snippet, but if you add n->{...} I believe it becomes valid Java 8. \$\endgroup\$ – Magic Octopus Urn Sep 30 '16 at 20:27
  • \$\begingroup\$ Thanks. I'm trying to get used to see lambda in Java. With lambda you can probably golf some of the for-loop. But I don't know how. \$\endgroup\$ – Roman Gräf Sep 30 '16 at 20:29
  • \$\begingroup\$ Yeah, all that stuff is an afterthought in Java; you'd likely be better off learning it in Python :). \$\endgroup\$ – Magic Octopus Urn Sep 30 '16 at 20:32
  • \$\begingroup\$ Unless I'm missing something, this doesn't support more than two inputs \$\endgroup\$ – pinkfloydx33 Oct 1 '16 at 16:51
  • \$\begingroup\$ If you compute the GCD, you can golf much more: int g(int a,int b){return b<1?a:g(b,a%b);}. LCM can then become int l(int[]a){int l=1;for(int n:a)l=l*n/g(l,n);return l;}, for a total of 99 bytes. \$\endgroup\$ – Olivier Grégoire Oct 3 '16 at 9:47
2
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MATL, 3 bytes

&Zm

This uses the builtin function with array input.

Try it online!

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2
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Brachylog, 17 bytes

,.#>=g:?z:%a#=h0,

Try it online!

Explanation

,.#>=               Output is a strictly positive integer
     g:?z           Zip the Output with the Input
         :%a        Compute Output mod I for each I in the Input
            #=h0,   All results must be equal to 0
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2
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Perl 6, 10 bytes

{[lcm] @_}

basically the same as:

sub ( *@_ ) { @_.reduce: &infix:< lcm > }
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2
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J, 11 bytes

>./&.(_&q:)

There is a solution for 3 bytes using the LCM builtin.

*./

Explanation

>./&.(_&q:)  Input: array of integers A
      _&q:   Get the prime exponents of each integer in A
>./&         Reduce by maximum on the lists
   &. _&q:   Convert the list of exponents back to an integer

*./  Input: array of integers A
  /  Reduce using
*.     LCM
\$\endgroup\$
2
\$\begingroup\$

CJam, 18 17 16 bytes

1 byte saved thanks to Martin Ender.

Incrementing until the LCM is found.

q~0{)_2$f%:+}g\;

Try it online

\$\endgroup\$
  • 1
    \$\begingroup\$ I'm not entirely familiar with CJam, but the reusability rule is for functions, not full programs. If your 17-byte solution is a full program that consistently works across runs, it's fine. \$\endgroup\$ – Mego Oct 2 '16 at 17:31
2
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Racket 13 bytes

lcm is a built-in function in Racket:

(apply lcm l)

Testing:

(define (f l)
   (apply lcm l))

(f (list 7 2)) 
(f (list 8 1)) 
(f (list 6 4 8)) 
(f (list 8 2 1 10)) 
(f (list 9 6 2 1 5))
(f (list 5 5 7 1 1)) 
(f (list 4 13 8 8 11 1))
(f (list 7 2 2 11 11 8 5))
(f (list 1 6 10 3 4 10 7))
(f (list 5 2 9 10 3 4 4 4 7)) 
(f (list 9 7 10 9 7 8 5 10 1))

Output:

14
8
24
40
90
35
1144
3080
420
1260
2520
\$\endgroup\$
  • \$\begingroup\$ Ahh. How can you use that syntax. I always gave up when I tried to learn Racket. \$\endgroup\$ – Roman Gräf Oct 1 '16 at 17:22
  • 1
    \$\begingroup\$ First word in brackets is a procedure name, rest are its arguments. If an argument is a procedure, it has to be in its own brackets. Values (non-procedures) are written without brackets. I find it to be an excellent general-purpose language with additional advantage of stress on functional programming. Being derived from Lisp, one also gets a sense of covering that area of programming. \$\endgroup\$ – rnso Oct 1 '16 at 18:19
  • \$\begingroup\$ I find coding keywords and language to be easier in Racket & Scheme than Lisp. \$\endgroup\$ – rnso Oct 1 '16 at 18:26
  • \$\begingroup\$ Yes, but did I said I understand Lisp? I more like languages like Jelly or Java. \$\endgroup\$ – Roman Gräf Oct 1 '16 at 18:27
  • 1
    \$\begingroup\$ Main syntax difference between Java and Racket is f(a, b) vs (f a b), x+y+z vs (+ x y z), x == y vs (eq? x y) and x=2 vs (define x 2), or if already defined, (set! x 2). Also no need to declare types like public static void or int char string etc. Hope that gets you interested in Racket again. \$\endgroup\$ – rnso Oct 2 '16 at 3:53
2
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R, 36 bytes (not builtin)

v=scan();i=1;while(any(i%%v))i=i+1;i

Takes the input. Then tests each positive integer by taking the mod.

\$\endgroup\$
  • \$\begingroup\$ I believe you need a cat around your last i \$\endgroup\$ – Giuseppe Oct 13 '17 at 20:10
  • \$\begingroup\$ @Giuseppe when I run it, the value prints fine. \$\endgroup\$ – user5957401 Oct 13 '17 at 20:19
  • \$\begingroup\$ see the discussion here, but I suppose ec=T is fine for +4 rather than +5 for cat(). \$\endgroup\$ – Giuseppe Oct 13 '17 at 20:22
  • 1
    \$\begingroup\$ regardless, this can be golfed down some v=scan();while(any((F=F+1)%%v)){};F with cat() or ec=T making it 40 or 39 bytes, respectively. And +1, very nice approach. \$\endgroup\$ – Giuseppe Oct 13 '17 at 20:23
1
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Pyth, 9 bytes

.U/*bZibZ

A program that takes input of a list on STDIN and prints the result.

Try it online or verify all test cases

How it works

.U/*bZibZ  Program. Input: Q
.U         Reduce Q by (implicit input fill):
   *bZ      Product of current and next value
  /   ibZ   divided by GCD of current and next value
           Implicitly print
\$\endgroup\$
1
\$\begingroup\$

Haskell, 10 bytes

foldr1 lcm

Usage example: foldl1 lcm [5,2,9,10,3,4,4,4,7] -> 1260.

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1
\$\begingroup\$

C#, 50+18 = 68 bytes

50 bytes for method defintion, +18 bytes for LINQ import.

using System.Linq;int L(int[]n,int i=1)=>n.All(x=>1>i%x)?i:L(n,i+1);

Pretty much the same as a lot of other answers. Counts up recursively until it finds the LCM. I was a bit surprised this didn't get a StackOverflowException, so I also have a non-recursive version which is actually just 1 byte longer.

using System.Linq;n=>{for(int i=1;;i++)if(n.All(x=>1>i%x))return i;};

Ungolfed:

using System.Linq;            // Import LINQ
int L(int[] n, int i = 1) =>  // Function declaration
    n.All(x => 1 > i % x)     // Check if each x in n divides i
        ? i                   // And if so return i
        : L(n, i + 1)         // Otherwise increment i and recurse
;
\$\endgroup\$
1
\$\begingroup\$

Pip, 10 bytes

W$+o%g++oo

Uses the "try every number until one works" strategy. Try it online!

            o is preinitialized to 1, g is list of cmdline args
   o%g      Mod o by each arg
 $+         Sum (truthy if any nonzero, falsy if all zero)
W           Loop while that expression is truthy:
      ++o     Increment o
         o  Autoprint o
\$\endgroup\$
1
\$\begingroup\$

PHP, 42 74 bytes

for(;($p=++$f*$argv[1])%$argv[2];);echo$p;

straight forward:
loop $f from 1 upwards; if $f*$a divides through $b without a remainder, the LCM is found.


I totally had overread the at least ... here´s the code for any number of parameters:

for(;$i<$argc;)for($p=$argv[$i=1]*++$f;++$i<$argc&$p%$argv[$i]<1;);echo$p;

Loop $f from 1 upwards while inner loop has not run to $argc.
Loop $i from 2 to $argc-1 while $f*$argv[1] divides through $argv[$i] without a remainder.
both loops broken: print $f*$argument 1.

\$\endgroup\$
1
\$\begingroup\$

Prolog (SWI), 46 bytes

l([],1).
l([X|Y],P):-l(Y,Q),P is X*Q/gcd(X,Q).

Try it online!

Another solution, 59 bytes:

l(A,X):-between(1,inf,X),forall(member(Y,A),X mod Y=:=0),!.
\$\endgroup\$
1
\$\begingroup\$

Python 3, 83 bytes

import math,functools as i
t=lambda t:i.reduce(lambda a,b:int(a*b/math.gcd(a,b)),t)
\$\endgroup\$
  • \$\begingroup\$ Welcome to PPCG! \$\endgroup\$ – Laikoni Jul 10 '18 at 17:36
  • \$\begingroup\$ You may want to include a link to an online testing site like Try it online! so it's easier for others to verify your answer. \$\endgroup\$ – Laikoni Jul 10 '18 at 17:40
1
\$\begingroup\$

Brachylog v2, 8 bytes

{×↙Xℕ₁}ᵛ

Try it online!

It's funny just how directly this maps on to the definition given in the challenge.

{     }ᵛ    Each element of
            the input
 ×          multiplied by
  ↙X        some arbitrary and inconsistent integer
    ℕ₁      is a natural number,
       ᵛ    which is the same for each element,
            and is the output.

A suspiciously slow but significantly shorter solution:

Brachylog v2, 5 bytes

f⊇p~d

Try it online!

Takes input through the output variable and gives output through the input variable. Rips right through the first four test cases but I'm still waiting on the fifth... Ordinarily, I'd still make it my primary solution and just trust that it works correctly, but I don't know why it hasn't confirmed that 90 is the LCM of 9, 6, 2, 1, 5 when I gave it 90 twenty minutes ago.

(Edit: It confirmed the answer after no more than 16 hours, and generated it alongside the LCM of 5, 5, 7, 1, 1 after about two days.)

         The output variable
   ~d    with duplicates removed
  p      is a permutation of
 ⊇       a sublist of
f        the factors of
         the input variable.

And another completely different predicate that accidentally more-or-less translates Fatalize's Brachylog v1 solution:

Brachylog v2, 10 bytes

;.gᵗ↔z%ᵛ0<

Try it online!

This was salvaged from a solution I'd made for this challenge before I realized the output wasn't restricted to being an integer.

 .            The output
; gᵗ↔z        paired with each element of
              the input,
      %ᵛ      when the first element of each pair is taken mod the second, is always
        0     zero.
              Furthermore, the output
         <    is strictly greater than
        0     zero.
\$\endgroup\$
0
\$\begingroup\$

Pyth - 7 6 bytes

No builtin.

*F{sPM

Try it online here.

\$\endgroup\$
  • 4
    \$\begingroup\$ Fails on [4], or anything else with a repeated prime factor. \$\endgroup\$ – isaacg Sep 30 '16 at 20:08

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