Project Euler Problem 5

Problem 5

30 November 2001

 

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is

evenly divisible

by all of the numbers from 1 to 20?

 

Answer: 232792560

 

 

Solution in C#

 

 

using System;

namespace ConsoleApplication1 {
    class Program {
        static void Main(string[] args) {
            var n = 20;
            for (var i = 19; i >= 2; i--) {
                if (n % i != 0) { n += 20; i = 20; }
            }
            Console.WriteLine(n);
            Console.ReadKey();
        }
    }
}

Solution in Perl

sub problem5 {
 my $n = 20;
 for ( my $i = 19 ; $i >= 2 ; $i-- ) {
  if ( $n % $i != 0 ) { $n += 20; $i = 20; }
 }
 print $n;
}

problem5();

Python

def problem5():
    n = 20
    i = 19
    while (i >= 2):
        if n % i != 0: n += 20; i = 20
        i -= 1
    print n

 

Go

 

func problem5() {
 n := 20
 for i := 19; i >= 2; i-- {
  if n%i != 0 {
   n += 20
   i = 20
  }
 }
 fmt.Println(n)

}

func main() {
 problem5()
}

 

 

C++

 

void problem5() {long long n = 20;for (int i = 19; i >= 2; i--) {if (n % i != 0) { n += 20; i = 20; }}cout << n << endl;}

 

 

Bash (very slow)

 

function problem5 {
    local n=20
    local i=19
    while ((i >= 2)); do
        if ((n % i != 0)); then ((n += 20)); ((i=20)); fi
        ((i--))
    done
    echo $n
}problem5

 

 

Javascript

 

function gcd(a,b) {return b==0?a:gcd(b,a%b)}
function lcm(a,b) {return a*b/gcd(a,b)}

 

var r=1;
for(var i=2; i<=20; i++) r=lcm(r,i);
r

 

Racket

 

(define (problem5)
  (display (apply lcm (range 1 21))))
(problem5)

 

 

Haskell

 

main = print $ foldr1 lcm [1..20]

 

* Clojure

 

(defn gcd2 [a b]
  (cond (zero? b) a
        :else (recur b (rem a b))))
(defn lcm2 [a b]
  (/ (* a b) (gcd2 a b)))
(defn problem5 []
  (println (reduce lcm2 (range 1 21))))

 

* newLISP

 

(define (lcm a b) (/ (* a b) (gcd a b)))
(apply lcm (sequence 1 20) 2)

 

 

Common Lisp

 

(let ((r 1))
  (loop for i from 2 to 20 do
       (setq r (lcm r i)))
  (print r))

Julia:

lcm(1:20)