Problem 239. Project Euler: Problem 5, Smallest multiple

Created by Doug Hull

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940 Solutions
397 Solvers

Last Solution submitted on Jan 14, 2021

Last 200 Solutions

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2 Comments

2 Comments

Aurelien Queffurust on 20 Apr 2012

FYI: we cannot execute "matlabpool open" on the labs. ;)

Yunpeng Teng on 4 Dec 2020

DON'T TALK WUDE

function y = euler005(x)
y=1;
for i=2:x
y=lcm(i,y);
end
end

Solution Comments

Solution 1688897

1 Comment

1 Comment

Roche de Guzman on 8 Dec 2018

Great problem!

Solution 41338

3 Comments

3 Comments

blaat on 14 Feb 2012

What it does: it finds the maximum number of occurrences for each prime number smaller than x in the factorisations of the numbers 1:x (e.g., for x = 10, the maximum number of occurrences for 2 is 3, since 2*2*2 = 8). If the product of all prime factors taken to the power of their maximum # of occurrences is then taken, the smallest number that is divisible by 1:10 is obtained. So for x = 10: 2 * 2 * 2 * 3 * 3 * 5 * 7 = 2520.

Alfonso Nieto-Castanon on 14 Feb 2012

nice

Manuel Garcia on 5 Nov 2020

cool

Solution 27118

1 Comment

1 Comment

Franck Dernoncourt on 3 Feb 2012

This code works but the server is too slow...

Problem Recent Solvers397

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Problem 239. Project Euler: Problem 5, Smallest multiple

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Problem 239. Project Euler: Problem 5, Smallest multiple

Solution Stats

Last 200 Solutions

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Solution Comments

Problem Recent Solvers397

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