Problem 239. Project Euler: Problem 5, Smallest multiple

Created by Doug Hull

Solve Later

{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetMode":"","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetMode":"","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>What is the smallest positive number that is evenly divisible by all of the numbers from 1 to input number?</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Thank you to</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"http://projecteuler.net/problem=5\"><w:r><w:t>Project Euler Problem 5</w:t></w:r></w:hyperlink></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

Solve

Solution Stats

825 Solutions
336 Solvers

Last Solution submitted on Oct 22, 2020

Last 200 Solutions

Problem Comments

1 Comment

1 Comment

Aurelien Queffurust on 20 Apr 2012

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

Solution Comments

Solution 1688897

1 Comment

1 Comment

Roche de Guzman on 8 Dec 2018

Great problem!

Solution 41338

2 Comments

2 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

Solution 27118

1 Comment

1 Comment

Franck Dernoncourt on 3 Feb 2012

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

Problem Recent Solvers336

Problem Tags

projecteuler

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Problem 239. Project Euler: Problem 5, Smallest multiple

Solution Stats

Last 200 Solutions

Problem Comments

Solution Comments

Problem Recent Solvers336

Suggested Problems

More from this Author52

Problem Tags

Community Treasure Hunt

Problem 239. Project Euler: Problem 5, Smallest multiple

Solution Stats

Last 200 Solutions

Problem Comments

Solution Comments

Problem Recent Solvers336

Suggested Problems

More from this Author52

Problem Tags

Community Treasure Hunt

Players