Project Euler Problem 5

[Philosophe]

Overview
This VI shows how to solve Project Euler Problem #5.

 
Description
Project Euler Problem #5:

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?

More information about Project Euler can be found @ www.projecteuler.net 

Note: This code implements a brute force method. May take several seconds to complete when "Numeric Control" > 15
Requirements

• LabVIEW 2013 (or compatible)

Steps to Implement or Execute Code

1. Run VI
2. Select Numeric Control

Additional Information or References

 **This document has been updated to meet the current required format for the NI Code Exchange.**

[StaceyG]

You do like the brute force, don't you. Part of the point of Project Euler is elegant and efficient solutions. Let's break this problem down a bit. What does it mean that a number X is divisible by all the number 1..N?

First, X will be divisible by all of the primes in 1..N. For instance, 2520 is divisible by 2, 3, 5, and 7, the primes in 1..10.

Second, X will need to be multiply divisible by these primes, up to a certain factor. By this I mean that, for example, 2520 must be divisible by 8, which is 2*2*2.The certain factor is (in this example) 10.

So, we can construct a more elegant and efficient solution:

1. Start with a value of 1.

2. Loop over the numbers 2..N.

3. If our current iteration is a prime, multiply the prime by itself until it is just under N. Multiply our result by this value.

4. Skip non-primes.

I'll post my solution shortly (after I figure out how.....)

[StaceyG]

Ok, I can't figure out how to upload my VI.