Machine Learning Team Lead at White Hat, Founder/Director at Farset Labs
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
Was not happy with limiting the specific factors of 3 and 5, and limit of 1000 so i implemented a generalised solution.
def solution(fs: List[Int], max: Int ) = ( 1 until max ).filter( f => fs.exists( n => f % n == 0)).foldLeft(0)(_+_)
Where fs is a List of integer factors and max is obv the highest limit. eg(Answer is in white font, so select the line “Answer” to see it
print("Answer: ",solution(List(3,5),1000)) >Answer: <span style="color: #ffffff;"><strong>233168</strong></span>
Thanks to @lichtsprung for the testing help.
The general aim is to complete a few dozen of these in Scala, then switch to Java, or C/C++, or something else, and loop around until i have a complete library of euler problems for all of the languages i end up using; Nothing like having long term goals!