## Project Euler Problem 1

Part 1,2

Project Euler problem 1 wants a person to find the sum of all numbers less than 1000 that are divisible by 3 and 5. I was able solve this problem pretty easily in Scala. My solution is very similar to how I would solve the problem in Java. Quickly glancing at the code, you can see there is less of it. Keywords in Java like **Public** and **static **aren’t in here. It may not see like much, but it adds up.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 | object Euler001 { def main(args: Array[String]) { var a = 0 var sum = 0 while(a < 1000) { if(a%3 == 0 || a%5 == 0) sum = sum + a a = a + 1 } println(sum) } } |

Using a while loop solved the problem. I have more variations of the solution I will post later. If you are trying to learn a new programming language, projects like Euler is a great way to do this. For now just trust me, I will write a follow up article on why.

## Colin Dellow 1:24 pm

onDecember 14, 2011 Permalink | Log in to ReplyPfft, that’s far too readable.

def reduce(i: Int) = if(i % 3 == 0 || i % 5 == 0) i else 0

(0 /: (0 until 1000)) { _ + reduce(_) }

Much better!

## Xavier 1:36 pm

onDecember 14, 2011 Permalink | Log in to ReplyPfff, that’s far too brutal:

3 * 333 * 334 / 2 + 5 * 199 * 200 / 2 – 15 * 66 * 67 / 2 is much better! =)

## Eliot 5:00 pm

onDecember 14, 2011 Permalink | Log in to ReplyI never thought about it that way. How did you come up with that approach?

## Xavier 6:57 am

onDecember 15, 2011 Permalink | Log in to ReplyWell, you don’t need Scala or any other programming language, good old math can do it for you

solution

= sum of multiples of 3 < 1000

+ sum of multiples of 5 < 1000

sum of n first numbers = n*(n+1)/2

sum of n first multiples of x = x*n*(n+1)/2

there are 333 multiples of 3 < 1000 (last one is 333 * 3 = 999)

there are 199 multiples of 5 < 1000 (last one is 199 * 5 = 995)

there are 66 multiples of 15 < 1000 (last one is 66 * 15 = 990)

I know that many people with coding skills will try to brute-force every project-euler problems… but some of those problems actually don't need to be brute-forced.

## Eliot 1:23 pm

onDecember 15, 2011 PermalinkVery good point brute force is not needed for this one. Thanks for the explanation.

## Didier 11:09 am

onJanuary 11, 2013 Permalinkok for good old math but where your magical numbers 333, 199, 66 are coming from. is there a mathematical formula to calculate how many multiples of x there are between a and b ??

## Colin Dellow 1:41 pm

onDecember 14, 2011 Permalink | Log in to ReplyTo explain my solution, I recast your imperative solution in a functional form. The two mutable variables you used — sum and a — were replaced with a foldLeft and a sequence.

What was a comes from (0 until 1000) — a Seq[Int] that includes the numbers 0 <= i accumulator + reduce(current) }

Which says: Initialize accumulator to 0. Iterate over the numbers for 0 until 1000, updating accumulator as needed.

You can then use Scala’s magic _ syntax sugar to make it more concise. Whenever you’re defining a function whose parameters are known (in this case, the accumulator = accumulator + reduce(current) is actually a function whose type is defined by the foldLeft function), if you use each parameter exactly once, each subsequent use of _ refers to the next parameter.

So, you could rewrite as:

(0 until 1000).foldLeft(0) { _ + reduce(_) }

## Kim 2:07 pm

onDecember 14, 2011 Permalink | Log in to Replynoobs…;)

1 to 1000 filter { x => x % 3 == 0 || x % 5 == 0 } sum

## ohboy 2:47 pm

onDecember 14, 2011 Permalink | Log in to ReplyPlease take some time to learn the language before posting these kinds of “solutions”. What you’ve done was just translate a Java solution and say it’s Scala, very very bad coding style. Kim’s solution is how you should be thinking about this in Scala.

## ohboy 3:46 pm

onDecember 14, 2011 Permalink | Log in to ReplyTo elaborate on what is condireded “bad style” in your code:

mutable variables: Scala promotes functional style coding, where everything is immutable

loop over a range/collection: again functional style transformations are preferred, look at reduce, fold, map, filter, sum, etc.

(and lose those semicolons, they’re evil

## Eliot 6:21 pm

onDecember 14, 2011 Permalink | Log in to ReplyI removed the semi-colons. Thank you for your insight. I appreciate your opinion. I feel that posting Java-Scala hybrid solutions help people understand the new concepts in Scala. I never stated use this code in production, or this is “good style.” I stated this is a solution I came up with.

## Chinmay 8:45 pm

onApril 30, 2013 Permalink | Log in to Replyprintln((1 until x).filter(a => a % 3 == 0 || a % 5 == 0).foldLeft(0)(_+_))