Project Euler – Problem 12 Solution

Yan Cui

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Problem

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …

Let us list the factors of the first seven triangle numbers:

1: 1

3: 1,3

6: 1,2,3,6

10: 1,2,5,10

15: 1,3,5,15

21: 1,3,7,21

28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

Solution

open System

let triangleNumber(n:int64) = [1L..n] |> Seq.sum

let findFactorsOf(n:int64) =
    let upperBound = int64(Math.Sqrt(double(n)))
    [1L..upperBound] 
    |> Seq.filter (fun x -> n % x = 0L) 
    |> Seq.collect (fun x -> [x; n/x])

let naturalNumbers = Seq.unfold (fun x -> Some(x, x+1L)) 1L

let answer =
    naturalNumbers
    |> Seq.map (fun x -> triangleNumber(x))
    |> Seq.filter (fun x -> Seq.length(findFactorsOf(x)) >= 500)
    |> Seq.head

My solution here is simple, first create the triangleNumber function that calculates the triangle number for a given natural number.

I then created a modified version of the findFactorsOf function I first used in the problem 3 solution to find ALL the factors of a number including 1 and itself.

Again I used Seq.unfold to build up a sequence of natural numbers and to find the answer I iterate through the natural numbers and for each get its triangle number; find the factors of the triangle number; and check if the number of factors is at least 500.

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2 thoughts on “Project Euler – Problem 12 Solution”

  1. mine is a bit more optimal, cuz I don’t start summing from 1 all the time but from previous triangle number, I also have a printf to see the process

    let p12 =
    let factors number =
    let ff = seq {
    for divisor in 1 .. (float >> sqrt >> int) number do
    if number % divisor = 0 then
    yield divisor
    yield number / divisor
    }
    ff|> Seq.distinct
    let rec loop i tri fc =
    printf “%A %A %A \n” i tri fc
    match fc with
    | x when x > 500 -> tri
    | x -> loop (i+1) (tri+i+1) ((factors (tri+i+1)) |> Seq.length)
    loop 1 1 1

  2. also, when you find factors you don’t have to go from 1 to upperbound but from 1 to sqrt(upperbound)

    and when you get n%i = 0 you add to he factors collection both n and i

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