Project Euler – Problem 47 Solution


The first two consecutive numbers to have two distinct prime factors are:

14 = 2 x 7

15 = 3 x 5

The first three consecutive numbers to have three distinct prime factors are:

644 = 2² x 7 x 23

645 = 3 x 5 x 43

646 = 2 x 17 x 19

Find the first four consecutive integers to have four distinct primes factors. What is the first of these numbers?


let hasDivisor(n) =
    let upperBound = bigint(sqrt(double(n)))
    [2I..upperBound] |> Seq.exists (fun x -> n % x = 0I)

let isPrime(n) = if n = 1I then false else not(hasDivisor(n))

let naturalNumbers = Seq.unfold (fun state -> Some(state, state+1I)) 1I

// define the sequence of prime numbers
let primeSeq = naturalNumbers |> Seq.filter isPrime |> Seq.cache

// recursive function to find the prime denominators for a number n
let rec getPrimeFactors denominators n =
    if n = 1I then denominators
        let denominator = primeSeq |> Seq.filter (fun x -> n % x = 0I) |> Seq.head
        getPrimeFactors (denominators @ [denominator]) (n/denominator)

// curry the getPrimeDenominators function to start with an empty list
let primeFactors = getPrimeFactors []

// function to get the number of distinct prime factors a number has
let distinctPrimeFactorsCount n = primeFactors n |> Seq.distinct |> Seq.length

// define the sequence of numbers with exactly 4 distinct prime factors
let seq = naturalNumbers |> Seq.filter (fun n -> distinctPrimeFactorsCount n = 4) |> Seq.cache

let answer =
    |> Seq.windowed 4
    |> Seq.filter (fun l -> Seq.max (l) - Seq.min (l) = 3I)
    |> Seq.head
    |> Seq.head

Whilst on the surface this isn’t a very difficult problem to solve, the biggest challenge I had was in making sure the solution runs in a reasonable time hence note the various places where I used Seq.cache to help improve the performance of this code.


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