The Problem

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n ).

If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

The Solution

First, we have to find all the divisors of a given n.

let intsqrt i = int(sqrt(float i))

let findDivisors n =
    let upperBound = intsqrt n
 
    [1..upperBound]
    |> Seq.filter (fun d -> n % d = 0)
    |> Seq.collect (fun d -> [d; n/d])
    |> Seq.filter (fun d -> d <> n)
    |> Seq.distinct

Like some of our prime-finding code, we only have to check for divisors up to the square root of n. If d is a divisor less than the square root of n, then both d and n/d are divisors of n. findDivisors checks all numbers up to the square root of n for divisibility, and adds any divisors it finds along with n/d. Lastly, it filters out n itself and uses Seq.distinct to make sure it only returns unique divisors.

Next, we need to find out if a given a and b are "amicable" according to the problem definition.

let sumDivisors n = findDivisors n |> Seq.sum

let d =
    seq { 1..9999 }
	|> Seq.map (fun n -> n,sumDivisors n)
	|> Map.ofSeq

let areAmicable (a,b) = d.[a] = b && d.[b] = a

The sumDivisors function simply sums the divisors we found above.

Next, we build a map of divisor sums. For the numbers 1 to 9999, n is the key and the sum of its divisors is the value.

With that in hand, we can determine if a given a and b areAmicable using the test given in the problem definition.

And now we can look for the answer.

seq {
    for a in 1..9999 do
        for b in 1..9999 do
            if a <> b then yield a,b
}
|> Seq.filter areAmicable
|> Seq.sumBy fst

First, we build a sequence of all possible pairs of numbers from 1 to 9999, skipping the case when a and b are equal. Then we filter out the pairs that areAmicable and sum the a values to get the answer.

Other Posts in This Series