The 21st problem on Project Euler states,

"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."

I'll first define a function, \(d(n)\).

`d(n) = [ k for k in 1:n-1 if n % k == 0 ] |> sum`

Next, we'll need some driver code to check all of the numbers under \(10,000\).

```
N = 10_000
checked = falses(3N)
pairs = Pair{Int, Int}[]
for a in 2:N
if checked[a] == false
checked[a] = true
b = d(a)
checked[b] = true
if d(b) == a && a ≠ b
push!(pairs, Pair(a, b))
end
end
end
reduce((acc, iter) -> acc + iter[1] + iter[2], pairs, init=0)
> 0.209513 seconds (146.66 k allocations: 8.028 MiB, 3.76% gc time, 6.53% compilation time)
>
> 31626
```

In this code snippet, we create an array of booleans using `falses`

so we can avoid double checking numbers we've already checked. If the `checked`

array shows a number hasn't yet been checked, then we check to see if \(d(a) = d(b)\) where \(a \neq b\). If both criteria pass, then we add the pair into our array of pairs.

Finally, we use the `reduce`

function to sum them up yielding the final result.

© Daniel Marvin. Last modified: July 22, 2021.