Evaluate the sum of all the amicable numbers under 10000.

Permalink: http://projecteuler.net/problem=21

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.

We can do it *bruteforcely*, but we won't.

Let start with the function `d`

. We call it `sum-of-proper-divisors`

```
(defn sum-of-proper-divisors [n]
(let [base (filter #(zero? (mod n %)) (range 2 (Math/sqrt n)))]
(reduce + 1 (concat (map #(/ n %) base) base))))
```

Bad approach to find all divisors: to iterate on all numbers from 1 to `n`

and check if it is divisor or not.
Instead, we iterate from 1 to `sqrt(n)`

, find one divisor and
calculate another symmetric divisor.

For example, take number `10`

.
We iterating from `2`

to `(sqrt 10)`

. Number `2`

is divisor, so we divide `(/ 10 2)`

, get `5`

which is also divisor. And so on.

This is predicate checks where two numbers amicable or not:

```
(defn amicable? [a b]
(and (not (= a b))
(= a (sum-of-proper-divisors b))
(= b (sum-of-proper-divisors a))))
```

Obvious enough, just consider that we skip case where `(= a b)`

.

The last part is to use our functions to obtain result.

```
(reduce +
(let [sums (vec (map sum-of-proper-divisors (range 1 10000)))]
(for [i (range 1 10000)]
(if (amicable? i (nth sums (dec i))) i 0))))
```

Just generate all sums, and check if another number `i`

produce the amicable pair. Done!

**P.S.** Take a look at the last snippet. `vec`

is really important there.