Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

Permalink: http://projecteuler.net/problem=23

To define abundant number letâ€™s revisit `sum-of-proper-divisors`

from the Clojure Euler: Problem 021

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

Using that function, `abundant?`

predicate just follows definition.

```
(defn abundant? [n]
(> (sum-of-proper-divisors n) n))
```

If you look carefully at `sum-of-proper-divisors`

it is buggy.

```
(sum-of-proper-divisors 25) => 1 ;; instead of 6
```

Square numbers are not handled correctly.
To fix that we add one more list to `concat`

.

```
(defn sum-of-proper-divisors [n]
(let [divs (filter #(zero? (mod n %)) (range 2 (Math/sqrt n)))]
(reduce + 1 (set (concat
(let [isq (int (Math/sqrt n))] ;; added
(if (= n (* isq isq)) [isq] [])) ;; added
divs
(map #(/ n %) divs))))))
```

Double check

```
(sum-of-proper-divisors 25) => 6
```

Great! Next step is to find all integer numbers which can not be written as the sum of two abundant numbers.

Precalculate all abundant numbers:

```
(def abundant (filter abundant? (range 12 28124)))
```

Now we can build all numbers which can be written as two abundant numbers and find their sum.

```
(->> (for [i abundant j abundant
:let [num (+ i j)]
:when (< num 28124)]
num)
(distinct)
(reduce +)
(- (reduce + (range 1 28124))))
```

This solution works ~44 sec, too much.

More clever approach is to iterate from all integers upto 28124 and check if some number can be written as a sum of two abundant numbers.

The trick is simple: we building sorted set of abundant numbers

```
(def abundant (into (sorted-set) (filter abundant? (range 12 28124))))
```

For every integer number `n`

we subtract one abundant number and check if `abundant`

set contains result.
There is a good function some that makes our life easier.

```
(defn abundant-sum? [n abundant]
(some #(abundant (- n %)) abundant))
```

Finding the sum

```
(->> (range 1 28124)
(remove #(abundant-sum? % abundant))
(reduce +))
```

It works ~26 sec.

Another optimization is to limit number of abundants in
`abundant-sum`

function.

```
(defn abundant-sum? [n abundant]
(some #(abundant (- n %))
(take-while #(< % n) abundant)))
```

Time reduced to ~14 sec. Pretty good.

**P.S.** Buggy version of `sum-of-proper-divisor`

solved the previous problem correctly, but it was just a luck. Make sure you test your functions.