Evaluate the sum of all the amicable numbers under 10000.
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
(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
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
We iterating from
(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.