added: my solutions in python and in erlang to the divisor challenge

challenges/easy/divisor-of-a-number/solutions/valentinadagata/main.erl

@@ -0,0 +1,21 @@
+-module(main).
+-export([start/0, d/1, sumdivs/1]).
+
+d(0) -> [];
+d(1) -> [];
+d(N) -> lists:sort([] ++ divisors(1,N,math:sqrt(N))).
+
+% if the square root is lower than the first num return empty
+divisors(F,_N,L) when F > L -> [];
+% if the remainder of N and the first is not 0, add 1 to the first number
+divisors(F,N,L) when N rem F =/= 0 -> 
+    divisors(F+1,N,L);
+% if the number is divisible for the first, add them to the list of divisors
+divisors(F,N,L) ->
+    [F, N div F] ++ divisors(F+1,N,L).
+
+sumdivs(N) -> lists:sum(d(N)).
+
+
+start() ->
+  io:fwrite("~w~n", [d(30)]).

challenges/easy/divisor-of-a-number/solutions/valentinadagata/main.py

@@ -0,0 +1,23 @@
+"""
+Divisors of a number
+Count the number of divisors of a positive integer `n`.
+
+divisors(4)  // => 3 (1, 2, 4)
+divisors(5)  // => 2 (1, 5)
+divisors(12) // => 6 (1, 2, 3, 4, 6, 12)
+divisors(30) // => 8 (1, 2, 3, 5, 6, 10, 15, 30)
+"""
+def divisors(n):
+    # create a list with all the numbers from 1 to n
+    num = list(range(1, n + 1))
+    div = []
+    # for each number, if n % num == 0: append to the divisors list div
+    # else delete the number from the num list
+    while len(num) > 0:
+        if n % num[-1] == 0:
+            div.append(num[-1])
+        num.pop()
+    #print(div)
+    return len(div)
+
+print(divisors(30))