mathe_fünfte_klasse.clj

;;
;; Lösung in Clojure
;;

(defn make-func [x] 
  (partial #(= (mod %2 %1) (- %1 1)) x))

(defn make-funcs []
  (cons 
    #(zero? (mod % 7)) 
    (map make-func (range 2 7))))

(filter (apply every-pred (make-funcs)) (range 1 1001))

;; gibt (119 539 959)

mathe_fünfte_klasse.py

# Lösung zu dieser Aufgabe in Python
# x % 2 = 1
# x % 3 = 2
# x % 4 = 3
# x % 5 = 4
# x % 6 = 6
# x % 7 = 0
#
# mit 1 <= x < 1000

from functools import partial

funcs = [partial((lambda n, x: x % n == n-1), n) for n in range(2, 7)]
funcs.append(lambda x: x % 7 == 0)
 
for i in range(1, 1001):
    if all(f(i) for f in funcs):
        print(i, end=" ")

# gibt:
# > 119 539 959