Se aplica de 3 ori Teorema impartirii cu rest: Ā
n=24Ā·cā+5 ā n-5=24Ā·cā ā n-5=Māā (multiplu de 24) (1)
n=40Ā·cā+5 ā n-5=40Ā·cā ā n-5=Māā (multiplu de 40) (2)
n=56Ā·cā+5Ā ā n-5=56Ā·cāĀ ā n-5=Mā
ā (multiplu de 56) (3) Ā
Din relatiile(1),(2) si (3) Ā ā n-5=[24;40;56]ān-5=840ān=845.