Kongruenciaval megoldhato, de joval tobb macera, mint mondjuk a mai technikai szinten beirni 3 sort a python interpreterbe:
>>> x = (1<<1992)-1
>>> divisors = [i for i in range(1, 100) if x % i == 0]
>>> len(divisors)
17
>>> len(str(x))
600
>>> divisors
[1, 3, 5, 7, 9, 13, 15, 17, 21, 35, 39, 45, 51, 63, 65, 85, 91]
>>> x
448488552841505673528450469210032025008717852378396562686579194072564945856394653850257611842385261686892954937104477645877780844087130988838254436002598900543476520098876093752690543547812675279019246054935493143678185404868704546669207132970571695198847642805321870347439959396966061131000641492071612036500659800720541037509294862530914292581776470704800791862135257150476594266547973795942452669207838277325767258308678827571890789686926497076331599351180046564667803842147156402969070145377463213888204787436429748157612446950906462925450296030791504591545034378214229573507677553758534574800895
A also par sor mar csak kivancsisag :)
Persze lehet meg ennel is egyszerubb modja.