We're given RSA values for n, e, and ct:
```n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317e = 65537c = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681```
[factordb](http://factordb.com/) fully factorizes n. However, it's not just a standard p*q (i.e. two prime) result (as the challenge descrption hints at)
In python:
```# pip3.8 install factordb-pycli>>> from factordb.factordb import FactorDB>>>>>> def check_factordb(N):...     factorized = FactorDB(N)...     factorized.connect()...     factor_list = factorized.get_factor_list()...     print(factor_list)...     return factor_list...>>> a = check_factordb(7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317)[2208664111, 2214452749, 2259012491, 2265830453, 2372942981, 2393757139, 2465499073, 2508863309, 2543358889, 2589229021, 2642723827, 2758626487, 2850808189, 2947867051, 2982067987, 3130932919, 3290718047, 3510442297, 3600488797, 3644712913, 3650456981, 3726115171, 3750978137, 3789130951, 3810149963, 3979951739, 4033877203, 4128271747, 4162800959, 4205130337, 4221911101, 4268160257]>>> len(a)32```
There's 32 prime factors! So we need to then do RSA with each of these values as a prime.
In order to calculate phi, we must now do:
```phi = (prime0 - 1) * (prime1 - 1) * ...  * (prime31 - 0)```
In java:
```        BigInteger phi = new BigInteger("1");        for (int i=0; i