Baby Rsa
The encryption used a small e, so find the e-th root of ct to decrypt it.
e = 3
n = 21240130069302595435883573568292543584653982426668643904196630885984119007899960150162877143271928662185885422702123670222165981446412189843665571992895649937195036232374014356896167929469467494531756153911013832353810970941919101050971790197002016280790620714887304192321101311465703150098410331176735899796484284165771555960758054286754565310439163189954842301676099617954811528874343372426916478057819577132937062857039063351856289801979923260408285890418889829381378968646646737194160697920287161229178345666260994127087040393511692642122516019055570881253021165130706539874713965212158253699181636631222365809257
ct = 80505397907128518326368510654343095894448384569115420624567650731853204381479599216226376345254941090872832963619259274943986478887206647256170253591735005504plain, _ =gmpy.root(80505397907128518326368510654343095894448384569115420624567650731853204381479599216226376345254941090872832963619259274943986478887206647256170253591735005504, 3)
print(number.long_to_bytes(int(plain)))The decrypted message was the flag short_and_to_the_point.