Welcome to my series of writeups for n00bzunit3d 2024 capture-the-flag competition. In this post, we look at the crypto/RSA challenge.

The challenge provides us with the encrypted flag, public key values e and n.

e = 3
n = 135112325288715136727832177735512070625083219670480717841817583343851445454356579794543601926517886432778754079508684454122465776544049537510760149616899986522216930847357907483054348419798542025184280105958211364798924985051999921354369017984140216806642244876998054533895072842602131552047667500910960834243
c = 13037717184940851534440408074902031173938827302834506159512256813794613267487160058287930781080450199371859916605839773796744179698270340378901298046506802163106509143441799583051647999737073025726173300915916758770511497524353491642840238968166849681827669150543335788616727518429916536945395813

We pass the information to any online RSA decyptor such as dCode and it gives us the flag.