By converting the hexadecimal numbers to their full-fledged decimal values, and entering them somewhere like here (and here) we can find the prime factors of which the n consists: 0x53a121a11e36d7a84dde3f5d73cf = 38456719616722997 Now let’s see if we can decrypt the data part.Īpparently, crypto nerds Alice and Bob failed to choose a large enough n, allowing us to recover their private RSA keys. So we have a sequence number, some data and a signature. U0VRID0gNDsgREFUQSA9IDB4MmMyOTE1MGYxZTMxMWVmMDliYzlmMDY3MzVhY0w7IFNJRyA9IDB4MTY2NWZiMmRhNzYxYzRkZTg5ZjI3YWM4MGNiTDs=ĭecoding this gives us the string SEQ = 4 DATA = 0x2c29150f1e311ef09bc9f06735acL SIG = 0x1665fb2da761c4de89f27ac80cbL Upon inspection of the packet capture, we notice every packet from Alice (192.168.0.13) to Bob (192.168.0.37) contains a base64-encoded payload. It looks mostly like garbage but maybe you can figure something out. He captured some traffic between crypto nerds Alice and Bob. Miller suspects that some of his students are cheating in an automated computer test. The first challenge I solved on 2015, hosted by FluxFingers, was Creative Cheating. Write-up of 2015’s Creative Cheating challenge.
0 Comments
Leave a Reply. |
Details
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |