EasyCTF_2018: RSA_v

Category: Cryptography Points: 200 Description:

Bob is extremely paranoid, so he decided that just one RSA encryption is not enough. Before sending his message to Alice, he forced her to create 5 public keys so he could encrypt his message 5 times! Show him that he still is not secure... rsa.txt.

Write-up

This challenge revolves around the underlying principle of RSA and how encrypting a message 5 times with increasing exponents ultimately leads to complete confidentiality exploitation. The very base of the formula goes along something like this,

c = m^e mod n

However, by encrypthing it multiple times with the same n but differing e you get,

c1 = m^e1 mod n
c2 = c1^e2 mod n
c3 = c2^e3 mod n
c4 = c3^e4 mod n
c5 = c4^e5 mod n

Which when compiled looks like

c5 = ((((m^e1)^e2)^e3)^e4)^e5 mod n

As powers simply multiply up, where 2^(3^2) is just 2^6, we can calculate the effective e.

e = e1 * e2 * e3 * e4 * e5
  = 27587468384672288862881213094354358587433516035212531881921186101712498639965289973292625430363076074737388345935775494312333025500409503290686394032069

Now that we have the effective e we can produce our proper challenge,

e = 27587468384672288862881213094354358587433516035212531881921186101712498639965289973292625430363076074737388345935775494312333025500409503290686394032069
n = 9247606623523847772698953161616455664821867183571218056970099751301682205123115716089486799837447397925308887976775994817175994945760278197527909621793469
c = 7117565509436551004326380884878672285722722211683863300406979545670706419248965442464045826652880670654603049188012705474321735863639519103720255725251120

e seems like a really big number, similar to smallRSA of PicoCTF, let's try the same exploit, using Wiener's Attack, as we used back then! Literally ripped almost the same script from back then.

# ./solve.py 
easyctf{keblftftzibatdsqmqotemmty}

Therefore, the flag is easyctf{keblftftzibatdsqmqotemmty}.

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