TBTL CTF 2024
solver by replican and gr3yr4t
archive dump chall : https://github.com/blockhousetech/TBTL-CTF/tree/master/2024/
Butterfly
flag from indexed db : U2FsdGVkX19wWL7itIL7TZcLTP/e1ulrZolI9AHTA8OBGOCodbZKdOxPF41rGV9C+X7PZPt9ISJKQMpTl+Fwew==
{"code":"CryptoJS.AES.decrypt(CIPHERTEXT, KEY).toString(CryptoJS.enc.Utf8)"} ( kita dapetin dari session ) kita asumsi bahwa flagnya nanti memakai function ini tinggal cari secretnya
secret ada di session lgsg aja
flag : TBTL{th15_1S_n0t_53CUR3_5T0r4G3}
Talk to you
lfi on the page , first we use ../flag.txt and the site said the flag in database.sqlite so we change the parameter and we get the flag
TBTL{4Typ1c41_d4T4B453_u54g3}
Wikipedia signatures
The exploits here are always around the fact that you can give texts that have the same modulus with N as the goal
Here however they forbid all payloads of the form pt + a*N
However we can simply do (N-pt)
As we know that (N-pt)^e = -C (mod N)
So taking -(-C) (mod N) gives us what we need and supplying that decrypts it successfully
TBTL{r3p347_4f73r_m3-d16174l_516n47ur3_15_n07_3ncryp710n}
Flagcheck
Rnd For Data Science
csv injection payload: ,delimiter,#
TBTL{d4T4_5c13nc3_15_n07_f0r_r0ck135}
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