TBTL CTF 2024

solver by replican and gr3yr4t

archive dump chall : https://github.com/blockhousetech/TBTL-CTF/tree/master/2024/

Chall
Category
Total Solved

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

CryptoJS.AES.decrypt('U2FsdGVkX19wWL7itIL7TZcLTP/e1ulrZolI9AHTA8OBGOCodbZKdOxPF41rGV9C+X7PZPt9ISJKQMpTl+Fwew==', 'secret key is very secure').toString(CryptoJS.enc.Utf8)

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

curl "https://tbtl-talk-to-you.chals.io/?page=database.sqlite" --output file

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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