ctf-writeup/2023/DownUnderCTF 2023/complementary

complementary

Two halves make a whole.

About the Challenge

We got 2 files, complementary.py and output.txt. Here is the content of complementary.py

flag = open('./flag.txt', 'rb').read().strip()
m1 = int.from_bytes(flag[:len(flag)//2])
m2 = int.from_bytes(flag[len(flag)//2:])
n = m1 * m2
print(n)

The code reads the flag.txt file, splits it into two parts, converts each part into an integer, multiplies them together, and prints the result. The exact purpose depends on the content of flag.txt. And here is the content of output.txt

6954494065942554678316751997792528753841173212407363342283423753536991947310058248515278

How to Solve?

We can use factordb to find the value of m1 and m2. Here is the command I used to find the value of m1 and m2

factordb 6954494065942554678316751997792528753841173212407363342283423753536991947310058248515278

And the output was

2 3 19 31 83 3331 165219437 550618493 66969810339969829 1168302403781268101731523384107546514884411261

Now multiply 2 x 3 x 19 x 31 x 83 x 3331 x 165219437 x 550618493 x 66969810339969829 and the result was 5952648940406176619300725964928725160189798. And we got 2 big numbers right now

m1 = 5952648940406176619300725964928725160189798
m2 = 1168302403781268101731523384107546514884411261

I create another program like this

m1 = 5952648940406176619300725964928725160189798
m2 = 1168302403781268101731523384107546514884411261

# Combine m1 and m2 as bytes
flag_bytes = (m1.to_bytes((m1.bit_length() + 7) // 8, 'big') +
              m2.to_bytes((m2.bit_length() + 7) // 8, 'big'))

# Decode the bytes to a string
flag = flag_bytes.decode('utf-8')

print(flag)

The code takes two integers, m1 and m2, converts them into a single byte sequence, and then decodes that byte sequence into a UTF-8 string. The print statement displays the resulting a flag.

DUCTF{is_1nt3ger_f4ct0r1s4t10n_h4rd?}