TSGCTF2020 Beginner’s Crypto

1 minute read


  • Solved after the CTF ended :(
    • I needed to see this post sooner…
    • One of the answers includes an arithmetic version of left and right shift.
  • I need to practice more crypto…

Using rewritten version that doesn’t read input from file.

First assert

assert(len(x))<=50); where x is a string (presumably).

This means that each unit of length of x is a byte (char). However, to easily deal with integers, we can transform this constraint.

The below is pseudocode!

len(str((int) x)) <= 2**(50*8) = 2**400

Second assert

assert(str(int.from_bytes(x, "utf-8"), byteorder='big') << 10000).endswith(...)

Assuming we already did the (int) conversion to change x to xint, we get


An interesting property about left shift is that x<<y = x*(2**y). Also, this is not relevant, but x>>y = x/(2**y)!

This means that our assertion is:


Putting it together

Let’s call our given ending sequence R. We will also call L = len(str(R)).

We know that

Eq 1: (xint * 2**10000) % 10**L = R.

And knowing L=175, (xint * 2*10000) % 10**175 = R.

So, let’s try to reverse this!

To reverse this, we would generally multiply both sides by the inverse of 2**10000 % 10**175. However, it is not possible to find the inverse because the gcd of 2**10000 and 10**175 is not 1! reasoning

But, we can massage the right hand terms of Eq 1 a bit (find the gcd and divide, basically).

(xint * 2**10000)%10**175

= ((xint * 2**10000)%10**175)%5**175)

= (xint*2**10000)%5**175)

We can find the inverse of 2**10000 % 5**175!

Then, we multiply by the inverse:

(xint * 2**10000 * inverse = R * inverse) % 5**175

(xint = solution) % 5**175

Keep in mind that 5**175 > 2**400, meaning the solution we get will not be cut off in a weird way. After getting the solution, we can convert the integer to bytes, then to a string. Plug into the challenge and make sure that we pass both assertions!

I tried to use Sage… but couldn’t wrap my head around it… I ended up just using Python. The script is included in the write-up repository.