『pwnable.kr』 Collision Problem solving

Wow… I had a feeling, but the difficulty’s on another level.

Thats absolutely lunatic, ōxō;;
~~That’s absolutely lunatic, ōxō;;~~

ε-(´・｀)

Perhaps because it involves a system hack,

it’s a whole level above the usual web-based

war games on Wargame.kr in terms of difficulty.

I ended up referring to someone else’s

solution while working through it.

And since posting the actual

source code would make the thread messy,

I’ll place it in a text file so it can simply be copy-pasted.

Anyway, let’s get started right away.

Right, if I’m having a punt at roughly translating
the gist of this in my own way,
it’s something like this:
So, you’re on about MD5 hash collisions then,
are we? (···) Right,anyway.

Right then, first things first,
open up yer Linux terminal.
Then you just wanna connect using

1	ssh col@pwnable.kr -p2222

Now,the password right?
It won’t actually show up as you’re typing it
– don’t mind that,just bash it in,hit Enter,
and bam, you’re logged in. Easy peasy!

Right then next up,you just wanna whack in

the ls command to see what files you’ve got lurking around.

Now col you’d probably reckon that’s
some sort of executable or maybe even a Python file, right?

But the actual one we really need to be bothered with,

the crucial bit, is col.c

so if you have a proper gander at the main bit,

the password, right, that needs to be 20

characters long. And here’s the crucial part:
the hash code (0x21DD09EC) and

the password return value,
they’ve both gotta be

absolutely spot on, exactly the same.
No messing about with that, understood

That
bash check_password(const char* p)
Bit, Yeah?
What it’s doing is, it’s taking chunks of the data
like, 4 bytes at a time – that’s the size of an int, innit?
– and it’s then accumulating those into ‘res’ five times.

So, the thinking is,
if you just add the exact same value five times over you should,

I reckon, get that hash code it’s after. Simple as that, really.

Derivation of results

Figure.1) For and res data expressed by cutting int type 4 bytes each

Right, so that’res’value,yeah?

That’s just the result of piling up
whatever came out of the ‘for’loop,
five times over.
And if you then,like,divvy that’res’value by five,
it turns out that the data

so if you then like,multiply this lot by five again
you’ll find you’re short by 0x4.So,
the assumption we’re making here is

that the data is essentially four lots of
0x6C5CEC8 and then one bit of 0x6C5CECC added in.

Figure. 2> Data expressed in little endian

then,since it’s all about little endian,yeah?
All you need to do is just feed the program that data
in hexadecimal,in the exact sequence.
That’s the one.

then, so you've gone and chucked the value in hexadecimal

using Python for that, yeah?

Now, here’s the kicker, mind you:
when the program picks that up as an input
it’ll jolly well recognise it as a char.
So, you just gotta watch out for that. Proper careful,like.