2021: corctf: supercomputer

2021/corctf/supercomputer/README.md

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+# supercomputer
+
+by [haskal](https://awoo.systems)
+
+crypto / 457 pts / 85 solves
+
+>I ran this code with a supercomputer from the future to encrypt the flag, just get your own and
+>decryption should be simple!
+
+provided files: [supercomputer.py](supercomputer.py) [output.txt](output.txt)
+
+## solution
+
+
+first, do some source review. there is not a lot of it
+```python
+def v(p, k):
+    ans = 0
+    while k % p == 0:
+        k /= p
+        ans += 1
+    return ans
+```
+this seems to be computing the number of times `p` is a factor of `k`
+
+```python
+p, q, r = getPrime(2048), getPrime(2048), getPrime(2048)
+print(p, q, r)
+n = pow(p, q) * r
+```
+this is taking a very very large power and multiplication (don't actually run this it'll never
+finish lol -- it was run on a supercomputer from the future but it can't be run on our computers)
+
+```python
+a1 = random.randint(0, n)
+a2 = n - a1
+assert a1 % p != 0 and a2 % p != 0
+t = pow(a1, n) + pow(a2, n)
+print(binascii.hexlify(xor(flag, long_to_bytes(v(p, t)))))
+```
+
+and here we generate a random number, subtract it from n, take the power and then use the
+factor-counting function to xor the flag
+
+now i don't know how to math at all so here i decided to run this with some much smaller primes and
+see what happens (dynamic analysis ftw). try replacing `p, q, r` with combinations of 3, 5, 7, and
+11.  you'll notice immediately that the result is always `2q`. so let's try it
+
+```python
+[ins] In [1]: import ast
+
+[ins] In [2]: with open("output.txt") as f:
+         ...:     [p,q,r] = [int(x) for x in f.readline().strip().split(" ")]
+         ...:     enc = ast.literal_eval(f.readline())
+
+[ins] In [3]: from pwn import xor
+
+[ins] In [4]: from binascii import unhexlify
+
+[ins] In [5]: enc = unhexlify(enc)
+
+[ins] In [6]: v = 2 * q
+
+[ins] In [7]: xor(enc, v.to_bytes(v.bit_length()//8+1, "big"))
+Out[7]: b'corctf{1_b3t_y0u_d1dnt_4ctu411y_d0_th3_m4th_d1d_y0u?}\ncorctf{1_b3t_y0u_d1dnt_4ctu411y_d0_th3_m4th_d1d_y0u?}\ncorctf{1_b3t_y0u_d1dnt_4ctu411y_d0_th3_m4th_d1d_y0u?}\ncorctf{1_b3t_y0u_d1dnt_4ctu411y_d0_th3_m4th_d1d_y0u?}\ncorctf{1_b3t_y0u_d1dnt_4ctu411y_d0_th3_m4'
+```
+
+yea

2021/corctf/supercomputer/output.txt

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+20936670545375210972091706288423179494163425035286134775773514440843943493090886819895346572945288304582498268271507942037581752184819846906869395551921930704321251130746547888224652316226957634541702883599286787839982090615950687496752999645558331533314682453610929822041558882012483238149288762974740347582024050756443700107245858419316423473568526347559377124536218894368962796664914408327949348396038507355935608178392088898784474582354438590711083089253977971653913217304360725716982473871023235180637867588860233011122300656470435644430602412710493441965130162664981423496370539240693045312454250776393871037539 19872523115298089612152987731023453644084277408261276810219001288407280019889227914287760742936580023163800626696116882213533508813201232707621762739857924392306902336092739272758773377952936022982446120177174082641600741522817135305633293579042208014735900229922142564590095968054337719254632703676737069746032384348392244892496672044899073391936273280270753785076044108870166304800552404013519058026991588856235381264192387525832530187004466616791531223421070547342377071358044704265893255021275811622959301157507095984825182110574434699593886509171425701861331576642311553357835312334349976576969220483604368671153 18342695102288954165224207958150786487860883752676419020596228714991017967256173183699487408637445601341687447489432163178271335469203559084363600703497940503946684342504933131623546315643648637992201226732630680112575643707020017139390225257319697353426087369722671485915571962910153169877358046375850132351117527591675467417925944135644417622440847857598273517926844822766083086147088819776687612745404553608100705862181700054385028096749375873889019995159762301115707945396140178370414857973922007665218670792403129624089144668480280115489465764431016721028424152163659378120333071194425845370101841510224643446231
+b'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'

2021/corctf/supercomputer/supercomputer.py

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+from Crypto.Util.number import getPrime, long_to_bytes
+from pwn import *
+import random, binascii
+
+flag = open('flag.txt').read()
+
+def v(p, k):
+	ans = 0
+	while k % p == 0:
+		k /= p
+		ans += 1
+	return ans
+
+p, q, r = getPrime(2048), getPrime(2048), getPrime(2048)
+print(p, q, r)
+n = pow(p, q) * r
+
+a1 = random.randint(0, n)
+a2 = n - a1
+assert a1 % p != 0 and a2 % p != 0
+
+t = pow(a1, n) + pow(a2, n)
+print(binascii.hexlify(xor(flag, long_to_bytes(v(p, t)))))
+
+
+