[Alex CTF 2017][Writeup][CR3: What is this encryption?]

Sloved this problem when contest ended
Hint

Fady assumed this time that you will be so n00b to tell what encryption he is using
he send the following note to his friend in plain sight :
p=0xa6055ec186de51800ddd6fcbf0192384ff42d707a55f57af4fcfb0d1dc7bd97055e8275cd4b78ec63c5d592f567c66393a061324aa2e6a8d8fc2a910cbee1ed9
q=0xfa0f9463ea0a93b929c099320d31c277e0b0dbc65b189ed76124f5a1218f5d91fd0102a4c8de11f28be5e4d0ae91ab319f4537e97ed74bc663e972a4a9119307
e=0x6d1fdab4ce3217b3fc32c9ed480a31d067fd57d93a9ab52b472dc393ab7852fbcb11abbebfd6aaae8032db1316dc22d3f7c3d631e24df13ef23d3b381a1c3e04abcc745d402ee3a031ac2718fae63b240837b4f657f29ca4702da9af22a3a019d68904a969ddb01bcf941df70af042f4fae5cbeb9c2151b324f387e525094c41
c=0x7fe1a4f743675d1987d25d38111fae0f78bbea6852cba5beda47db76d119a3efe24cb04b9449f53becd43b0b46e269826a983f832abb53b7a7e24a43ad15378344ed5c20f51e268186d24c76050c1e73647523bd5f91d9b6ad3e86bbf9126588b1dee21e6997372e36c3e74284734748891829665086e0dc523ed23c386bb520

During contest I know this is RSA but I didn’t know How to slove it!!! :sad:
After contest I read about RSA and sloved it :cry:

I got p q e and c
c: Cypher text need decrypt
e: public key in RSA
so I need d is a private key to decypt c
I find d step by step below

1. n = p*q
2. phi(n) = (p - 1) * (q - 1)
3. using Extended Euclid to find d
4. plain text = c^d % n

Here is my python code

p=0xa6055ec186de51800ddd6fcbf0192384ff42d707a55f57af4fcfb0d1dc7bd97055e8275cd4b78ec63c5d592f567c66393a061324aa2e6a8d8fc2a910cbee1ed9
q=0xfa0f9463ea0a93b929c099320d31c277e0b0dbc65b189ed76124f5a1218f5d91fd0102a4c8de11f28be5e4d0ae91ab319f4537e97ed74bc663e972a4a9119307
e=0x6d1fdab4ce3217b3fc32c9ed480a31d067fd57d93a9ab52b472dc393ab7852fbcb11abbebfd6aaae8032db1316dc22d3f7c3d631e24df13ef23d3b381a1c3e04abcc745d402ee3a031ac2718fae63b240837b4f657f29ca4702da9af22a3a019d68904a969ddb01bcf941df70af042f4fae5cbeb9c2151b324f387e525094c41
c=0x7fe1a4f743675d1987d25d38111fae0f78bbea6852cba5beda47db76d119a3efe24cb04b9449f53becd43b0b46e269826a983f832abb53b7a7e24a43ad15378344ed5c20f51e268186d24c76050c1e73647523bd5f91d9b6ad3e86bbf9126588b1dee21e6997372e36c3e74284734748891829665086e0dc523ed23c386bb520
phi = (p - 1) * (q - 1)
y0=0
y1=1
d = 0
#Extented Euclid
while phi > 0:
    r = phi % e
    if (r == 0):
        break
    t = phi // e
    y = y0 - y1 * t
    phi = e
    e = r
    y0 = y1
    y1 = y
if (e > 1):
    print("cannot find d")
else:
    d = y
n = p * q
plain = pow(c, d, n)
print str(hex(plain))[2:-1].decode("hex")

Flag is ALEXCTF{RS4_I5_E55ENT1AL_T0_D0_BY_H4ND}

[AlexCTF 2017][Writeup][RE4: unVM me]

Hint

If I tell you what version of python I used … where is the fun in that?

Download here

Check it using file command. I got

unvm_me.pyc: python 2.7 byte-compiled

_. “I know version of python bro!!!” ;)

Next I decompile .pyc file using pycdc. I got

# Source Generated with Decompyle++
# File: unvm_me.pyc (Python 2.7)

import md5
md5s = [
    0x831DAA3C843BA8B087C895F0ED305CE7L,
    0x6722F7A07246C6AF20662B855846C2C8L,
    0x5F04850FEC81A27AB5FC98BEFA4EB40CL,
    0xECF8DCAC7503E63A6A3667C5FB94F610L,
    0xC0FD15AE2C3931BC1E140523AE934722L,
    0x569F606FD6DA5D612F10CFB95C0BDE6DL,
    0x68CB5A1CF54C078BF0E7E89584C1A4EL,
    0xC11E2CD82D1F9FBD7E4D6EE9581FF3BDL,
    0x1DF4C637D625313720F45706A48FF20FL,
    0x3122EF3A001AAECDB8DD9D843C029E06L,
    0xADB778A0F729293E7E0B19B96A4C5A61L,
    0x938C747C6A051B3E163EB802A325148EL,
    0x38543C5E820DD9403B57BEFF6020596DL]
print 'Can you turn me back to python ? ...'
flag = raw_input('well as you wish.. what is the flag: ')
if len(flag) > 69:
    print 'nice try'
    exit()
if len(flag) % 5 != 0:
    print 'nice try'
    exit()
for i in range(0, len(flag), 5):
    s = flag[i:i + 5]
    if int('0x' + md5.new(s).hexdigest(), 16) != md5s[i / 5]:
        print 'nice try'
        exit()
        continue
print 'Congratz now you have the flag'

:laughing: flagis a string with length = 65
Using Online MD5 reverse page for md5s[0] I got ALEXC
:haha: Remember flag format is ALEXCTF{[A-Za-z0-9_]*}
So, format of 5 character next is TF{xx
I brutefore using sample_string = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789_"
and got TF{dv
Try with 5 character last xxxx} with sample_string
I got 6v3k}
I see dv and 6v3k is lowwercase character. So maybe all character is lowercase :smirk:
with 10 elements of md5s not use yet I use new sample_string="abcdefghijklmnopqrstuvwxyz0123456789_"
And here is my code

import md5

sgen = "abcdefghijklmnopqrstuvwxyz0123456789_"
md5s = [
    #0x831DAA3C843BA8B087C895F0ED305CE7L,
    #0x6722F7A07246C6AF20662B855846C2C8L,
    0x5F04850FEC81A27AB5FC98BEFA4EB40CL,
    0xECF8DCAC7503E63A6A3667C5FB94F610L,
    0xC0FD15AE2C3931BC1E140523AE934722L,
    0x569F606FD6DA5D612F10CFB95C0BDE6DL,
    0x68CB5A1CF54C078BF0E7E89584C1A4EL,
    0xC11E2CD82D1F9FBD7E4D6EE9581FF3BDL,
    0x1DF4C637D625313720F45706A48FF20FL,
    0x3122EF3A001AAECDB8DD9D843C029E06L,
    0xADB778A0F729293E7E0B19B96A4C5A61L,
    0x938C747C6A051B3E163EB802A325148EL]
    #0x38543C5E820DD9403B57BEFF6020596DL]

for i1 in sgen:
    for i2 in sgen:
        for i3 in sgen:
            for i4 in sgen:
                for i5 in sgen:
                    s = i1 + i2 + i3 + i4 + i5
                    hash_num = int('0x' + md5.new(s).hexdigest(), 16)
                    for i in range(0, len(md5s)):
                        if  hash_num == md5s[i]:
                            print(s + ":" + str(i))

:wait: And :tada:
I got the flag :joy:
Flag is ALEXCTF{dv5d4s2vj8nk43s8d8l6m1n5l67ds9v41n52nv37j481h3d28n4b6v3k}