Masrt's Blogs

Hacktober CTF

Sun, 18 Oct 2020 00:00:00 +0000

Shoeless Hellhole

Crypto

Points: 400

Solution

We were given a hex string and the clue a Shoeless Hellhole. I actually solved part of it, when I visited the ghosttown website for the first time!

1c0b4f9a4c18130dc631186fd7356bd1e7ca8f75623382bca6aaa60d9482785d

There was something regarding Shoeless Hellhole on a calculator

ghostown

A quick google search gives a blog about highschool students using the calculator to write some messages.

1337 DUh!

So SHOELESS HELLHOLE should become 54037355 43774073!! But it was clearly said in those blogs that the message is readable when the calculator is read upside down… So in order for the message to be read SHOELESS HELLHOLE when upside down, we must type 37047734 55373045 !!

See, it works!!

So, this must be key to the decrypting the hex-string! I tried xor and such at first but it didnt help.

Notice the length of the key here… its 16, so if we take that in hex we get a 8 byte key! and what takes a 8 bytes key?? DES

cyberchef soln just coz its easy here

So we get VQW6+F7 Kings Park, New York … using this as the flag doesnt work! The location points to a un-named building in the Kings Park. You can see more images in the maps but just clicking the image tab in the google search gives you the result!

google search

Its the same brown building, flag{Kings Park Psychiatric Center}, those authors really made us feel nuts!

spiderlabs-ctf

Sun, 09 Aug 2020 00:00:00 +0000

Software Defined Radio!

Introduction

This category makes use of skillset, reversing-engineering captured signals!

I made use of Inspectrum, as the challenge said to decode the signal, but it can also be done in many other ways!

Let’s check the first challenge as a tutorial for the last two!

You turn me on and off [150]

Description:

We captured alien radio data… Can you crack their message? sdr-challenge1.cfile

p.s. download inspectrum

So, after the hassle of setting up a new software and not messing up your system, open the challenge file with inspectrum!

$ inspectrum sdr-challenge1.cfile

Scroll the cursor, till you get to the captured packects!

Leave the settings as default, the main purpose here is to make the signal “noise free”, and clear, so we can adjust the settings to do that!

But as it seems the default settings suits the best!

Now, in the Time Selection Section, set Symbols to 25 or so and enable the cursors! And then move and resize the region that appears so that the smallest region aligns itself to the length of the smallest signal block!

Somewhat like this closely!

Then increase the no of Symbols and re-align them as per the blocks, do this till the entire signal data is covered with the cursors!

Now its easy peasy from here!

Right-click the signal-data and then do Add Derived Plot > Amplitude Plot!

A red horizontal line will appear!

Just drag the red line, over the top of the signal data till you see a nice amplitude graph, niceness depends upon your alignment

Now, you maybe we wondering why we are doing all this! See, our main motive is to extract the binary data encoded within the signal! the smallest signal block represents a 1 and a blank space represents a 0 … the initial bits go like this 1000110 which decode to ascii-F , get it?

Now, you are free to extract bits manually but this way is just better! Although I did extract it manually for the first two challs until I discovered this method! :(

So finally, right-click on the plot and do Extract Symbols > To stdout

Now open the terminal, and you will the extracted bits although not accurate its all that we need!

Now fire up python and enjoy!

>>> c='0.296352, -0.999998, -0.999996, -0.999996, 0.322055, 0.312156, -0.999999, -0.999981, 0.29718, 0.295499, -0.999999, 0.308109, 0.321133, -0.999996, -0.999997, -0.999993, 0.301978, 0.316911, -0.999998, -0.99999, -1, -1, 0.298925, -0.999997, 0.304455, 0.318662, -0.999992, -0.999991, 0.305682, 0.327311, 0.316653, -0.999996, -0.999993, 0.309524, -0.999999, 0.310872, 0.30265, -0.999999, 0.314892, -0.999997, 0.317225, 0.316971, -0.999999, -0.999998, 0.315871, 0.321062, -0.999992, -0.999999, -0.999999, 0.318565, 0.322979, -0.999996, -0.999996, -0.999985, -0.999992, -0.999998, 0.293918, 0.314347, -0.999996, -0.999999, 0.304823, 0.294362, -0.999994, -0.999998, -1, 0.284555, 0.315224, -0.999997, 0.280219, -0.999999, -0.999999, -0.999996, -1, 0.283892, 0.306363, -0.999995, 0.302248, 0.302009, 0.303467, -0.999994, 0.321145, 0.296669, -0.999994, -0.999993, 0.308316, 0.307696, -0.999999, -0.999985, 0.307899, 0.288028, -0.999993, -0.999998, -0.999999, -0.999994, 0.308951, -0.999984, -0.999998, 0.304631, 0.291852, -0.999991, -0.999983, 0.282598, 0.319025, -1, -0.999994, 0.306636, 0.296733, -0.999999, -0.999995, 0.295223, -0.999998, -0.999999, 0.298153, 0.309762, -0.999996, -0.999998, -0.999996, 0.292907, 0.305967, -0.999993, 0.295379, 0.291825, -0.999996, -0.999997, 0.283395, -0.999996, 0.297688, -0.999993, 0.305137, 0.303011, -0.999984, -0.999998, 0.288985, -0.999986, 0.289759, -0.999993, -0.999995, 0.283813, 0.293202, 0.293763, -0.999996, -0.999992, 0.295041, -0.999999, -0.999996, 0.289967, 0.282267, -0.999994, -0.999997, 0.28599, 0.312664, -0.999992, 0.29214, 0.300696, -0.999993, -0.999998, 0.301849, -0.999997, 0.307317, -0.999999, -0.999983, 0.295189, 0.298169, -0.999998, 0.303839, 0.293838, -0.999999, -0.999991, 0.297174, 0.281549, -0.999981, -0.999998, -0.999999, 0.304161, 0.29304, -0.999989, -0.999996, 0.284885, 0.293954, -0.999996, -0.999993, 0.29524, 0.282804, -0.999993, -1, 0.29767, 0.284739, -1, -0.999984, 0.291954, -1, -0.999985, 0.295192, 0.288656, -0.999986, -1, -1, 0.277962, 0.286773, -0.999995, -0.999995, 0.29657, 0.281652, 0.284367, -1, -0.999999, 0.272971, -1, -0.999996, 0.301735, 0.276792, -1, -0.999993, 0.271849, 0.27387, -1, 0.273795, 0.260515, -0.999998, -1, 0.276843, -0.999998, 0.261844, -0.999998, 0.270989, 0.267367, -0.999998, -1, 0.276232, -1, 0.268199, -0.999999, -0.999993, 0.273989, 0.279627, -0.999999, 0.268841, 0.2762, -0.999998, -0.999994, -0.999985, 0.286941, 0.270063, -1, 0.274396, -0.999992, -0.999997, -0.999984, -0.999993, 0.283191, 0.275578, 0.28392, -0.999996, -0.999996, -0.999994, -0.999995, 0.288057, 0.302771, -0.999999, -0.999991, -0.999997, 0.294791, 0.300922, -0.999997, -0.999992, 0.313433, 0.320796, -0.999997, -0.999998, -0.999999, 0.304164, -1, -0.999997, 0.314363, 0.319958, 0.315523, -0.999996, -0.999999, 0.295627, -0.999996, 0.329332, 0.318756, -1, -0.999998, 0.313605, -0.999997, -0.999995, -0.999997, 0.315632, 0.314954, -0.999988, -1, -1, 0.29864, 0.32669, 0.30328, -1, -0.999999, -0.999999, -0.999997'
>>> c=c.split(', ')
>>> b=''
>>> for i in c:
		b+=str(round(float(i))+1)
>>> b
'100011001101100011000010110011100101101011001100011000001100110001101000011011101100110011000010011001100110010011000110110010101100101001110010011001101100101001101100110001100110011001100100110001100111001001100110110010101100101001101100011010000111000011000110011000100111001011001000110001110000'

#you can decode this using cyberchef or modify it in python! prepend a zero to final result; 
#as first byte has only 7 bits! there are also so trailing zeros, which we can remove!

#or play with python
>>> bytes.fromhex(hex(int('0'+b[:-5],2))[2:])
b'Flag-f0f47fa32cee93e6c32c93ee648c19dc'

Now lets fly through the others shall we?

Let’s Get Scwhifty! [300]

Description:

The aliens are on to us… They’ve started wearing tin foil hats and switched things up with their radio transmission. Can you crack their latest message?

sdr-challenge2.cfile

authors have a phd in loopholes and blackholes

Things are a bit different here… you will notice that its not like the previous plot this time, but rather a continous plot!

Increase the FFT size and Zoom by 3 notches to see things more clearly!

Now lets experiment a different method here, since the blocks are a bit small to perfectly align with the cursors, go to Add Derived Plot > Add Frequency plot!

Then right-click the plot and do Add Derived Plot > Add Threshold plot!!

You will see something like this!

Now you can easily align the cursor with the help of this!

We dont need an amplitude graph now, just right-click the threshold graph and do Extract Symbols > To Stdout

This time the output on the terminal will be exact bits! Cheers!

>>> c='0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0'
>>> c=c.split(', ')
>>> b=''
>>> for i in c:
		b+=i
>>> b
'010001100110110001100001011001110010110101100101011001100011010101100001001100010011011001100100001100010110000100110000001100100110010101100011001100100011010101100110011000100011001000110100001100110110011000111001001100010110010001100101011001100011010001100110011000110011100100110011001101100010'

#some trailing bits again... cyberchef is better tbh, but lazy to do screencaps 

>>> bytes.fromhex(hex(int(b[:-4],2))[2:])
b'Flag-ef5a16d1a02ec25fb243f91def4fc936'

Don’t flip out, Thomas! [500]

Description:

The aliens’ capabilities have advanced and they’ve decided to improve the reliability of their transmissions. Can you crack their message?

sdr-challenge3.cfile

aliens are fools

This last one is extactly as the first chall with some changes!

First you will notice that the signal is almost double the lenth, hmm!

The graphs seems much nicer but the output is ridiculous! During the CTF, my output was near -1's and 0's but one half of the output this time was near 12's, not a problem per say but WEIRD!!

>>> c='-0.99997, 11.1106, 11.1694, -0.999992, -0.99992, 11.0514, -0.999968, 11.2205, -0.999963, 11.2827, 11.3478, -0.999706, 11.2109, -0.999982, -0.999999, 11.3926, -0.999948, 11.4896, 11.3279, -0.999991, 11.3649, -0.999993, -0.999934, 11.4388, 11.4549, -0.999987, 11.5114, -0.999974, -0.999995, 11.457, -0.999936, 11.4815, -0.999993, 11.4435, 11.5223, -0.999986, 11.5477, -0.999972, -0.999988, 11.3776, -0.999974, 11.5101, -0.99999, 11.3224, -0.999983, 11.3388, 11.4577, -0.999906, -0.999991, 11.4539, 11.3973, -0.999994, 11.3508, -0.999942, -0.999912, 11.4042, -0.999982, 11.4069, 11.126, -0.999917, 11.39, -0.99999, 11.402, -0.999971, -0.999905, 11.3848, -0.999994, 11.4153, 11.4869, -0.999947, -0.999994, 11.3809, 11.407, -0.999992, 11.5127, -0.999999, -1, 11.2166, 11.4048, -0.999997, -0.999987, 11.3736, -0.999984, 11.4122, 11.4557, -0.999969, 11.3333, -0.999862, -0.999958, 11.2766, -0.99996, 11.442, 11.3356, -0.999952, 11.4744, -0.999998, -0.999985, 11.3108, 11.2808, -0.99999, 11.4343, -0.999953, -0.999961, 11.5343, -0.99996, 11.2761, -0.999791, 11.4361, 11.462, -0.999798, 11.4301, -0.999996, -0.999568, 11.4521, -0.999577, 11.3659, 11.3863, -0.999829, 11.2894, -0.999966, -0.99977, 11.2104, -0.999741, 11.34, 10.9381, -0.999562, 10.8276, -0.999992, -0.998843, 11.4343, 11.1713, -0.998019, 11.2633, -0.999994, -0.996435, 11.5627, -0.996624, 12.0348, -0.99681, 11.3207, 12.5427, -0.999923, -0.99685, 12.7359, -0.995529, 12.6011, -0.991811, 11.4391, 11.8511, -0.981237, 11.3679, -0.999999, -0.974221, 11.5376, 8.69522, -0.97395, 7.13133, -0.999956, -0.966855, 4.86812, -0.933531, 11.2898, 2.72975, -0.776728, 1.46008, -0.99995, -0.407322, 0.277943, 0.105821, 11.248, -0.459846, -0.999998, 1.42713, 11.4811, -0.84345, -0.99995, 3.63274, 11.4124, -0.936897, 5.25712, -0.940375, -0.999943, 6.75393, -0.944075, 7.80711, -0.951536, 8.42594, 11.6765, -0.965547, 8.55706, -0.963816, -0.999956, 9.526, 11.5646, -0.98561, 9.87272, -0.991985, -0.999885, 10.5751, -0.997044, 10.509, 11.4934, -0.999509, 10.9338, -0.999938, -0.999993, 11.3851, -0.999757, 11.2488, 11.6994, -0.999778, 11.7194, -0.99954, -0.999955, 11.6482, -0.999371, 11.7608, -0.999642, 11.8469, 11.3558, -0.999358, 11.7235, -0.999738, -0.999985, 11.7062, -0.999893, 11.5553, 11.4792, -0.999949, 11.7136, -0.999989, 11.3926, -0.999937, -0.999962, 11.3578, -0.99995, 11.512, -0.999984, 11.2242, -0.999968, 11.3714, -0.999968, 11.1957, 11.3752, -0.999876, 11.3421, -0.999881, -0.999984, 11.4418, -0.999987, 11.3295, -0.999989, 11.3639, -0.999995, 11.1448, -1, 11.2525, -0.999947, 11.4662, 11.4968, -0.999956, 11.2253, -0.999972, 11.2004, -0.999967, -0.999922, 11.3457, -0.999953, 11.3532, -0.999999, 11.3621, -0.99998, 11.3495, -0.999971, 11.402, 11.4859, -0.999956, 11.28, -0.999987, -0.999999, 11.3755, 11.5724, -0.99999, -0.999987, 11.5551, 11.5318, -0.999985, -0.999998, 11.5093, -0.999975, 11.4682, 11.3485, -0.99994, 11.442, -0.999937, 11.4189, -0.999951, -0.999969, 11.4334, -0.999974, 11.3993, -0.999999, 11.4988, -0.999929, 11.5455, -0.999947, 11.467, 11.5224, -0.999968, 11.5196, -0.999897, 11.3603, -0.999869, -1, 11.476, -0.999996, 11.7765, -0.999954, 11.4227, -0.999948, 11.6779, 11.5175, -0.999988, 11.5862, -0.999993, -0.999983, 11.3233, -0.999983, 11.3699, -0.999747, 11.4706, 11.53, -0.999929, 11.501, -0.999981, -1, 11.4644, -0.999996, 11.4545, 11.5542, -0.999924, 11.5259, -0.999954, -0.999991, 11.4125, -0.99999, 11.4324, 11.4998, -0.999974, 11.3682, -1, -0.999994, 11.3739, -0.999999, 11.5215, 11.5162, -0.999993, 11.3841, -0.999984, -0.999991, 11.4099, 11.701, -0.999991, -0.999904, 11.417, 11.5382, -0.999938, -0.999831, 11.5091, -0.999947, 11.4452, 11.4956, -0.999997, 11.6209, -0.999965, -0.999908, 11.6406, 11.4881, -0.999985, -0.999937, 11.5793, 11.5048, -0.999999, -0.999977, 11.6667, -0.999864, 11.6195, 11.4243, -0.999984, 11.453, -0.99999, -0.999942, 11.9443, 11.7193, -0.99994, 11.5859, -0.999991, 11.8026, -0.99998, -0.999891, 11.5995, 11.9154, -0.999911, 11.8573, -1, -0.999973, 11.766, -0.999995, 11.8198, -0.999977, 11.8024, 11.7497, -0.999933, -0.999969, 11.9119, -0.999981, 11.8596, 11.8561, -0.999906, 11.7955, -0.999977, -0.999998, 11.8521, -0.999998, 12.1264, 12.0249, -0.999994, 11.9876, -0.999986, -0.999973, 12.0215, -0.999913, 12.0875, -0.999947, 12.1385, 11.8405, -0.999831, 11.9156, -0.999943, -0.999982, 11.965, -0.999893, 11.9324, 11.9148, -0.99995, 11.8908, -0.999996, -0.99998, 11.9732, -0.999998, 12.0759, 11.8144, -0.999946, 11.9711, -0.999971, -0.999971, 11.8938, -0.99994, 12.0442, -0.999996, 11.9205, 12.0126, -0.999986, -0.999998, 12.1517, 11.9, -0.999919, 11.8862, -0.999991, -0.99992, 11.889, -0.999941, 11.9077, 12.0352, -0.999969, -0.999997, 11.861, -0.999978, 11.9759, -0.999988, 11.9281, -0.999982, 11.869, 11.9888, -0.99996, 11.9993, -0.999988, -0.999932, 11.9825, -0.999939, 12.0927, -0.999998, 12.143, -0.999904, 11.9318, -0.999978, 12.001, -0.999992, 12.2496, 12.12, -0.999982, 12.2498, -0.999989, -0.999907, 12.0789, -0.999907, 12.1036, 12.1294, -0.999961, 12.188, -0.999997, -0.999947, 12.0596, -0.999857, 12.0133, 12.2547, -0.999989, 12.09, -0.999975, 11.9732, -0.99996, -0.999942, 12.1539, -0.999841, 12.1453, 11.9095, -0.999982, -0.999971, 12.2394, -0.999996, 12.2216, 11.9789, -0.999989, 12.1954, -0.999992, -0.999994, 12.1189, -0.99988, 12.271, 12.1448, -0.999989, -0.999974, 12.1851, -0.99999, 12.2271, -0.999991, 12.2379, 12.1345, -0.999996, 12.2838, -0.999985, -0.999987, 12.4144, 12.3354, -0.999957, 12.3025, -0.999971, 12.0337, -0.999995, -0.999999, 12.4081, -0.99999, 12.3376, 12.1878, -0.999874, 12.175, -0.999995, -0.999967, 12.3091, 12.3558, -0.999985, -0.999974, 12.3708, -0.999985, 12.1641, -0.999961, 12.3227, -0.999963, 12.4121, 12.2726, -0.999891, 12.4122, -0.999961, -0.999889, 12.2769, 12.2579, -0.99996, -0.999738, 12.4337, -0.999352, 12.1021, 12.2841, -0.99999, -0.999939, -0.999931, -0.999946, -0.999954, -0.999984, -0.999998, -0.999911'

>>> c=c.split(', ')
>>> b=''
>>> for i in c:
		a=round(float(i))
		if a==-1:
			b+='0'
		else:
			b+='1'

>>> b
'0011010100101101001011001011010100101101010100110010110100101011010011001011001101001011010010110010110101001011010010110100101100101101010011010100101100101101001011010011001100101101010010110010110100101101001011010100101101001010110101010100101101010101010010101101010101001011001100110100101011010101010010101101010100101101010010110100101101001011010010110011001101001011001100110100101100101011001011010100110100101101001011010100101101001011010010110101001100101101001101010100101101010101010010110100101101001010110100110100101101001101010010110010101101001011001101010100101100110100'
>>>len(b)  #removing extra bits
592

Now you will notice that this bit-stream is exactly twice the length of bit-stream of the FLAG which is 37 bytes, 37*8*2= 592!

So, the bits must be encoded somehow! Its not that hard! By observing the bits of the first-byte of the flag, F i.e, 01000110 we can deduce that:

1 --> 10,11
0 --> 00,01

Let’s end this shall we?

>>> flag=''
>>> for i in range(0,592,2):
	m=b[i:i+2]
	if m=='01' or m=='00':
		flag+='0'
	else:
		flag+='1'

		
>>> bytes.fromhex(hex(int(flag,2))[2:])
b'Flag-3c3b6ecfc808588c3557bf31d0392744'

Thanks for reading!!

Related post and a pdf that could help!!

Inspectrum_Reverse_with_DSpectrumGUI

opensource-sdr-movement

Castors-CTF

Sun, 31 May 2020 00:00:00 +0000

Arithmetic

Solution

We needed to solve simple arithmetic operations but we have little time and they get troll-some! Sounds like a challenge description heh?

Use Sympy, its the best for these types of challenges!

Here my code:

from pwn import *
import re
from sympy import solve,Symbol,sympify

r=remote('chals20.cybercastors.com',14429)

print(r.recv())
r.sendline('')

mappings={'plus':'+','divided-by':'//','minus':'-','multiplied-by':'*','one':'1','two':'2','three':'3','four':'4','five':'5','six':'6','seven':'7','eight':'8','nine':'9','zero':'0'}
cn=0
while True:
	data=r.recv().decode()
	if 'castors' in data:
		print(data)
		break
	data=data.split(' ')
	print(data)
	for i in range(len(data)):
		if data[i] in mappings:
			data[i]=mappings[data[i]]
	data=' '.join(data)
	#print(data)
	num=re.findall(r'\d+',data)
	eqn=data[data.index(num[0]):data.index('?')]

	eq='x='+eqn
	#print(eqn)
	sympy_eq = sympify("Eq(" + eq.replace("=", ",") + ")")
	ans=solve(sympy_eq)
	print(ans)
	r.sendline(str(ans[0]))
	r.recvuntil('\n')

castorsCTF(n00b_pyth0n_4r17hm3t1c5}, pretty nooby!

Jigglypuff’s Song

crypto

Point: 479

Description:

Can you hear Jigglypuff’s song?

Solution:

This was a really easy chall… but the random pixel stream on the image kinda confused me! Half of the time, I just kept staring at the image and did nothing!

All we had to do was a basic steganography attack! Using LSB to conceal data in images doesn’t alter the image much, but MSB does a lot. As was the case here! Once that clicked, it was as easy as a sunny day!

from PIL import Image

im=Image.open('chal.png')
pixels=im.load()

#print(im.size,im.mode)

search=[]

s=''
for i in range(im.height):
	for j in range(im.width):
		a=pixels[j,i]
		r=bin(a[0])[2:].zfill(8)[0]
		g=bin(a[1])[2:].zfill(8)[0]
		b=bin(a[2])[2:].zfill(8)[0]
		s+=r+g+b

print(bytes.fromhex(hex(int(s,2))[2:]))

castorsCTF{r1ck_r0ll_w1ll_n3v3r_d3s3rt_y0uuuu} ::The Damn Author trolls you twice… once in the trailing bits part and the entire concealed message also contains Rick’s song twice!

Bagel Bytes

crypto

Point: 444

Description:

nc chals20.cybercastors.com 14420

Solution:

It was a imple ECB-oracle attack nothing much! All you had to do was notice that!

from pwn import *
import binascii

r=remote('chals20.cybercastors.com',14420)



def get_data(payload):
	r.recvuntil('choice: ')
	r.sendline('2')
	r.recvuntil('> ')
	r.sendline(payload)
	r.recvuntil(':\n')

	data=r.recvuntil('\n')[:-1]

	return data

flag=''
for i in reversed(range(0,48-len(flag))):
	payload='A'*i
	data=get_data(payload)

	print(len(data),i)
	for i in range(0,len(data),32):
		print(data[i:i+32])

	chars='castorsCTF}{I_L1k3_mua_1203456789abcdefghijklmnopqrstuvxyz?ABCDEFGHIJKLMNOPQRSTUVWXYZ'
	for j in chars:
		print('checking',j)
		check=payload+flag+j
		reply=get_data(check)
		if reply[:96]==data[:96]:
			flag+=j
			break

	print(flag)

castorsCTF{I_L1k3_muh_b4G3l5_3x7r4_cr15pY} -i- eazy-peasy!

Amazon

crypto

Points: 489

Description:

Are you watching the new series on Amazon?

Solution:

The hint was infront of us for this one! Amazon–> Amazon Prime –> Primes! Ascii representation of each character was multiplied by consecutive prime numbers!

a=['2', '3', '5', '7', '11', '13', '17', '19', '23', '29', '31', '37', '41', '43', '47', '53', '59', '61', '67', '71', '73', '79', '83', '89', '97', '101', '103', '107', '109', '113', '127', '131', '137', '139', '149', '151', '157', '163', '167', '173', '179', '181', '191', '193', '197', '199', '211', '223', '227', '229']
b=['198', '291', '575', '812', '1221', '1482', '1955', '1273', '1932', '2030', '3813', '2886', '1968', '4085', '3243', '5830', '5900', '5795', '5628', '3408', '7300', '4108', '10043', '8455', '6790', '4848', '11742', '10165', '8284', '5424', '14986', '6681', '13015', '10147', '7897', '14345', '13816', '8313', '18370', '8304', '19690', '22625']

for i in range(len(b)):
	print(chr(int(b[i])//int(a[i])),end='')

castorsCTF{N0_End_T0d4y_F0r_L0v3_I5_X3n0n}, how did the author expect us to find this before space were given in the ciphertext!

0x101 Dalmations

crypto

Points: 496

Description:

Response to Amazon: Nah, I only need to be able to watch 101 Dalmatians.

Solution:

It was exact the Amazon chall with one extra step! You couldn’t have possibly solved it without solving Amazon!

All we have to do was take the primes multiply them by the ascii representation of the chars, and take its modulus with 0x101!

(char*prime)%0x101

It just bruteforce to reverse it! Could have used modular inverses but eh!

p=['2', '3', '5', '7', '11', '13', '17', '19', '23', '29', '31', '37', '41', '43', '47', '53', '59', '61', '67', '71', '73', '79', '83', '89', '97', '101', '103', '107', '109', '113', '127', '131', '137', '139', '149', '151', '157', '163', '167', '173', '179', '181', '191', '193', '197', '199', '211', '223', '227', '229','239', '241' ,'251' ,'257', '263', '269','271' ,'277' ,'281' ,'283' ,'293' ,'307' ,'311','313','317']

a=['c6', '22', '3d', '29', 'c1', 'c5', '9c', 'f5', '85', 'e7', 'd7', '0e', '46', 'e6', '21', 'e7', 'dd', '8d', 'db', '43', 'a0', '34', '77', '04', '7f', '32', '13', '8c', 'c9', '01', '65', '78', '5f', 'c0', '14', '8e', '33', 'bf', 'bc', '02', '21', '79', 'e1', '5d', 'd3', '46', 'e0', 'ca', 'ee', '72', 'c2', '26', '38']
m=257
flag=''
for i in range(len(a)):
	for j in range(30,150):
		if (j*int(p[i]))%m==int(a[i],16):
			#print(j)
			#print(chr(j))
			flag+=chr(j)
	#print('\nlolol')

print(flag)

castorsCTF{1f_y0u_g07_th1s_w1th0u7_4ny_h1n7s_r3sp3c7}. I needed a lot of hints, so no respect for me I guess!

Noob_CTF (Pre)

Sat, 16 May 2020 00:00:00 +0000

Fake IP

Web

Points: 100

Description:

Elliot found that someone might have messed up with challenge creation. Check, if they provided the right IP for this challenge.

site

Solution:

The link takes you a site… let’s check the source code.

On the last line we have this assets/js/main.js source link! Clicking it and reaching the end of that page, we see a comment!

++++++++++[>+>+++>+++++++>++++++++++<<<<-]>>>>++.-----.++++++++++.------.<<++.>>+.++++++.-----------.++++++.++++++++++++.<<.>>------------------.+++++++++++++++++.-------------.<<.>>++++++++++.+++++++.-----------------.+++++++++++++.+++.---------.-------.-.<<.>>++++++++++.+.++++++++.<<++++++++++++++.--------------.>+++++++++++++++++.++++++++++++++.+++++++.---------.>--------.--.<++.<.>>+++++++.-----.<<.>>+++++.<+++.---.<.>>+++.--------.+++.------.<-.<.>>+++.<++.<.>.-----.>----.<++++.<.>----------------------------.+++++++.<+++++++.>>++++++++.<<+++++++.--------------.>>-----------.++++++++++++..----.+++.<----------------------.-----------..>----------------.+++++.-------.+++++++++++..<-.>++.+..-------------.-.+++++++++++++++++.-----.++++++++++++.<.>-----.---------------.--.+++++.<+.++.-.++.--.+++.---.-.

Now this is clearly Brainfuck language. Decoding it in tio.run we get,

fake flags are overuled now. Welcome to the world of fake IP's. https://chall.noobarmy.tech/102030/

Great! So we now have the correct address! Going to this site and veiwing the source code we see three comments as:

part1 - 5BjMEvLjTs
part 2- e2Y0a2VfZmw0Z3NfYXIz
part 3- ?ZRo.D)#R3CboF.Cc$3SI/

These seem like base64-ish encoded stuff… At first I concatenated them and used Base85 etc but it wasn’t. Decode them seperately!

Part1 is Base58, 2 is Base64 and the 3rd part is Baase85 encoded. Decoding them we get the final flag as noobctf{f4ke_fl4gs_ar3_s0_m1ddl3_cl4ss}

Bottled Something(?)

Misc

Points: 300

Description:

Mr Holmes had a suspicion on Moriarty for the creation and spread of COVID-19, So when he started inspecting Moriarty’s mails, He found that Moriarty sent this txt file to one of his closest friend. Maybe there’s a vaccine in this thing? Take a look.

Sometimes opening files in browser helps. ;)

Hint:There’s a link on this page that’ll be helpful.

Page

Flag format: noobctf{flag_here}

Solution:

Get the file and opening it you will see a large base64 encoded string… Now we don’t exactly need to visit that page but a hint is what we get. In the last section of that page we see a challenge were a zipfile was created from as base64 encoded string.

Don’t use that website linked use this one-liner to do it!

cat Study_Encodings_dude.txt | base64 -d > abc.zip

Unzip the file and you will see loads of PNG images! At first I tried loads of other things with these but then I tried the other hint of opening them in the browser!

Now as you keep viewing each of them, you will come across 3 such images which are not infact PNGs but ‘Animated PNGs’… namely,

220x-sherlockwiki_studypink.png
239x-name-sherlock.png
300x-staricase.png

These “Animated PNGs” have a frame or 2 in which a QR CODE shows up. Scan them by whatever means you can. I myself wasted few screenshots to capture the frame and it was done that way!

noobctf{Ah_Animated_PNGs_are_kinda_fun} is the vaccine?!

An0nym0us G1rl_fri3nd

Forensics

Points: 200

Description:

r3curs1v3_pr0xy friends believes that his girlfriend is cheating on him.Somehow he managed to get a document from his girlfriend computer. But he found that it is encrypted. So he asked me to decrypt the file and get the data for him.Please help me to get the document decrypted.

Hint:Common sense may help you. Hint:Search for some common password on internet. It may help you somewhere.

Solution:

This was actually fairly easy! The docx file given is encrypted. We use,

office2john.pl Office.docx > hash.txt

To get the hash of the file. Now using simply john hash.txt gives us the passwor as Disney. This CTF had easy passwords. Now here where I hit the road block… Its kinda funny, I use Google Docx and My MS OFFICE is not activated for fuck’s sake they don’t let you copy the contents of the file. I used abiword simple but fine!

So yeah… open the file and you willa Love Story and lots of spaces and tabs in the lower sections. Hmm… must be stegsnow. Copy the contents of the docx to a text file and try stegsnow with -C, it doesn’t work!

A while later the second hint was released i.e, to use common passwords. So use password, password123 etc and you will find 12345678 works. p.s. I bruteforced it!

stegsnow -p 12345678 -C snow.txt

noob{r3curs1v3_pr0xy_adm1r3_y0ur_kn0wl3dg3} Thanks!

Television Transmission

Forensics

Points: 300

Description:

Google “Slow Scan Television Transmission”. You will learn something new for sure.

Hint:Do you know about LSB? Remember don’t get trolled :)

Solution:

Another wierd one huh? Mr. Proxy! Well, as the title and everything says we ought to use SSTV for this. Going that route we get trolled by a smiley face laughing at us… so what to do?

The hint says lsb… set lets look for a lsb-tool on .wav files. We find stegolsb!

Using this tool was bit of a hassle but I got it!

stegolsb wavsteg -r -i audio.wav -o output.txt -n 1 -b 100

The wavsteg is used for .wav file steganography! Running it you get the flag, noob{lsb_c4n_h4v3_s0m3_ju1cy_d4t4}. I leant…

Web Steg

Misc

Points: 300

Description:

Combination of web and stegnography is amazing.You have to think out of the box.

link

Hint:Check the deliver to page.

Solution:

This chall had the worst name ever, as discussed with the author! We are given a link to site whose source code doesn’t gives much!

Initially I thought I had to do something with those the various image link available but it was a dead-end. Then one of the .css files had a image link but it was useless as well.

A lot dicussion with the author later and slamming my head a lot, the dform.html link had something to do with it!

As you can see the are a no. of whitespaces or tabspaces even after where the text ends. This means only one thing… stegsnow. You seem to like it a lot author!

copy the text in the source-code of dform.html and put it into a text file! Use,

stegsnow -C troll

noob{w3b_and_f0r3ns1c_c0mb1n4t10n}. I have my doubts 400%!

Deoxyribonucleic

Crypto

Points: 300

Description:

I always wonder how we match to our ancestor?

Not that E4sy! Think out of the BOX :)

Hint:Heard About CTF KATANA???

Solution:

Doing this just bcoz I am crypto guy! I knew about this but thought about using the DNA-codons approach… then they released the hint to CTF_Katana and I was sure!

The thing is we have this binary to DNA code,

A-00
T-11
C-10
G-01

So we need to decode the binary strings given in the text file as per this intruction!

data=<REDACTED>
data=data.split(' ')

code={'00':'A','11':'T','10':'C','01':'G'}

#converting the binary string to DNA code!
for i in data:
	c=code[i[:2]]+code[i[2:4]]+code[i[4:]]
	data[data.index(i)]=c

#print(' '.join(data))

dna_alpha={}
alphabet='abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890 .'
#print(len(alphabet))

#translating the DNA code to english alphabet
for i in range(0,64):
	bi=bin(i)[2:].zfill(6)
	c=code[bi[:2][::-1]]+code[bi[2:4][::-1]]+code[bi[4:][::-1]]  
	#[::-1] is important if you use the conversion from binary approach!
	dna_alpha[c]=alphabet[i]

#print(dna_alpha)

plain=''
for i in data:
	plain+=dna_alpha[i]

print(plain)

noob{n00b_4rmy_601n6_70_r0ck_0n3_d4y}… solved this in the penultimate minute LMAO!

zh3r0-CTF (Pre-CTF)

Thu, 07 May 2020 00:00:00 +0000

is this the robo loanguage?

Crypto

Points:495

Description:

I have found some thing can you tell me what’s inside it.

Solution:

We are give a file which contains this:

--.- .-- -..- ... -.-- ... -... -- .- .-- ..... .-.. .. --. .-- -. .- --. ..-. .--. .. .... -. .--- -... ...-- -... .-.. -.-. -. .-. ...-

3,4,8,9,12,15,16,17,20,24,25,28,29,30,32

Decoding the morse code gives us a base64-like string!

QWXSYSBMAW5LIGWNAGFPIHNJB3BLCNRV

But decoding this, gives us junk! Now if we take the numbers given and play with the the character in those index we might get something!

We actually had to turn the letters at those indices to lowercase

a='QWXSYSBMAW5LIGWNAGFPIHNJB3BLCNRV'
b=[3,4,8,9,12,15,16,17,20,24,25,28,29,30,32]
c=[a[i].lower() if i+1 in b else a[i] for i in range(len(a))]
print(''.join(c))

We now have the correct base64… decoding it gives this Alla fine l'hai scoperto, which seems like another language but lets try translating it!

yoo! its translates to In the end you found out

Hence the flag is zh3r0{In_the_end_you_found_out}

AES

Crypto

Points: 567

Description:

You people kept asking about AES. So here it is

0d152c0ffefd78da21f04c769546b29b

enc(Key)=”dnmjrqsqfsfnfqgt”

Hint: no 13 may help you

Solution:

I’m mad!!! was sleeping after having enough of brainfuck in general, when they released the hint for this one!

Anyway, it’s AES decryption using ECB mode since we don’t have any IV here. As the challenge depicts the key in encoded! … well, 13! ROT13!

so ROT13(key)==qazwedfdsfsasdtg … and you don’t need my help but still!

from Crypto.Cipher import AES

c='0d152c0ffefd78da21f04c769546b29b'
Key=b"qazwedfdsfsasdtg"

cipher=AES.new(Key,AES.MODE_ECB)
a=cipher.decrypt(bytes.fromhex(c))
print(a)

You will get zh3r0{AES_1s_C00000l} as flag!

Ultra Strong RSA

Crypto

Points: 920

Description:

A simple RSA maybe be easy to crack, what if we use multilayered encryption, where we encrypt the flag in multiple methods. To crack an advanced RSA Script, study the script well before even trying to crack it.

Not always tools help.

HAPPY HACKING :)

Solution:

We are given a python script and a textfile…which has RSA encrypted text in it!

from base64 import b64encode as b64e
from Crypto.Util.number import *
from Crypto.PublicKey import RSA
from flag import flag, hint

pt=bytes_to_long(flag)
ht=bytes_to_long(hint)
p=getPrime(2048)
q=getPrime(2048)
r=getPrime(2048)
s=getPrime(2048)
x=getPrime(1024)
y=getPrime(1024)
e=65537
if x>y:
	key=x-y
else:
	key=y-x

key=b64e(str(key))

n=x*y
n1=p*q
n2=r*s
n3=n1*n2

if p>q:
	a=p-q
else:
	a=q-p
b=r+q
c=p+r

enh=pow(ht,e,n)
ct=pow(pt,e,n3)

opt="N : "+str(n)+"\ne : "+str(e)+"\nenc(Hint) : "+str(enh)+"\n\nN1 : "+str(n1)+"\nN2 : "+str(n2)+"\nN3 : "+str(n3)+"\nCipher Text : "+str(ct)+"\n\n\nkey : "+str(key)+"\n"
pvt="a = "+str(a)+"\nb = "+str(b)+"\nc = "+str(c)+"\n"
test="p="+str(p)+"\nq="+str(q)+"\nr="+str(r)+"\ns="+str(s)+"\nx="+str(x)+"\ny="+str(y)+"\n"
#print(opt,pvt)


ob=open("out.txt","w")
ob.write(opt)
ob.close()
ob=open("hint.txt","w")
ob.write(pvt)
ob.close()
ob=open("test.txt","w")
ob.write(test)
ob.close()

At first the hint which is important is encrypted with (e,n). Info about n is in key which is x-yory-x! Simple qudractic here… x**2+key*x-n=0 will give us x,y!

from Crypto.Util.number import inverse,long_to_bytes
from sympy import solve,Symbol

hint_cp=<REDACTED>
N1=<REDACTED>
N2=<REDACTED>
N3=<REDACTED>
N=<REDACTED>
KEY=<REDACTED>
C=<REDACTED>
e=65537

y=Symbol('y')
Y=solve(y**2+KEY*y-N,'y')[1]
X=N//Y

tot_h=(X-1)*(Y-1)
d_h=inverse(e,tot_h)

hint=long_to_bytes(pow(hint_cp,int(d_h),N))
#print(hint)
#hint:https://pastebin.com/Ss2RBhVN

We get a link, and the data for the next part which is similar to the first one! Quadratics here too!

a = <REDACTED>
b = <REDACTED>
c = <REDACTED>
#a=q-p,b=r+q,c=r+p

q=Symbol('q')
P=solve(q**2+a*q-N1,'q')[1]

Q=a+P
R=c-P

S=N2//R

tot=(P-1)*(Q-1)*(R-1)*(S-1)

d=inverse(e,tot)

m=pow(C,int(d),N3)

print(long_to_bytes(m))

Finally you get zh3r0{Y0u_h4v3_b34ten_RSA} !!!

BaseXOR

Crypto

Points: 481

Description:

The author seemed to find a strange string can u help him in decoding them

FQ0ACgAUVgcCSTAdAwpvGA0AFm8EVkonAVo6BkhCWBg=

Solution:

Decode the base64, you get this!

b"\x15\r\x00\n\x00\x14V\x07\x02I0\x1d\x03\no\x18\r\x00\x16o\x04VJ'\x01Z:\x06HBX\x18"

Now lets XOR the first few bytes with the flag-format zh3r0

We get oe3x0… this might possible be the key we need! Let’s XOR it with the whole thing!

Yoo! We get zh3r0{34zy_x0r_wh3n_k3y_15_50r7}!! Our flag!

Are you the Master? 2

Master

Points: 983

Description:

Enter key as the flag. (part1) folder in drive may help you

Solution:

We get another RSA encrypted file with the source code! It’s similar to the Ultra RSA chall but yet different! See here, we have a maths problem which can mislead us to may places!

So, the important info given is:

N1=p*q N2=r*s A=p+q+r+s B=(p-q)-(r-s)

Now after sneding few hours trying bi-quadractics, getting iotas and skimming through my old high-school notes, I got it!

See, this way we get a quadratic in sq which fun fact… also gives pr as its other root! So, when we get pr and sq, we can perform GCD on it with N1 and N2 and obtain p,q,r and s!!

from sympy import Symbol,solve
from Crypto.Util.number import *

N1=<REDACTED>
N2=<REDACTED>
E =65537
ct1 =<REDACTED>
ct2 =<REDACTED>

#p+q+r+s
A=<REDACTED>

#(p-q)-(r-s)
B=<REDACTED>

#t1t2=A+B+2-C

x_coeff=(A**2-B**2)//4-(N1+N2)


x=Symbol('x')
sol=solve(x**2-x_coeff*x+N1*N2,'x')

X,Y=sol[0],sol[1]

p=GCD(X,A)
q=GCD(Y,A)

tot1=(p-1)*(q-1)
d1=inverse(E,tot1)

m=pow(ct1,int(d1),A)
print(long_to_bytes(m))

zh3r0{Th1s_1s_4_K3y_gfh} the flag or a key?

Are you the Master? 3

Crypto

Points: 1199

Description:

You must me the master to get till here.

now give me the flag.

drive part 2 folder.

Solution:

In the drive folder we get a binary name iv and a img.txt. Now iv can only mean the initialization vector of an AES encrypted data!

The flag from Master_3 says its a key… so maybe thats all we need! Running the IV tho… prints the message It’s a string, so maybe we have to do strings on it to get the iv!

The H character her seems unecessary so let’s remove it! We get Here is you IV: ryq9rkoU+XhtIuoFAvujDQ==

Its seems we are ready to decrypt!!!

from Crypto.Cipher import AES
from base64 import b64decode

iv=b64decode('ryq9rkoU+XhtIuoFAvujDQ==')
key='zh3r0{Th1s_1s_4_K3y_gfh}'

dat=b64decode(open('img.txt','rb').read())

cipher=AES.new(key,AES.MODE_CBC,iv)

plaintext=b64decode(cipher.decrypt(dat))
#print(plaintext)
f=open('flag.png','wb').write(plaintext)

In the plaintext, we see corrupted PNG headers and also other corruptions. Lets fix those with ghex!

Now we see the flag on opneing the file!

zh3r0{You_are_master_Of_All_Arts} Cheers!

Houseplant CTF

Mon, 27 Apr 2020 00:00:00 +0000

CH3COOH

Crypto

Points:50

Description:

None

Hint: Short keys are never a good thing in cryptography.

Solution:

We are given a large passage of encrypted ciphertext. Hmm, what could it be… The title says CH3COOH… which is the chemical formula for Vinegar. Vinegar, Vinegar… Vigenere. Its a vignere cipher.

Using my self written code from my D3crypt0r, after saving the text to a file!

Vigenere cipher without a key is done by frequnecy analysis of letters in an alphabet… doing so we get the key tony… Now onto the decryption!

Simple, right! rtcp{vinegar_on_my_fish_and_chips_please} was the flag!

“fences are cool unless they’re taller than you” - tida

Crypto

Points:50

Description:

They say life’s a roller coaster, but to me, it’s just jumping over fences.

Solution:

We are given a ciphertext with english-letter alphabet, brakets and numbers in any jumbled order. Well, any hint… the title is suffice. fences here denote the railfence cipher.

I solved this too using my D3crypt0r… plz do check it out!

You can see clearly that solution 2; which means using 3 rails, we get a close enough answer that makes sense.

rtcp{ask_tida_about_rice_washing} is the flag!

Returning Stolen Archives

Crypto

Points:50

Description:

So I was trying to return the stolen archives securely, but it seems that I had to return them one at a time, and now it seems the thieves stole them back! Can you help recover them once and for all? It seems they had to steal them one at a time…

Hint: Well you sure as hell ain’t going to solve this one through factorization.

Solution:

We are given a python file which is probably… how the data in intercepted.txt is encrypted. The title gives a clear hint that R S A is used!

p = [redacted]
q = [redacted]
e = 65537
flag = "[redacted]"

def encrypt(n, e, plaintext):
  print("encrypting with " + str(n) + str(e))
  encrypted = []
  for char in plaintext:
    cipher = (ord(char) ** int(e)) % int(n)
    encrypted.append(cipher)
  return(encrypted)

n = p * q
ct = encrypt(n, e, flag)
print(ct)

As per the encrypting script… RSA encryption has been performed on each character of the plaintext. No other information on bitsize of the primes or n in general.

Hmm… looking at the intercepted.txt file… we see N is sufficiently large and there’s the hint that we can’t solve it through factorization! Hmm…

n=[redacted]
e=65537
ct=[redacted]

plaintext=''
for i in ct:
	for j in range(1,256):
		if pow(j,e,n)==i:
			plaintext+=chr(j)

print(plaintext)

Heh! All there was to do was take every ascii character and check whether it matches the corresponding ciphertext or not!

rtcp{cH4r_bY_Ch@R!} for peace-keeping sake!

Rivest Shamir Adleman

Crypto

Points: 338

Description:

A while back I wrote a Python implementation of RSA, but Python’s really slow at maths. Especially generating primes.

Hint: There are two possible ways to get the flag ;-)

Solution:

Another straight giveway… Rivest Shamir Aldeman is what RSA stands for #themakers! The chall.7z file given, had a lot more things than expected… easy to have your head spining!

Now encrypt.py and decrypt.py is where my eyes went first… coz ofc. Both files… didn’t give much, rather than getting the keys from from a file and performing encryption and decryption respectively. One key note is that a verification string was added ‘VERIFICATION-UpTheCuts-END\n’ , which could be important if we decrypt using the decrypt.py file.

OK, enough of this crap! The real deal is the generate_keys.py!!

First we see two primality test funcs… one a general one and other a Miller_Rabin Test! There are a few more funcs… but the real hint lies at line 115…

You see, while generating primes… this func generates two primes but it changes the first prime to a value read from component.txt … if a certain primes.json doesn’t exists otherwise it reads the primes from it!… Now since, there isn’t any primes.json we may assume that this is indeed the case!

So, its goes easy from here… we get one prime from component.txt, the public key from public-key.json and the ciphertext from secrets.txt.enc and play around.

import json
from Crypto.Util.number import *

c=int(open('secrets.txt.enc').read()[2:],16)
key=json.loads(open('public-key.json').read())

p=int(open('component.txt').read())
q=key['n']//p

tot=(p-1)*(q-1)
d=inverse(key['e'],tot)

m=pow(c,d,key['n'])
print(long_to_bytes(m))

There you go; we get the nice message as,

rtcp{f1xed_pr*me-0r_low_e?} is the flag!

Now as the hint says there is surely another way to find this flag. The flag reads fixed prime or low e?… I am pretty sure my method summarizes the fixed prime one! I wasn’t able to look into the other because of the ongoing contest but do give it a try! I will too!

Rainbow Vomit

Crypto

Points:535

Description:

o.O What did YOU eat for lunch?!

Hint: Replace spaces in the flag with { or } depending on their respective places within the flag.

Hint: Hues of hex

Hint: This type of encoding was invented by Josh Cramer.

Solution:

A quick search of Josh Cramer & hues of hex as the hint says gave enough us information about his encoding scheme

He used pixels of differnt colours arranged in 2x3 rectangles to encode the letters and numbers

Now this basically tells us what to do… lets look for a image to text decoder… oh wait, no … NO… NO!!. There isn’t any or one that I could find. So yea… fold up your sleeves and get ready to code! coz the image provided to us, as much as we can zoom would be painful to do it manually!

At 2000% zoom in a linux machine… we see this,

The image is 104X34 in dimensions and we can tell that there’s a 2 pixel width of border… that would make our data 100x30 in dimensions, which fits as mulitples of our base rectangle dimensions!

from PIL import Image

'''
taking input along lines
'''

# striping surrounding white pixels

'''
im=Image.open('output.png')

pixels=im.load()

im1=Image.new('RGB',(100,30))
pixy=im1.load()

for i in range(2,32):
	for j in range(2,102):
		pixy[j-2,i-2]=pixels[j,i]

im1.save('better.png')
'''

colour={"B":(0,0,0),
"W":(255,255,255),
"G":(128,128,128),
"r":(255,0,0),
"g":(0,255,0),
"b":(0,0,255),
"y":(255,255,0),
"c":(0,255,255),
"m":(255,0,255)
}

a=[colour["m"],colour["r"],colour["g"],colour["y"],colour["b"],colour["c"]]
b=[colour["r"],colour["m"],colour["g"],colour["y"],colour["b"],colour["c"]]
c=[colour["r"],colour["g"],colour["m"],colour["y"],colour["b"],colour["c"]]
d=[colour["r"],colour["g"],colour["y"],colour["m"],colour["b"],colour["c"]]
e=[colour["r"],colour["g"],colour["y"],colour["b"],colour["m"],colour["c"]]
f=[colour["r"],colour["g"],colour["y"],colour["b"],colour["c"],colour["m"]]
g=[colour["g"],colour["r"],colour["y"],colour["b"],colour["c"],colour["m"]]
h=[colour["g"],colour["y"],colour["r"],colour["b"],colour["c"],colour["m"]]
i=[colour["g"],colour["y"],colour["b"],colour["r"],colour["c"],colour["m"]]
j=[colour["g"],colour["y"],colour["b"],colour["c"],colour["r"],colour["m"]]
k=[colour["g"],colour["y"],colour["b"],colour["c"],colour["m"],colour["r"]]
l=[colour["y"],colour["g"],colour["b"],colour["c"],colour["m"],colour["r"]]
m=[colour["y"],colour["b"],colour["g"],colour["c"],colour["m"],colour["r"]]
n=[colour["y"],colour["b"],colour["c"],colour["g"],colour["m"],colour["r"]]
o=[colour["y"],colour["b"],colour["c"],colour["m"],colour["g"],colour["r"]]
p=[colour["y"],colour["b"],colour["c"],colour["m"],colour["r"],colour["g"]]
q=[colour["b"],colour["y"],colour["c"],colour["m"],colour["r"],colour["g"]]
r=[colour["b"],colour["c"],colour["y"],colour["m"],colour["r"],colour["g"]]
s=[colour["b"],colour["c"],colour["m"],colour["y"],colour["r"],colour["g"]]
t=[colour["b"],colour["c"],colour["m"],colour["r"],colour["y"],colour["g"]]
u=[colour["b"],colour["c"],colour["m"],colour["r"],colour["g"],colour["y"]]
v=[colour["c"],colour["b"],colour["m"],colour["r"],colour["g"],colour["y"]]
w=[colour["c"],colour["m"],colour["b"],colour["r"],colour["g"],colour["y"]]
x=[colour["c"],colour["m"],colour["r"],colour["b"],colour["g"],colour["y"]]
y=[colour["c"],colour["m"],colour["r"],colour["g"],colour["b"],colour["y"]]
z=[colour["c"],colour["m"],colour["r"],colour["g"],colour["y"],colour["b"]]

A0=[colour["B"],colour["G"],colour["W"],colour["B"],colour["G"],colour["W"]]
A1=[colour["G"],colour["B"],colour["W"],colour["B"],colour["G"],colour["W"]]
A2=[colour["G"],colour["W"],colour["B"],colour["B"],colour["G"],colour["W"]]
A3=[colour["G"],colour["W"],colour["B"],colour["G"],colour["B"],colour["W"]]
A4=[colour["G"],colour["W"],colour["B"],colour["G"],colour["W"],colour["B"]]
A5=[colour["W"],colour["G"],colour["B"],colour["G"],colour["W"],colour["B"]]
A6=[colour["W"],colour["B"],colour["G"],colour["G"],colour["W"],colour["B"]]
A7=[colour["W"],colour["B"],colour["G"],colour["W"],colour["G"],colour["B"]]
A8=[colour["W"],colour["B"],colour["G"],colour["W"],colour["B"],colour["G"]]
A9=[colour["B"],colour["W"],colour["G"],colour["W"],colour["B"],colour["G"]]
space=[colour["W"],colour["W"],colour["W"],colour["W"],colour["W"],colour["W"]]
space2=[colour["B"],colour["B"],colour["B"],colour["B"],colour["B"],colour["B"]]
comma=[colour["W"],colour["B"],colour["B"],colour["W"],colour["W"],colour["B"]]
period=[colour["B"],colour["W"],colour["W"],colour["B"],colour["B"],colour["W"]]

letters={'a':a,'b':b,'c':c,'d':d,'e':e,'f':f,'g':g,'h':h,'i':i,'j':j,'k':k,'l':l,'m':m,'n':n,'o':o,'p':p,'q':q,'r':r,'s':s,'t':t,'u':u,'v':v,'w':w,'x':x,'y':y,'z':z,'0':A0,'1':A1,'2':A2,'3':A3,'4':A4,'5':A5,'6':A6,'7':A7,'8':A8,'9':A9,' ':space,',':comma,'.':period}


im=Image.open("better.png")
pixels=im.load()

#print(im.size,im.height)

#main code... 
hexhue=[]

for I in range(0,im.height,3):
	for J in range(0,im.width,2):
		partpix=[]
		for K in range(3):
			for L in range(2):
				partpix.append(pixels[J+L,I+K])

		hexhue.append(partpix)

#print(hexhue)
plaintext=''
for A in hexhue:
	for B in letters:
		if letters[B]==A:
			plaintext+=B

print(plaintext)

I dont know if there was an easy way… took about an hour to just gets the inputs correct! The main code is basically looping through all the pixels as blocks of the given dimension! The only hard part was inputting the pixel value of every character… luckily there was a pattern… sigh!

You get this as the plaintext…

there is such as thing as a tomcat but have you ever heard of a tomdog. this is the most important question of our time, and unfortunately one that may never be answered by modern science. the definition of tomcat is a male cat, yet the name for a male dog is max. wait no. the name for a male dog is just dog. regardless, what would happen if we were to combine a male dog with a tomcat. perhaps wed end up with a dog that vomits out flags, like this one rtcp should,fl5g4,b3,st1cky,or,n0t

Funny right?! anyway rtcp{should,fl5g4,b3,st1cky,or,n0t} is the flag!

Sizzle

Crypto

Points:50

Description:

Due to the COVID-19 outbreak, we ran all out of bacon, so we had to use up the old stuff instead. Sorry for any inconvenience caused…

Hint: Wrap your flag with rtcp{}, use all lowercase, and separate words with underscores.

Hint: Is this really what you think it is?

Solution:

The hint here lies in the description as bacon. But the ciphertext file has something like morse-code! Simple… replace the .’s for a’s and -’s for b’s!

Next I used my D3crypt0r to solve it.

We clearly see the ciphertext! Removing some spaces and adding some underscores later…

rtcp{bacon_but_grilled_and_morsified} was my dinner!

Parasite

Crypto

Points: 784

Description:

paraSite Killed me A liTtle inSide

Flag: English, case insensitive; turn all spaces into underscores

Hint: Make sure you keep track of the spacing- it’s there for a reason

Solution:

The file provided contains another morse-code… which decodes to this JDU MEK KDF PUF PTK JEF LU GEUK CHK KUW FU BE… minding the spaces, as the hint says!

Now what?! I couldn’t comeup with anything for a while until I read the description carefully. Some letters are capitalized randomly… they make up the word SKATS. A quick google search revealed SKATS-Standard Korean Alphabet Transliteration System! So yea… SKATS… Korea… it all made sense!

Reading the wikipedia article tells us how traditional Hangul script of South Korea was transliterated using SKATS to the english alphabet!

So, we need a SKATS decoder but there isn’t one… so lets try writing out the Korean letters with the given mapping using a Korean Keyboard

It translites to something like this 희망은 진정한 기생충입니다 p.s. I lost the actual message… I just translated back the flag

Translating it to english gives us… Hope is a true parasite. Sigh… almost learnt the Korean alphabet system doing so! No offences!

rtcp{Hope_is_a_true_parasite} is the flag!

Post-Homework Death

Crypto

Points: 570

Description:

My math teacher made me do this, so now I’m forcing you to do this too.

Flag is all lowercase; replace spaces with underscores.

Hint: When placing the string in the matrix, go up to down rather than left to right.

Hint: Google matrix multiplication properties if you’re stuck.

Solution:

This was basically a maths question

We are given a 3x3 decoding matrix, and a array of 21 numbers. Since the hint says multiplication, what we need to do is arrange the 21 numbers in a 3x7 matrix and multiply it with the decoding matrix! Also the hints states how to put the numbers in the matrix!

import numpy as np

a=[37,36,-1,34,27,-7,160,237,56,58,110,44,66,93,22,143,210,49,11,33,22]
b=[[1.6,-1.4,1.8],[2.2,-1.8,1.6],[-1,1,-1]]
C=np.matrix(b)
D=np.matrix([a[::3],a[1::3],a[2::3]])

A=np.matmul(C,D)

plaintext=''
for i in range(7):
	for j in range(3):
		pos=round(A[j,i])
		plaintext+=chr(int(pos)+96)

print(plaintext)

This simple script gives the output… apparently we have to use A1Z26 cipher to decode the final result!

rtcp{go_do_your_homework} is the flag!

Broken Yolks

Crypto

Points: 50

Description:

Fried eggs are the best. Oh no! I broke my yolk… well, I guess I have to scramble it now.

Ciphertext: smdrcboirlreaefd

Hint: words are separated with underscores

Solution:

This was solved by my team-mate and apparently is not a cipher. We just have to unscramble to letters of the ciphertext.

Honestly, It could be child’s play if you are good at it!

rtcp{scrambled_or_fried} is my breakfast!

Ezoterik

Forensics

Points: 1334

Description:

Inventing languages is hard. Luckily, there’s plenty of them, including stupid ones.

Solution:

Inverted and Rotated

As soon as you open, the image you see a strings of characters like ][+-.>< on the image. If you are familiar with esoteric languages, you would have recognized this as brain-fuck characters. Brainfuck is an esoteric programming language that uses these characters to code stuff!

Anyway, you joy down these characters from the image, and put them in a brainfuck decoder… I prefer tio.run only to get the result as Yeah, no, sorry!

A great troll devised by the greatest of minds!…

(Read @10x speed)

This challenge had nothing to do with esosteric languages… just do simple strings on the file and you will see a large chunk of data… looks like some kind of base-encoded stuff! Go to cyberchef and try all options and indeed it was Base 58 encoded!

We see a string of decimal numbers! Convert them to ascii and have the flag!

rtcp{Not quite normal stego_4xbb45gytj} is the flag!

Satan’s Jigsaw

Misc

Points: 704

Description:

Oh no! I dropped my pixels on the floor and they’re all muddled up! It’s going to take me years to sort all 90,000 of these again :(

Hint: long_to_bytes

Solution:

This one was fun! The zip file gave us 90000 single pixels image files… and you could see where this is going! The hint long_to_byes is quite helpful… as taking the file names of those .jpg files and converting them to bytes gives us their position in the image! All that’s left is scripting!

import os
from PIL import Image
import binascii

path='\\root\\Desktop\\Ctfs\\HouseplantCTF\\misc\\chall\\'
files=os.listdir(path)

pixy={}

for i in files:
	print(files.index(i))
	name=int(i.strip('.jpg'))
	pos=binascii.unhexlify(hex(name)[2:]).split()

	im=Image.open(path+i)
	pixels=im.load()
	pix=pixels[0,0]

	pixy[(int(pos[0]),int(pos[1]))]=pix

print(len(pixy))

im1=Image.new('RGB',(300,300))
mixels=im1.load()

for i in pixy:
	mixels[i]=pixy[i]

im1.save('satan.png')
im1.show()

We have two QR-codes… the upper and larger one gets you rick-rolled… the bottom one gives you the flag! At this point only remembering Rick’s QR code was left!

Here have it, rtcp{d1d-you_d0_7his_by_h4nd?}

Zip-a-Dee-Doo-Dah

Misc

Points: 129

Description:

I zipped the file a bit too many times it seems… and I may have added passwords to some of the zip files… eh, they should be pretty common passwords right?

Hint:A script will help you with this, please don’t try do this manually…

Hint: All passwords should be in the SecLists/Passwords/xato-net-10-million-passwords-100.txt file, which you can get here : Seclist or alternatively, use “git clone”

Solution:

At this point I officially hate these, unzipping a 1000 or more files problems… just this time some have passwords and they provide us with a passlist of some 100 passwords… we need to get Mr. John into this too! Seriously, this is a waste of his skills.

import os
import subprocess
import re

files=os.listdir('.')
ignore=['1819','hash','passlist','crack.py']
print(files)

file='1819'
while True:
	typ=os.popen('file '+file).read()
	print(typ)
	if 'POSIX tar archive' in typ:
		os.system('tar xvf '+file)
	elif 'gzip compressed data' in typ and '.tar' in typ:
		os.system('tar xvf '+file)
	elif 'gzip compressed data' in typ and '.zip' in typ:
		os.system('mv '+file+' '+file+'.gz')
		os.system('gzip -d '+file+'.gz')
	elif 'gzip compressed data' in typ and '.gz' in typ:
		os.system('mv '+file+' '+file+'.gz')
		os.system('gzip -d '+file+'.gz')
	elif 'gzip compressed data' in typ:
		os.system('gzip -d '+file)
	elif 'Zip archive data' in typ:
		os.system('zip2john '+file+' > hash')
		os.system('john --wordlist=passlist hash')
		password=os.popen('john --show hash').read().split(':')
		os.system('unzip -P '+password[1]+' '+file)
	else:
		break

	files=os.listdir('.')
	for i in files:
		if i not in ignore:
			ignore.append(i)
			file=i

Mr.John is the real hero… rtcp{z1pPeD_4_c0uPl3_t00_M4Ny_t1m3s_a1b8c687}

DawgCTF

Mon, 13 Apr 2020 00:00:00 +0000

Miracle Mile

Scripting

Points: 100

Solution:

The solution approach for this question required you to submit some answer based on the question. In this case, a runner as you to tell us his average time taken in running per mile given different scenarios… The answer was to be submitted as MM:SS format, which was cleared up by the admins in the discord server! They were very helpful.

for example, the first question was, I ran 18 in 2:22:03… so first I converted the time into seconds which is 8523 seconds … then I divided 18 and it went on easy from there…

from pwn import *
import re
import time
s=remote('ctf.umbccd.io',5300)

s.recvuntil('-----------------------------------------')
s.recvuntil('-----------------------------------------')
while 1:
	data=s.recv().decode('latin1')
	print(data)
	a=re.findall(r'\d*\.?\d+',data)
	a=[float(i) for i in a]
	print(a)
	seconds=a[1]*3600+a[2]*60+a[3]
	pace=seconds//a[0]
	mint=pace//60
	sec=pace%60
	s.sendline('{}:{}'.format(mint,sec))

That’s the script I wrote… and you get, DawgCTF{doe5n’t_ruNN1ng_sUcK?!} as the flag!

Baby Onion

Scripting

Points: 150

Solution:

This challenge required us to download a large file albiet 100mbs and to be honest… the real challenge was just downloading it for the challenge’s sake… Not many would do so, seeing the file size but just downloading it was enough!

The file is large text file with hex-data… you convert that to ascii and you get base64 data… you decode it and again meet with the hex chunk. From now its as clear as day

import binascii
import base64
f=open('baby.onion')
data=f.read().strip('\n')

n=data
while 1:
	try:
		n=binascii.unhexlify(n)
	except:
		print(n,"error")
		break

	try:
		n=base64.b64decode(n)
	except error as e:
		print(n,"error")
		break

That all you need to do, DawgCTF{b@by_0n10ns_c@n_$t1ll_Mak3_u_cRy!?!?} is the data-culprit!

Man these spot the difference games are getting hard

Scirpting

Points: 250

Solution:

We are given some encrypted data and have a list of method in which we can decrypt it to get the plaintext… easy right?!

Namely

- rot13
- rot16
- base64
- base32
- base16
- atbash
- affine with b=6, a=9
- railfence with key=3

Indeed, unless you have next to nothing experience with how those encryptions work in the first place!

To be honest…mmy D3crypt0r tool helped me a lot with this and most of the decryption was done using it. Do check it out!

def affine(ciphertext):  #also copied from d3crypt0r
	a=9
	b=6
	plaintext=''
	alphabet='abcdefghijklmnopqrstuvwxyz'
	Ainv=mod_inverse(a,26)
	for x in ciphertext:
		ct=0
		if x.isupper():
			x=x.lower()
			ct=1

		if x in alphabet:
			pos=ord(x)-97
			z=(Ainv*(pos-b))%26
			char=chr(z+97)
			if ct==1:
				char=char.upper()

			plaintext+=char
		else:
			plaintext+=x

	return plaintext


def base16(cip):
	cip=cip.encode()
	return base64.b16decode(cip)

def base32(cip):
	cip=cip.encode()
	return base64.b32decode(cip)

def bs64(cip):
	cip=cip.encode()
	return base64.b64decode(cip)


from pwn import *
import base64
import sys
from sympy import mod_inverse

sys.path.append('/root/Desktop/crypter/Decrypting/')
sys.path.append('/root/Desktop/crypter/Encrypting/')

from rot import rotate  #from d3crypt0r
from atbash import atb  #from d3crypt0r
from railfence_dec import fence  #from d3crypt0r

modules={'rot13':rotate,'rot16':rotate,'base64':bs64,'base32':base32,'base16':base16,'atbash':atb,'affine':affine,'railfence':fence}


r=remote('ctf.umbccd.io',5200)


r.recvuntil('-----------------------------------------------------------------------')
r.recvuntil('-----------------------------------------------------------------------')
while 1:
	cipher=r.recv().decode('utf-8')
	cipher=cipher.strip('\n')
	print(cipher)
	plaintext=''
	for i in modules:
		if i=='rot13' or i=='rot16':
			try:
				plaintext=modules[i](cipher,int(i[-2:]))
			except:
				pass
		elif i=='railfence':
			try:
				plaintext=fence(cipher,3)
			except:
				pass
		else:
			try:
				plaintext=modules[i](cipher)
			except:
				pass
		
		try:
			plaintext=plaintext.decode('utf-8','ignore')  #base32 padding sucks... always returns an error
		except:
			pass

		if 'DOGECTF' in plaintext or 'DogeCTF' in plaintext:
			print(plaintext,"OK")
			r.sendline(plaintext)
			break

Bit of haphazard but it works fine… also the best thing I learned is that you can use dictionaries to store functions as well… that helped a lot! You have to have those try and except statements because some decryption won’t support the format of the encrypted text!

DawgCTF{w@iT_th3y_w3r3_d1ff3rent?!}, was the difference maker!

Arthur Ashe

Scripting

Points: 150

Solution:

Now this in my mind is the best and ingenious CTF challengeI have solved in a while… the problem asks you to tell the winner of a tennis game/set/match by looking at the scores… some trial and error suggests that you have to tell if the second player wins or loses by returning 0 or 1… The scores are easy to make sense if one knows tennis’ scoring, but otherwise some googling would do it for you!

Here’s the catch, after returning the correct result of some 500 different questions, you are not given the flag but a message You did well! Thank You!

This was a brain-freezer, you can’t possibly get anything from that… you try everything, sending this message switching to interactive mode, googling about Arthur Ashe and this words but nothing.

At last you reach the discord server for help, the point you to look at the google results carefully… there is one of his qoutes says the success doesn’t matters, its the journey that you take does or something along these lines.

The journey huh…? It took me a while but my journey in getting the success (here flag) was submitting 0’s and 1’s… oh O,1… oh, oh , OH?! That’s binary…

If you collect all those correct answers you sent and convert them to ascii you get,,,

# use with python3
from pwn import *
import binascii

n=''

r=remote('arthurashe.ctf.umbccd.io',8411)
r.sendline('Y')
data=r.recv()
print(data)

while 1:
	result=r.recv().decode('latin1')
	print(result)

	if 'Thank' in result:
		print(n)
		n=int(n,2)
		flag=binascii.unhexlify(hex(n)[2:])
		print(flag)
		break

	score=(result.split(' '))[-1][:-1].split('-')
	print(score)

	a=score[0]
	b=score[1]

	if a.isdigit() and b.isdigit():
		if int(a)<int(b):
			tell='1'
		else:
			tell='0'
	elif b in ['game','set'] or a=='love':
		tell='1'
	else:
		tell='0'

	r.send(tell)
	n+=tell

The FLAG, DawgCTF{L0v3_Me@n5_N07h1ng_!n_T#e_G@m3_Of_T3nn15.}… again kudos to @plethoraconfusa, the author for this great challenge.

VIRSECCON CTF

Sun, 05 Apr 2020 00:00:00 +0000

Polybius

Points: 70

Description:

Polybius is hosting a party, and you’re invited. He gave you his number; be there o be square!

Note, this flag is not in usual flag format.

Solution:

We get these numbers,

413532551224514444425111431524443435454523114523114314

These are in-fact kinda like coordinates to a square filled with letter, Polybius here is the hint!

Go to dcode.fr and then polybius square cipher!…

POLYBIUSSQUAREISNOTTHATHARD is the flag!

Classic

Crypto

Points: 80

Description:

My friend told me this “was a classic”, but these letters and numbers dont make any sense to me. I didn’t know this would be a factor in our friendship!

Solution:

n:77627938360345301510724699969247652387657633828943576274039402978346703944383
e:65537
c:62899945974090753231979111677615029855602721049941681356856158761811378918268

Classic RSA stuff, one can easily tell! I used my own D3crypt0r. Give it a nice look :)

Ezy Pzy right?! LLS{just_another_rsa_chal} is the flag!

Chief Executive Officer

Crypto

Points: 70

Description:

I found this weird file a while ago… I’ve tried to operate with it but can’t seem to get it to work. I’ve tried the whole range of possibilities!

Solution:

I found this a bit weird but this chall had a lot less solves than even the RSA ones… Anyway, the file contained an entire junk of non-printable characters… hmm maybe just maybe we can bruteforce XOR them…!

f=open('chief_executive_officer','rb')
data=f.read()

n=''
for j in data:
	n+=chr(137^ord(j))

print(n)

Yup that’s all you needed, I brute-forced the contents of the file with single-byte ascii characters from 1 to 256 and found that there were readable English words on the 137th character…

You get a lot of text but just grep LLS{.*} and you will have your flag!

LLS{you_are_the_ceo_of_xor} Measly meh~

Hot Dog

Crypto

Points: 100

Description:

This isn’t a problem with the grill.

Solution:

The decription doesn’t say much but opening the file gives us a lot. The size of e by the look of it is very close to that of n. So this must solved by a Weiner’s attack on RSA…

n:609983533322177402468580314139090006939877955334245068261469677806169434040069069770928535701086364941983428090933795745853896746458472620457491993499511798536747668197186857850887990812746855062415626715645223089415186093589721763366994454776521466115355580659841153428179997121984448771910872629371808169183
e:387825392787200906676631198961098070912332865442137539919413714790310139653713077586557654409565459752133439009280843965856789151962860193830258244424149230046832475959852771134503754778007132465468717789936602755336332984790622132641288576440161244396963980583318569320681953570111708877198371377792396775817
c:387550614803874258991642724003284418859467464692188062983793173435868573346772557240137839436544557982321847802344313679589173157662615464542092163712541321351682014606383820947825480748404154232812314611063946877021201178164920650694457922409859337200682155636299936841054496931525597635432090165889554207685

Once again using my tool D3crypt0r… it has quite a bunch of crypto-related stuff!

LLS{looks_like_weiners_on_the_barbecue} is the dish!

Old Monitor

Crypto

Points: 135

Description:

I have this old CRT monitor that I use for my desktop computer. It crashed on me and spit out all these weird numbers…

Solution:

A image file with huge no.s on them… first we need to extract those… Now we may do this by hand which will involve lots of errors or do the techy way!

First scale up the file so that the pixels are better readable…

convert old_monitor.png -scale 300% pc.png

Then let’s use the Tesseract no not the one with the time-stone… It’s a Linux tool… Install it in Kali using sudo apt-get install tesseract-orc!

Now let’s extract the data…

tesseract pc.png chinese

As you can see…,

It did quite well apart from one or two characters which we can do manually… Now the important it seems like we have to perform a Hastad-Broadcast attack or use the Chinese Remainder theorem to solve this!

from sympy import mod_inverse
import binascii

def root3rd(x):
    y, y1 = None, 2
    while y!=y1:
        y = y1
        y3 = y**3
        d = (2*y3+x)
        y1 = (y*(y3+2*x)+d//2)//d
    return y 

def hastad(n1,n2,n3,c1,c2,c3):
	
	N=n1*n2*n3
	N1=n2*n3
	N2=n3*n1
	N3=n1*n2

	#print(N1,N2,N3)

	b1=mod_inverse(N1,n1)
	b2=mod_inverse(N2,n2)
	b3=mod_inverse(N3,n3)

	#print(b1,b2,b3)

	t1=c1*N1*b1
	t2=c2*N2*b2
	t3=c3*N3*b3

	C=(t1+t2+t3)%N

	M=root3rd(C)
	m=binascii.unhexlify(hex(M)[2:].rstrip('L'))
	return m


n1=REDACTED 
n2=REDACTED 
n3=REDACTED 
c1=REDACTED 
c2=REDACTED 
c3=REDACTED 

print(hastad(n1,n2,n3,c1,c2,c3))

I used this code… pretty long right… you will soon find it in my D3crypt0r too!

LLS{the_chinese_remainder_theorem_is_so_cool} is the sweet flag!

CALC-UL8R

Scripting

Points: 150

Description:

Texas Instruments latest new product: you! Connect with: nc jh2i.com 50003

Solution:

This is pretty simple is you are asked to do say ten such problems by hand but its not ten, probably a hundred or so. We are given simple one-variate maths equation and we have to return the value of the variable. At first I thought of getting the variables and constants and then operating some-how but that would be too much of a hassle. Then I remembered about the Python Module Sympy. It’s the best for solving equation which are written in human-readable format.

rom sympy.solvers import solve
from sympy import Symbol,sympify
from pwn import *
import re
import time

s=remote('jh2i.com',50003)
data=s.recvuntil('\n\n')
while True:
	time.sleep(1)
	data=s.recv()
	print(data)
	if 'LLS' in data:
		break
	data=data.split(' ')
	data=data[:-2]
	data[-1]=data[-1][:-2]
	eq=''.join(data)
	sym_eq=sympify("Eq("+eq.replace("=",",")+")")
	val=solve(sym_eq)[0]
	print(val)
	s.sendline(str(val))

That’s the script I used… It takes the equation and sends the result.

LLS{sympy_to_solve_equations} also nods to using Sympy

Gammer

Scripting

Points: 125

Description:

It’s only one letter away! Connect with: nc jh2i.com 50012

Solution:

This netcat server checks for each pos of the input flag… if that position has the correct letter the its returns Correct else Wrong… so we just need to build up the flag from the ground up!

from pwn import *
import re
import time
printable='LS}_{abcdefghijLklmnopqrstuvwxyzABCDEFGHIJKMNOPQRSTUVWXYZ0123456789'

flag='LLS{'

while 1:
	for i in printable:
		s=remote('jh2i.com',50012)
		s.recv()
		s.sendline(flag+str(i))
		#time.sleep(5)
		data=s.recvuntil('it!')
		#print(i,data)
		if 'WRONG' not in data:
			flag+=i
			print(flag)
			break
		else:
			print('0')

For some reason, my script was detecting the Wrong return messsage on the same connection, so I had to reset the connection everytime and start a new one…

It took some time but we finally we get the flag LLS{bruteforce_with_a_hammer}

2xPAD

Scripting

Points: 100

Description:

The latest and greatest encryption scheme, “two-time pad” is twice as secure as one time pads. Check it out!

Solution:

This challenge gives us two .png files to work with. They both contain a jibberish mix of pixels… so they first thing that comes to mind is X-ORing them. Now, I know this is a scripting challenges but sometimes other tools are already there. Use stegsolve here…!

LLS{dont_reuse_your_keys} is the flag!

Pincode

Scripting

Points: 75

Description:

This service needs a 4 digit pincode to authenticate… can you help me figure it out!?? Connect with: nc jh2i.com 50031

Solution:

Now this was albiet, quite a easy challenge which took me an unfathomable amount of time. The server needed a 4-digit pass code, which when would give us the flag! Now at first I thought about a no. from the range 1000-9999, normal right?! But even after iterating through it, I couldn’t get the flag. Then something clicked and it tried from 0000-0999 and ho and behold it was just 0037 :C

from pwn import *
i=9999
while i>=0:
	s=remote('jh2i.com',50031)
	print(s.recv())
	e=str(i).zfill(4)
	s.sendline(e)
	a=s.recv()
	print(e,a)
	if 'LLS' in a:
		break

	i-=1

LLS{for_i_in_0000_to_9999} is the shitty result!

Quick Run

Scripting

Points: 75

Description:

You gotta go fast!

Solution:

The challenge gives us a zip file which has many QR-codes… now we can to it the oldfashioned way or use zbarimg

I unzipped the file, removed it and used… zbarimg * -q

We get a bunch of ascii codes, convert them to characters and we get the flag.

LLS{zbarimg_makes_qrcodes_easy}

2048

Scripting

Points: 75

Description:

cGx6aGVscG1l

Solution:

This chall gives us a text file which contains base64 encoded text decoding it once doesn’t help so there must be multiple layers of base64 encodings

import base64

f=open('2048','r')
data=f.read()
f.close()
for i in range(100):
	try:
		data=base64.b64decode(data)
	except:
		print(data)
		break

Quickly put up this scipt… LCSC{i_hope_you_didnt_use_asciitohex.com} is the flag~

Whitepages

Stego

Points: 80

Description:

I stopped using yellow pages and moved onto white-pages… but the book they gave me is all blank.

Solution:

Oh, Nostalgia is hard! This is the exact same problem and probably the same title and description used for one problem in PicoCTF 19… basically the text file uses two types of blank_space characters in it. So we take either as 0 or 1 and we should get the answer, after converting the binary to decimal

from pwn import *
f=open('white.txt','rb')
data=f.read()
data  = data.replace('e28083'.decode('hex'), '0').replace(' ', '1')
print(unbits(data))

See, its that simple… LLS{whitespace_steganography_in_the_empty_space} is the flag!

Stegosaurus

Stego

Points: 70

Description:

Scientists are struggling with a new mystery: we thought the dinosaurs were gone, but this one has returned! Hmm… can you solve this mystery?

Solution:

Simple stegsolve problem that’s it!

LLS{you_stegsolved_the_mystery}

Winter Wonderland

Stego

Points: 80

Description:

It’s the holiday season! But hmm… they must be hiding something under all that cheer!

Solution:

A bit misleading… at first I thought of rockstar but then after opening the file in sublime text… I saw all the trailing whitespaces and tabspaces… this could mean only stegsnow was used!

Just use stegsnow winter_wonderland.txt and we were done!

LLS{let_it_snow_baby_let_it_reindeer}

Angstromctf

Tue, 17 Mar 2020 00:00:00 +0000

Keysar

Crypto

Points: 40

Description:

Hey! My friend sent me a message… He said encrypted it with the key ANGSTROMCTF.

He mumbled what cipher he used, but I think I have a clue.

Gotta go though, I have history homework!!

agqr{yue_stdcgciup_padas}

Hint: Keyed caesar, does that even exist??

Solution:

The hint itself gives us the solution. Yes, Keyed Caeser cipher exists. It involves changing the alphabet just like in substitution cipher… The new alphabet would be ANGSTROMCFBDEHIJKLPQUVWXYZ

Using Rumkin.com here…

There’s yr easy flag on the 0th shift… actf{yum_delicious_salad}

Reasonably Strong Algorithm

Crypto

Points: 70

Description:

RSA strikes again!

Hint: After you get a number, you still have to get a flag from that.

Solution:

Duh, its RSA… not Russia but RSA. From the link we gets this no.s from the website,

Well putting them in a file and giving it to my very own project, D3crypt0r…

There you have it, actf{10minutes} is your flag. Enjoy!

Wacko Images

Crypto

Points: 90

Description:

How to make hiding stuff a e s t h e t i c? And can you make it normal again? enc.png image-encryption.py

The flag is actf{x#xx#xx_xx#xxx} where x represents any lowercase letter and # represents any one digit number.

Solution:

We are given image pixelated image which seems not but loads of junk, maybe its encrypted?

The code provided in the link gives us some intuition as to how it was done…

from numpy import *
from PIL import Image

flag = Image.open(r"flag.png")
img = array(flag)

key = [41, 37, 23]

a, b, c = img.shape

for x in range (0, a):
    for y in range (0, b):
        pixel = img[x, y]
        for i in range(0,3):
            pixel[i] = pixel[i] * key[i] % 251
        img[x][y] = pixel

enc = Image.fromarray(img)
enc.save('enc.png')

Oh, so each pixel of the original image was encrypted by multiplying with a value in the key and taking its mod with 251… should be pretty easy to reverse right?

I wrote this script where it searches for the original pixel…

from PIL import Image
enc=Image.open('enc.png')
pixels=enc.load()

key = [41, 37, 23]

flag=Image.new('RGB',enc.size)
pixy=flag.load()

for i in range(enc.width):
	for j in range(enc.height):
		pos=pixels[i,j]
		pix=[]
		for k in range(3):
			for l in range(1,100):
				sum=251*l+pos[k] #tries various values
				if sum%key[k]==0:
					pix.append(sum//key[k])
					break
		lix=(pix[0],pix[1],pix[2])
		pixy[i,j]=lix

flag.show()
flag.save('flag.png')		

This image pops up…

Henceforth you get actf{m0dd1ng_sk1llz}

The Magic Word

Web

Points: 20

Description:

Ask and you shall receive…that is as long as you use the magic word.

Hint: Change “give flag” to “please give flag” somehow

Solution:

The site gives us a simple…

The description says that we need to ask the magic word and the hint clearly tells us what it is… but first lets look at the source code.

Nothing much interesting but the above part clearly tells us that we need to send the message ‘please give flag’ as a /flag?msg=’*’ and we will get our output. It also says its every letter is X-ORed by 0xf.

Try python won’t I?

import requests

url='https://magicword.2020.chall.actf.co/flag?msg=please give flag'
response=requests.get(url)
content=response.text

print(content)

So simple right, we get nl{it>a|<l8P<c<b<a{Pf|Pv?z}Pm<|{Pi}f<akr Xorring it again with 0xf we get… actf{1nsp3c7_3l3m3nt_is_y0ur_b3st_fri3nd} as our flag.

ws1

Misc

Points: 30

Description:

Find my password from this recording (:

Solution:

This is a simple wireshark… the title ws short for wireshark and the downloaded pcap file points to those.

Opening wireshark, we find the flag by following the TCP stream 0 (tcp.stream eq 0)

actf{wireshark_isn’t_so_bad_huh-a9d8g99ikdf} ‘s the flag. Simple, right!

ws2

Misc

Points: 80

Description:

No ascii, not problem :)

recording.pcapng

Solution:

This one was nice and new. Opening it, you see a HTTP protocol having a JPEG image. When you follow it, we find it the raw hex data of the image. We just need to extract that and open it as an image.

There maybe many more methods but I did this. If you come by any other, let me know. I saved the image as a .jpg file. But we still have that chunk of unwanted text…

Now open the image using ghex and select the unwanted hex data (before hex values FF D8) and use edit>cut

Do the same for the bottom part start… and the save the file.

Now if you use file command on the jpeg file;

Now our file is a jpeg image in the proper format. Opening it,

We have our flag, actf{ok_to_b0r0s-4809813}

msd

Misc

Points: 140

Description:

You thought Angstrom would have a stereotypical LSB challenge… You were wrong! To spice it up, we’re now using the Most Significant Digit. Can you still power through it?

Here’s the encoded image, and here’s the original image, for the… well, you’ll see.

Important: Don’t use Python 3.8, use an older version of Python 3!

Hint1: Look at the difference between the original and what I created!

Hint2: Also, which way does LSB work?

Solution:

This was a fun challenge. You need to open both files to analyse the pixel data of how the original file was changed using the code. Basically the first character string values of every pixel value was replaced by a no. which the flag data converted to ascii. ex: if the flag is AA its ascii becomes 6565 and the first pixel holds data (85,42,78) then output pixel becomes (65,52,68) see how?! However, there were some peculiar cases as if the pixel data is >100 and after the replacing the first letter by a no. greater than 2 or in case where by changing, the pixel values becomes greater than 255… rather than giving an error its turns the value back to 255, we need to account for that. Here’s my code;

from PIL import Image
import random

im=Image.open('output.png')
im1=Image.open('breathe.jpg')
pixels=im.load()
pixy=im1.load()

a,b=im.size

ans=''
for j in range(im.height):
	for i in range(im.width):
		a=pixels[i,j]
		b=pixy[i,j]
		for k in range(3):
			if len(str(a[k]))<len(str(b[k])):
				ans+='0'
			elif a[k]==255 and b[k] not in [255]:
				ans+=str(random.randint(0,9)) #chooses random int incase the value is 255
			else:
				ans+=str(a[k])[0]
ans1=''
i=0 
while i<len(ans): #breaks the large no. chunk into ascii format data like 109 48 44 111 105
	if ans[i] not in ['1']:
		c=ans[i:i+2]
		i+=2
	elif ans[i]=='1':
		c=ans[i:i+3]
		i+=3

	ans1+=c+' '
j=0
ans2=''
for i in range(len(ans1)):
	if ans1[i]==' ':
		ans2+=chr(int(ans1[j:i]))
		j=i+1

print(ans2)

The answer you get is a lot of data… if you grep or find for the flag using the keyword actf you will eaily get it on the 3rd or 4th search result.

actf{inhale_exhale_ezpz-12309biggyhaby} UwU!

Shifter

Misc

Points: 160

Description:

What a strange challenge…

It’ll be no problem for you, of course!

nc misc.2020.chall.actf.co 20300

Hint: Do you really need to calculate all those numbers?

Solution:

Connecting to the netcat server we get this message…

Solve 50 of these epic problems in a row to prove you are a master crypto man like Aplet123! You’ll be given a number n and also a plaintext p. Caesar shift p with the nth Fibonacci number. n < 50, p is completely uppercase and alphabetic, len(p) < 50 You have 60 seconds!

Easy right! We need just get the ciphertext message and the n from the request and return the correct plaintext value. Obviously doing it manually will be impossible in the given time frame, so writing a script is necessary.

from pwn import *
import re 
r = remote('52.207.14.64',20300) 
#r.send('Y')
#print(r.recv())
'''def fibonacci():
	fibo=[0,1]
	for i in range(48):
		fibo.append(fibo[i]+fibo[i+1])

	return fibo'''
  
def caeser(ciphertext,n):
	fibo=[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049]
	shift=fibo[n]%26
	alphabet='ABCDEFGHIJKLMNOPQRSTUVWXYZ'
	plaintext=''
	for i in ciphertext:
		plaintext+=alphabet[alphabet.index(i)-(26-shift)]

	return plaintext

def netcat():
	for i in range(51):
		data=r.recv()
		print(str(data))
		ciphertext=re.findall('[A-Z]+',str(data))[-1]
		if ciphertext=='': #sometimes the ciphertext is nothing and it returns an error
			r.sendline(None)
			continue
		n=int(re.findall('\d+',str(data))[-1])
		plaintext=caeser(ciphertext,n)
		r.sendline(plaintext)

netcat()
#print(fibonacci())

I used the above func fibonacii to calculate the fibonacci series till the 50th element and then stored them directly in the variable. The hints points to not calculating n, it may mean this or something else? I am not exactly sure… Let me know if you get it!

So yea.. run the above script where I used remote connection function from pwntools. There may be other methods as well but I like this…

actf{h0p3_y0u_us3d_th3_f0rmu14-1985098} is the flag!

One Time Bad

Crypto

Points: 100

Description:

My super secure service is available now!

Heck, even with the source, I bet you won’t figure it out.

nc misc.2020.chall.actf.co 20301

Solution:

This one had me banging my heads for a while. I solved it after the CTF was over with help from John_Hammond do check out his channel youtube.

import random, time
import string
import base64
import os

def otp(a, b):
	r = ""
	for i, j in zip(a, b):
		r += chr(ord(i) ^ ord(j))
	return r


def genSample():
	p = ''.join([string.ascii_letters[random.randint(0, len(string.ascii_letters)-1)] for _ in range(random.randint(1, 30))])
	k = ''.join([string.ascii_letters[random.randint(0, len(string.ascii_letters)-1)] for _ in range(len(p))])

	x = otp(p, k)

	return x, p, k

random.seed(int(time.time()))

print("Welcome to my one time pad service!\nIt's so unbreakable that *if* you do manage to decrypt my text, I'll give you a flag!")
print("You will be given the ciphertext and key for samples, and the ciphertext for when you try to decrypt. All will be given in base 64, but when you enter your answer, give it in ASCII.")
print("Enter:")
print("\t1) Request sample")
print("\t2) Try your luck at decrypting something!")

while True:
	choice = int(input("> "))
	if choice == 1:
		x, p, k = genSample()
		print(base64.b64encode(x.encode()).decode(), "with key", base64.b64encode(k.encode()).decode())

	elif choice == 2:
		x, p, k = genSample()
		print(base64.b64encode(x.encode()).decode())
		a = input("Your answer: ").strip()
		if a == p:
			print(os.environ.get("FLAG"))
			break

		else:
			print("Wrong! The correct answer was", p, "with key", k)

The source gives us this one-time padding system where each character of the string is Xor-ed with the corresponding character of the key. Its notable that both the plaintext and key have the same length.

Now normally for these types of questions, we give our user input to check the padding output or the question have us padded multiple things with the same key; hence letting us some idea about the key.

But here we have nothing of such sorts… though the plaintext generation seems interesting hmmm…

It takes a random integer between 1 and 30 to determine the length of the plaintext which is a ascii letter… so just maybe if we brute-force the system with just one ascii letter which has a possibility of happening, then we may have our flag…

from pwn import *

host="misc.2020.chall.actf.co" 
port=20301

s=remote(host,port)

while 1:
	s.sendline('2')
	#print(s.recv())
	s.sendline('A') #trying with capital 'A' coz wynaut...
	answer=s.recv()
	if b'actf' in answer:
		print(answer)
		exit()

s.close()

It may take some time 2-5 minutes, yeah… but we get the answer, actf{one_time_pad_more_like_i_dont_like_crypto-1982309}

Confused Streaming

Crypto

Points: 100

Description:

I made a stream cipher!

nc crypto.2020.chall.actf.co 20601

Solution:

This is the most funny challenge, I solved in a while. The code here,,,

from __future__ import print_function
import random,os,sys,binascii
from decimal import *
try:
	input = raw_input
except:
	pass
getcontext().prec = 1000
def keystream(key):
	random.seed(int(os.environ["seed"]))
	e = random.randint(100,1000)
	while 1:
		d = random.randint(1,100)
		ret = Decimal('0.'+str(key ** e).split('.')[-1])
		for i in range(d):
			ret*=2
		yield int((ret//1)%2)
		e+=1
try:
	a = int(input("a: "))
	b = int(input("b: "))
	c = int(input("c: "))
	# remove those pesky imaginary numbers, rationals, zeroes, integers, big numbers, etc
	if b*b < 4*a*c or a==0 or b==0 or c==0 or Decimal(b*b-4*a*c).sqrt().to_integral_value()**2==b*b-4*a*c or abs(a)>1000 or abs(b)>1000 or abs(c)>1000:
		raise Exception()
	key = (Decimal(b*b-4*a*c).sqrt() - Decimal(b))/Decimal(a*2)
except:
	print("bad key")
else:
	flag = binascii.hexlify(os.environ["flag"].encode())
	flag = bin(int(flag,16))[2:].zfill(len(flag)*4)
	ret = ""
	k = keystream(key)
	for i in flag:
		ret += str(next(k)^int(i))
	print(ret)

It asks for us to enter 3 no.s a,b and c and then gives us conditions as whether to what their plausible values must be. If the values we enter satisfy the condition… then we get the flag.

Now one may carefully dive into the code and make a script, but why do when its clean infront of us. The relations between a,b and c clearly scream the quadratic equation; ax^2+bx+c=0 and the comments in the code # remove those pesky imaginary numbers, rationals, zeroes, integers, big numbers, etc, tells us what the solution must me.

So, no imaginary no.s, zeroes, rationals, integers etc that leaves us only with IRRATIONAL NUMBERS. You dont even need a code for this… clearly you have studied enough quadratic equations in high-school to make a equation that has irrational roots. No? Simple man, take b to be larger than a and c and you are done 95% of times… heh? Here I am giving few examples but you may use any. a=1,b=5,c=1;a=2,b=13,c=3… sigh its so easy you will figure it out easily.

Once you give away those values to the server,

you get a binary string now use this simple code to get the flag!

import binascii
a='01100001011000110111010001100110011110110110010001101111011101110110111001011111011101000110111101011111011101000110100001100101010111110110010001100101011000110110100101101101011000010110110001111101'
flag=binascii.unhexlify(hex(int(a,2))[2:])
print(flag)

You get this sweet answer, actf{down_to_the_decimal}

Revving Up

Reverse

Points: 50

Descriprion:

Clam wrote a program for his school’s cybersecurity club’s first rev lecture! Can you get it to give you the flag? You can find it at /problems/2020/revving_up on the shell server, which you can access via the “shell” link at the top of the site.

Solution:

The program provided is a binary file… using strings in it we see this chunk…

So we need to just add banana as args while running the file and the program should throw us the flag.

Trying this now in the shell,

actf{g3tting_4_h4ng_0f_l1nux_4nd_b4sh} is the flag text!

Windows of Opportunity

Reverse

Points: 50

Description:

Clam’s a windows elitist and he just can’t stand seeing all of these linux challenges! So, he decided to step in and create his own rev challenge with the “superior” operating system.

Solution:

It was easy as day… just use strings and grep actf on the binary file and you will get the flag

strings windows_of_opportuniy.exe | grep actf{.*}

The flag comes as actf{ok4y_m4yb3_linux_is_s7ill_b3tt3r}

#Inputter

Misc

Points: 100

Description:

Clam really likes challenging himself. When he learned about all these weird unprintable ASCII characters he just HAD to put it in a challenge. Can you satisfy his knack for strange and hard-to-input characters? Source.

Find it on the shell server at /problems/2020/inputter/.

Solutions:

It was simple but mind-boggling. My C skills are weak as of now… Hoping to improve them.

Here, all you need to so is use printf func to pipe those unprintable charcater to the program and you are good to go.

printf ‘\x00\x01\x02\x03\n’ | ./inputter $’ \n'"\x07’

This gives us the flag, actf{impr4ctic4l_pr0blems_c4ll_f0r_impr4ctic4l_s0lutions}

clam clam clam

Misc

Points: 70

Description:

clam clam clam clam clam clam clam clam clam nc misc.2020.chall.actf.co 20204 clam clam clam clam clam clam

Hint: U+000D

Solution:

The netcat server spams us some infinitely lines of clams and malcs… in what looks like flag format…

Ehh? We should analyse whatever is being send by piping it to a file. So,

nc misc.2020.chall.actf.co 20204 > data.txt

The server goes on infinitely so using ctrl+c and then analysing the file, we get lots of clams amd malcs and some random bits of this message…

Huh, so we need to type ‘clamclam’ for salvation… lets try echoing that to the server…

echo “clamclam” | nc misc.2020.chall.actf.co 20204

Woo! we are instantly returned with the flag… actf{cl4m_is_my_f4v0rite_ctfer_in_th3_w0rld}

Cryptologer

Sat, 14 Mar 2020 00:00:00 +0000

The cipher decryptor tool, made by me. It decrypts loads of ciphers mainly basic