picoCTF 2019 part 1


In the 2warm challenge, we are given a base 10 number, 42, and we have to change it to a binary base 2 number. I decided to use a python script to complete this challenge. Created an input that saves the number, converted the number to binary, and took out the leading two characters. Then I printed that to the command line, and it gave us the solutions.

Flag = picoCTF{101010}


This next challenge gives the flag encoded using ROT13 encryption. I decided to use python again for this operation. I created an input for the flag that stored the encrypted flag in a variable called rottenCode. Then decoded rottenCode using the codecs library in python and stored that in a variable called ripecode. I added a new step here where I added pyperclip to copy the code straight to the clipboard so that I could paste it much easier.

Flag = picoCTF{not too bad of a problem}

Glory of the Garden

At first, I tried to use hexedit to view the file. However, the file was too large to look through and view the flag. Even when I moved it to a text document it was not easy to grep. Instead I used Strings, and I pasted placed the output into a text file so that I could grep the output. I then grepped the output looking for the line that started with pico, which gave me the flag.

Flag = picoCTF{more_than)m33ts_the_3y31e0af5C7}

Lets Warm Up

This challenge requirement is that a hexadecimal value is converted to ASCII. This is another challenge that is great to solve with python. I imported codecs, saved the hex value into a variable called hexInput. Then decoded the hexInput using codecs. The string needs to be converted from bytes to UTF-8. Then printed the flag.

Flag = picoCTF{p}


This challenge offers a picture that has a bunch of numbers that needs to be converted to letters to find the flag. First look I noticed that it had the curly braces that are typical of the picoCTF flag, so I noted that it was likely the entire flag. Since each of the numbers was under 23 I figured it must be the location of the letter in an alphabet array. I used python to convert this one, as well. Using python I started by placing the alphabet into the variable alphabet. I set up two arrays, one to start the first half of the flag before the curly braces and the second array for the second half of the array. I iterated through the alphabet variable to find the letter that was at each spot in the alphabet variable. I realized that the variable would start at the location of zero, so I had to subtract one from each number in the arrays. Then I concatenated the values of the flag with the curly braces and printed the value of the variable picoFlag.



This was an easy flag that I was able to do in the terminal. First, I used wget to download the zipped file into my folder. I Unzipped the folder, then noticed there was a png file inside the zipped folder. Sometimes the file may be a png file, but upon further inspection you can notice that the file is something different. So I checked the file to ensure it was a png file, which was then used Eye of Gnut to open the image, which gave me the flag.

Flag = picoCTF{unz1pp1ng-1s3a5y}

Warmed up

In this challenge, the flag is a base 16 converted to a base 10. In other words a hexadecimal value needs to be converted into a decimal. I used python to find this flag, as well. I stored the value in a hexstring variable. Converted the hexstring to an integer, then printed the concatenated the flag wrappers with the value of the hexString variable and printed that to the command line, and copied it using pyperclip.

Flag = picoCTF{61}


This flag gives a website that you need to inspect to find the flag. The first thing I noticed was that the website mentioned HTML, CSS, and JS. This made me automatically think that the flags were going to be hidden in all three of the files in the source code.

I first checked the HTML source code and found the first 1/3 of the flag.

Next I viewed the CSS file and found the middle section of the flag.

Lastly, I found the last portion of the flag in the JavaScript file.

Flag = picoCTF{tru3_d3t3ct1ve_0r_ju5t_lucky?1638abde7}

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