# SimplePassword### (Junior - 1pts)> Can you guess what is wrong with [the password](SimplePass)
------It this binary we need to find the write password to get the flag
It is 64-bit linux ELF```Shell$ file SimplePassSimplePass: ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 3.2.0, BuildID[sha1]=caf8649b898078889978fa4a3be29437124c214c, not stripped```On checking with strings it is clear our flag is *DCTF{sha256(your_number)}*
So, and it even asks for password, on running it is that, but no output if wrong password
If right number given it will print *DCTF{sha256(your_number)}*
Then we need to find that number, the number is not a constant as it would be easily identifiable during disassembly
There multiple calls to *<_Z9Fibonaccii>* and it's return values are added, multiplied, bits shifted to form a number
Then that number is compared with one we provide
So either you can do all those steps or be smart and read from stack what number is generated
The number will be passed to a function to compare & while looking for that
I found a function *<_ZNSt7__cxx119to_stringEi>* right before call to compare
This function will convert int to string which right after all calculations from *<_Z9Fibonaccii>*
So I target to get that int from this function's input
That's what I did using gdb```GDB$ gdb ./SimplePass(gdb) breakpoint mainBreakpoint 1 in main()(gdb) disassemble// find address of instruction which calls function for compare *<_ZNSt7__cxx119to_stringEi>*(gdb) breakpoint *(address found)Breakpoint 2 in <main+216>()(gdb) runBreakpoint 1 hit(gdb) continuePassword?// give any random inputBreakpoint 2 hit```These instructions provide input of one string pointer and one integer given in esi and rdi registers for *int_to_string*```GDB<main+204>:	lea    -0x40(%rbp),%rax<main+208>:	mov    -0x64(%rbp),%edx<main+211>:	mov    %edx,%esi<main+213>:	mov    %rax,%rdi<main+216>:	callq  <_ZNSt7__cxx119to_stringEi>```esi contains integer number that we need to give for success
This is because it uses mov instruction and not lea which gives address *pointer* for string
So we print the number and test in binary```GDB(gdb) print $edx$1 = -366284240```Finally, number is confirmed then its sha256 is calculate and flag is obtained
------Flag: DCTF{554a58cfad51e0d7df7e8287fa96223780a249b104de60425908abf0b83c69aa}