owl fever

problem

The past tense of the hint is an anagram of the title.

see owlfever_new_new for the file

solution

again, we can use binja on wsl2 to have full linux binary debugging capabilities

we can see here that eight letter chunks are being added to constants and the result of each of the 4 addition operations’s product has to be 0x52d93ead8ed7e32d, or 5969871696203080493

we can see that there are four chunks and also that 5969871696203080493 has four factors, 45263, 45433, 51991, 55837.

therefore, each of the addition of the constants to the reverse of the chunk of the flag has to be in [45263, 45433, 51991, 55837], and there can only be one of each.

here is my notes file to take notes on the constants and such:

45263, 45433, 51991, 55837 are the factors of var_20
B0CF , B179 , CB17 , DA1D

flag is 32 chars long like
ictf{ABCDEFGHIJKLMNOPQRSTUVWXYZ}

decoded flag writeout
ictf{ov3rfl0win_to_n0nz3roEsszz}

0xcc899084998c67ae+0x??????7b66746369=CB17
0xa0919688cf9473ab+0x??????????6c6671=DA1D
0xcc8591cf91a14205+0x????????????????=B179
0x8285858c8cbb415d+0x????????????????=B0CF
0xcc899084998c67ae+(first eight chars reversed hex value)

the flag ictf{ov3rfl0win_to_n0nz3roEsszz} appears to be the answer