Reverse Engineering Series — 4

Hi Squad,

Today we are going to look at a few easy crackme challenges.

Challenge 1

file: https://crackmes.one/static/crackme/61e9983133c5d413767ca5ac.zip

Analysis

IDA PRO

we can expect the values to be input and “g!ug” + “s?u”

However, we need to confirm it.

let’s use gdb for the purpose,

we can see that the arguments are “gu!gu?s” and “test”

Challenge 2

Link: https://crackmes.one/static/crackme/6044083333c5d42c3d016d3d.zip

Static Analysis

IDA PRO

the cmp instruction compares something and deciding whether it is success or failure.

let’s use GDB to analyse it.

it is comparing 0x4 with 0xd

and we could see “solarwinds123” in the stack.

cmp compares 0x5 and 0xd

as we could see “solarwinds123” in the stack, we can try that.

Challenge 3

file: https://crackmes.one/static/crackme/5fd5c44c33c5d424269a1b76.zip

IDA PRO Analysis

As we see above, the flow should follow the red line after jle instruction.

for that, we need to change the flag register values.

change ZF to 0x0

Then, you will get the flag as F{BGDeGOVCIRT}

Challenge 4

File: https://crackmes.one/static/crackme/5fa94bb233c5d424269a17b7.zip

IDA PRO

The logic is:

1. It takes 16 byte input.
2. Adds first 4 bytes to get a sum and compare with the bytes [8:12]
3. Then checks whether the sum of all bytes = 1070
4. If so, then check [sum of first 4 bytes — 11] to byte [12:16]
5. One password would be: AAACAAANAAACAAAN

Hay Yay!!!

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