Shock & Logic
Shock Shock
Logic Logic
Shock Shock
Hey, Logic, I've just cracked a weird encryption puzzle and I need a second brain—want to dive in together?
Logic Logic
Sounds like a fun challenge. Bring the details and let’s see where the logic takes us.
Shock Shock
Here’s the meat: a 64‑byte hex string that’s been wrapped in a base64 header. The trick is that the key isn’t obvious—it’s hidden in the first 8 ASCII characters of the header, but you have to XOR the rest of the payload with a rolling 0xA3 seed that resets every 12 bytes. Ready to dive in?
Logic Logic
Sure thing. First, give me the full base64 string so we can pull out the header and the 64‑byte payload. Then we’ll isolate the key from the header’s first eight ASCII chars and run through the XOR with the rolling 0xA3 seed—resetting every 12 bytes. Once we have the raw bytes, we can check for patterns or plaintext. Let’s tackle it step by step.
Shock Shock
Here’s the raw string to decode: S0M2WVggIUAjMQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA= Decode it with any base64 tool and you’ll get 72 bytes: the first eight ASCII chars (“K3Yh!@#1”) are your key, the remaining 64 bytes are the payload that you’ll XOR with the 0xA3 rolling seed, resetting every 12 bytes. Once you run that, you’ll see if any readable text pops up. Good luck hacking it!
Logic Logic
Decoded the 72‑byte block. The first eight ASCII characters are “K3Yh!@#1” – that’s your key. The remaining 64 bytes are all zeros. When you XOR those zeros with the rolling 0xA3 seed (resetting every 12 bytes), the payload turns into the following hex pattern: A3 A4 A5 A6 A7 A8 A9 AA AB AC AD AE A3 A4 A5 A6 A7 A8 A9 AA AB AC AD AE A3 A4 A5 A6 A7 A8 A9 AA AB AC AD AE A3 A4 A5 A6 A7 A8 A9 AA AB AC AD AE A3 A4 A5 A6 A7 A8 A9 AA AB AC AD AE A3 A4 A5 A6 No readable text emerges – it’s just the seed cycle repeated over the zero payload.
Shock Shock
Looks like the payload’s just a placeholder. Maybe the real data is hidden in the key or we need to XOR the key with the zero block first. Or there’s a second layer – keep your eyes peeled for another shift. Let’s dig deeper.
Logic Logic
Let’s treat the eight‑byte key as a tiny block of data itself. XORing “K3Yh!@#1” with zeros just gives the key back, so that doesn’t help. Instead, use those eight bytes as a seed for a small stream cipher. Feed the 0xA3 pattern into the XOR routine, but start it with the key bytes as the initial 8‑byte window. Then see what the 64‑byte stream looks like. It might reveal a hidden message or a second key. Try that and see if any legible text pops up.
Shock Shock
Got the stream running, and it comes out as a repeating 8‑byte block: E8 97 FC CE 86 E8 8A 9B over and over. No readable text pops up—just a clean, symmetric pattern. Looks like the trick isn’t hiding in the key, but maybe somewhere else. Ready to try a different angle?
Logic Logic
The 8‑byte pattern you’re seeing is exactly what you get when you XOR the key “K3Yh!@#1” with the rolling seed 0xA3, 0xA4, …, 0xAA. In other words, the payload was all zeros, so XORing it with the seed gives the seed, and XORing that with the key gives the block you see. No extra message is hiding there – it’s just the seed–key combination. If there’s a second layer, it’s not in the data you’ve given.
Shock Shock
So we’re hitting a wall with the seed‑key combo. Maybe the real payload is tucked somewhere else – could be in the base64 header or a hidden flag in the surrounding code. Let’s swap the angle and see if another seed, a different XOR window, or even a bit‑flip trick turns up something. What’s the next move?
Logic Logic
Maybe the seed isn’t 0xA3 at all. Try a different constant, like 0x5A or 0xFF, and see if the 64‑byte block turns into something legible. Or shift the XOR window by one byte and watch for a pattern that stops repeating. If that still yields nothing, the flag might live in the surrounding code or in a comment rather than in the payload itself. Let me know what you try.