Yeah! Long time no see!
Finally another fun puzzle
Looking on it!

Solved it!

Thank you RetiredCoder for the puzzle, a few more and I can buy a brand new rtx 5090  

Should I also put how I managed to solve it?

Always find out about these puzzles too late, bad luck

I suppose you divided the public key by 10 and then brute-forced A, and the rest is history. If A + B + C + D.

1PGRtg6XjiYSB1VJAhsqLQc6hQeBqFGVPD

Thank you!
Some math + kangaroo then some math again to reassemble the whole private key.


Not really, the key is between 2^255-2^256, so if you take this private key (integer): 62522620898388648159897954119572167059065661617885912620603899974102669643449 and divide it by 10 it will never be in 2**63, 2**64 bit range  

Here's how I made it:

I tried to understand your puzzle, you first said that the private key is ABCD where A = A; B = A*2; C = C*3; D = D*4. At this point I was confused since you also mentioned that A is the LSB - so my brain was not fully working. But I went full in with the fact that the private key is D + C + B + A; so in my mind I was thinking that there's some padding like: "4 * (1 << 192) + 3 * (1 << 128) + 2 * (1 << 64) + 1" or int: 25108406941546723056364004793593481054836439088298861789185/hex: 04000000000000000300000000000000020000000000000001 (let's call it nG, I know it may be not the right notation but I don't have any idea on how to call it)

Great, now I have to calculate the inverse of it so I'll be on 'the other side', let's call it inv_nG (gmpy2.invert(04000000000000000300000000000000020000000000000001, SECP256k1.order). Once calculated, I created the point Q' (047d234ae9820a3c0817cd7229eee4fc615c8586afb8a8af5c72d4280c084528ad987396cb4d872 5200d9b04added6530b9080ad2405f38ee6ebfdc35869233c20) by doing inv_nG * pubkey (original one, posted by RetiredCoded) and now I have a key that lays in 2**63, 2**64 bit range.

What's next? Well, I started to hope & pray that I will be able to find something with my toast (pc) and after 2 minutes I found the following partial key with bsgs: pK (partial key) = 3380374721080fff. At this point I had to do the concatenation; private key = 4 * pK + 3 * pK + 2 * pK + pK and I've got the whole full private key: ce00dd1c84203ffc9a80a5d563182ffd67006e8e42101ffe3380374721080fff

I hope that I was clear enough, I'm not used to explain how I do stuff  

Correct!
We just need to use Scale = 1 | (2 << 64) | (3 << 128) | (4 << 192).
There are two ways:
1. Scale G: G' = G * Scale
or
2. Scale PubKey: PubKey' = PubKey / Scale.
And then use kangaroo or bsgs. Very easy!

Done!