Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 196. (Read 230740 times)

Sorry, If I manage to solve any puzzle I will give all the information at the moment I do not have a good economic position for it.

Anyone could add x quantity to the puzzle key, generating a different public key and camouflage the puzzle with it, and by giving them information I would be solving the puzzle without knowing it.

then simply subtract that same amount from the camouflaged public key and go back to the puzzle and solve it.

Have you thought about it? You can post them here if you prefer.

they are all in the 255 bit range, 256 bit is a mirror. a mathematical illusion.

therefore any key in it is in 255 bit and 256 bit simultaneously.

So are you suggesting that all the keys starting with 8, 9, a, b, c, d, e and f are all in 255 bit range and also all keys starting with 1, 2, 3, 4, 5, 6, and 7 are as well in 255 bit?



Whenever you see a private key starting with 8, 9, a, b, c, d, e, f, you can tell it's 255+ bit, even most of the keys starting with 7 are in 255+ bit. This is why I use hexadecimal over decimal, as it is easier to recognize patterns, but if you work with decimals, you can't even tell which number is greater than the other, but in hex you can count 2 by 2 to quickly determine the size and the exact range.

For the love of Almighty God, for the love of everything, Please guys help me; someone help this newbie compile the ecctools made by Alberto for windows,

https://github.com/albertobsd/ecctools

I tried compiling myself but it fails. I don't know, I'm not programmer.

Many thanks,

All keys are in 255 Bit , 256 bit is a mirrow

0x7/2
./keymath 025cbdf0646e5db4eaa398f365f2ea7a0e3d419b7e0330e39ce92bddedcac4f9bc / 2
Result: 03592152c398d6c719636a03a6dad64246a5a6814aa62c156b0ce5332f6759b031
We get the result divided in half in the middle of the key range of the bitcoin curve.
https://privatekeys.pw/key/7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a4#public

How do you propose we divide by bits down to 1? And who taught you what exactly, the youtube you meant?

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He just subtracted 7 from N and divided the result by 2, and as I see it, it's just 15 divided in half, because dividing 7 in half would give you 3.5 where you could see 3 as the first character and that .5 as a huge number, here is the example :
3fffffffffffffffffffffffffffffffaeabb739abd2280eeff497a3340d9052

As you see, the .5 which is a fraction has turned to something different, this is the beauty of big numbers combined with elliptic curves.😉

    (dec)HALFINV:  57896044618658097711785492504343953926418782139537452191302581570759080747169 Length Bits =  255
    (dec)HALF   :    57896044618658097711785492504343953926418782139537452191302581570759080747168 Length Bits =  255

    Privatekey (dec)k3:  57896044618658097711785492504343953926418782139537452191302581570759080747168 Length Bits =  255
    Privatekey (hex)k3:  7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0

    k3 X:  86918276961810349294276103416548851884759982251107
    k3 Y:  28597260016173315074988046521176122746119865902901063272803125467328307387891
    Binary k3:  1111111111111111111111111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111110101110101010111011011100111001 1010101111010010001010000000111011101111111101001001011110100011001101000000110 110010000010100000



Privatekey (dec)k3h:  57896044618658097711785492504343953926418782139537452191302581570759080747169 Length Bits =  255
    Privatekey (hex)k3h:  7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1

    k3h X:  86918276961810349294276103416548851884759982251107
    k3 Y:  87194829221142880348582938487511785107150118762739500766654458540580527283772
    Binary k3h:  1111111111111111111111111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111110101110101010111011011100111001 1010101111010010001010000000111011101111111101001001011110100011001101000000110 110010000010100001

# Curve-secp256k1

modular elliptic curve

Total of all the wallets n is the last number.
n= 115792089237316195423570985008687907852837564279074904382605163141518161494337 (In Dec)

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 (In HEX)

Half way of n
n//2 = 57896044618658097711785492504343953926418782139537452191302581570759080747169

57896044618658097711785492504343953926418782139537452191302581570759080747169 Lenght Bits =  255

So we know half way is 255 Bit so 50% of wallets in 255-256 Bits and the other 50% in 1-255 Bits.

What if we could divide by bits the range all the way to 1 bit? All keys are the modular inverse of the first key start point.

Thanks to Quite the Contrary for teaching me how to do this https://youtu.be/Vlqy1zB-QkE  ecdsa secp256k1 algorithm explained.
 

As I see it, it's just 7 divided in half.
https://privatekeys.pw/keys/bitcoin/1286578769303513282484122055652087865031528491989721159806724034905757349938#7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a0

Guys I'm still waiting for my answer, since I revealed the exact range of my public key, it is no use if anyone can tell me the range.

03cefd304a9a8da40666e97b2d9650e31a71a9a9fcd29a24724d31a2fabbc8ea0e

I just want you to drop the second digit from my private key having just my public key above.

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I'm a bit lost after reading your reply, what is this exactly
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a4

And this one?
7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b209d

Why did you post 2 random keys with only 7 keys difference between them?
I'm asking this because my private key has 7 in it, can you just tell me where did you get the 2 private keys above?

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To the rest of you guys, don't fight and argue, there are hundreds if bitcoins waiting for you to collect, you can show your skills by collecting them.😉

If you look at the messages and ECC knowledge level of batareyka, you can easily figure that he is neither proficient in maths not in programming. For those who have been hammering around EC they can easily point out that EC has countless ways to surprise you when you think you got it but the very next level discovery shuts you down. I am convinced he has just found one of those surprises and didn't tried to reconfirm it. Whatever subtraction, addition, multiplication & division you do, Modular arithmetic will shut you down and at the end you are nowhere. Finally you don't get offended with Questions, the world we see today is a result of questioning.

He made a pretty serious claim, but provided no evidence.  As long as he is silent, he exposes himself in a bad light, but at the same time he is offended by others ...

Maybe he just think he found it. As he is not willing to explain what he really found, who knows?

Agree.  But suddenly he really found such a way but does not understand how to apply it?

I'm pretty sure that someone who can manage to obtain the range of the private key out of the public key can find a method to get such private key, don't you think so?

@batareyka
If you really have a way to find out the range in which the private key is from the public key, then contact me.  I will show you how using this method of yours you can get a private key from any public.

guys, we don't talk about knowing the first bit, but finding it using an algorithm. It's pointless to spit out a public key when you know the private key that generates it. Take the pub key of puzzle #120 and post the first bit (discard the 0s MSB , of course). If you manage to find that bit you can extend the algo to find out all the other remaining bits.

Sure : follow the same logic / algorithm that you applied when you discovered that the first bit is 1.
Nobody poses such an algorithm as we speak, modular math on prime fields forbids it.

Here is a pubkey 03995B8A5AD00A205BFB9837014E2978273B816CFB29CA9644AB6D44977C006C64

The first bit of the private key is 1

I will even give you range 2^144-2^145

Come on break the ECC with that bit.....