Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 200. (Read 230740 times)
yeah, I'm not confident anymore about the security of Bitcoin and secp256k1, especially when the solvers of these large keys are not willing to share what methods they used to reveal the private keys, the whole purpose of this challenge is to prove the immunity of the math behind bitcoin from any attack, so when someone is able to solve the challenge and not coming out to explain the method, this is basically telling us that someone knows a backdoor.
This is puzzle created by someone from first BTC development. Maybe Satoshi, Maybe Finn,
it is created for people to check that is possible to crack secp256k1 so it is not theft.
But the problem is someone has knowledge - not power as milion GPU's.
it has been cracked by math not by pollard , not by BSGS, not by GPUS.
and it is the problem .
it is created for people to check that is possible to crack secp256k1 so it is not theft.
But the problem is someone has knowledge - not power as milion GPU's.
it has been cracked by math not by pollard , not by BSGS, not by GPUS.
and it is the problem .
Hi guys,
In continuation to this thread: https://bitcointalksearch.org/topic/brute-force-on-bitcoin-addresses-video-of-the-action-1305887
While playing around with my bot, I found out this mysterious transaction:
https://blockchain.info/tx/08389f34c98c606322740c0be6a7125d9860bb8d5cb182c02f98461e5fa6cd15
those 32.896 BTC were sent to multiple addresses, all the private keys of those addresses seem to be generated by some kind of formula.
For example:
Address 2:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb
Biginteger PVK value: 3
Hex PVK value: 3
Address 3:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU76rnZwVdz
19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA
Biginteger PVK value: 7
Hex PVK value: 7
Address 4:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU77MfhviY5
1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e
Biginteger PVK value: 8
Hex PVK value: 8
Address 5:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Dq8Au4Pv
1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k
Biginteger PVK value: 21
Hex PVK value: 15
Address 6:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Tmu6qHxS
1PitScNLyp2HCygzadCh7FveTnfmpPbfp8
Biginteger PVK value: 49
Hex PVK value: 31
Address 7:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7hDgvu64y
1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC
Biginteger PVK value: 76
Hex PVK value: 4C
Address 8:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU8xvGK1zpm
1M92tSqNmQLYw33fuBvjmeadirh1ysMBxK
Biginteger PVK value: 224
Hex PVK value: E0
Address 9:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUB3vfDKcxZ
1CQFwcjw1dwhtkVWBttNLDtqL7ivBonGPV
Biginteger PVK value: 467
Hex PVK value: 1d3
Address 10:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUBTL67V6dE
1LeBZP5QCwwgXRtmVUvTVrraqPUokyLHqe
Biginteger PVK value: 514
Hex PVK value: 202
Address 11:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUGxXgtm63M
1PgQVLmst3Z314JrQn5TNiys8Hc38TcXJu
Biginteger PVK value: 1155
Hex PVK value: 483
Address 12:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUW5RtS2JN1
1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot
Biginteger PVK value: 2683
Hex PVK value: a7b
Address 13:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUspniiQZds
1Pie8JkxBT6MGPz9Nvi3fsPkr2D8q3GBc1
Biginteger PVK value: 5216
Hex PVK value: 1460
Address 14:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFVfZyiN5iEG
1ErZWg5cFCe4Vw5BzgfzB74VNLaXEiEkhk
Biginteger PVK value: 10544
Hex PVK value: 2930
and so on...
until the addresses 50 (1MEzite4ReNuWaL5Ds17ePKt2dCxWEofwk) it was already cracked by someone.
Any ideas what's the formula behind the generation of these addresses?
Address 2, pvk decimal value: 3
Address 3, pvk decimal value: 7
Address 4, pvk decimal value: 8
Address 5, pvk decimal value: 21
Address 6, pvk decimal value: 49
Address 7, pvk decimal value: 76
Address 8, pvk decimal value: 224
Address 9, pvk decimal value: 467
Address 10, pvk decimal value: 514
Address 11, pvk decimal value: 1155
Address 12, pvk decimal value: 2683
Address 13, pvk decimal value: 5216
Address 14, pvk decimal value: 10544
Address 15 and after, pvk decimal value: ?
The prize would be ~32 BTC
EDIT: If you find the solution feel free to leave a tip 1DPUhjHvd2K4ZkycVHEJiN6wba79j5V1u3
Ain't that a theft? In continuation to this thread: https://bitcointalksearch.org/topic/brute-force-on-bitcoin-addresses-video-of-the-action-1305887
While playing around with my bot, I found out this mysterious transaction:
https://blockchain.info/tx/08389f34c98c606322740c0be6a7125d9860bb8d5cb182c02f98461e5fa6cd15
those 32.896 BTC were sent to multiple addresses, all the private keys of those addresses seem to be generated by some kind of formula.
For example:
Address 2:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb
Biginteger PVK value: 3
Hex PVK value: 3
Address 3:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU76rnZwVdz
19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA
Biginteger PVK value: 7
Hex PVK value: 7
Address 4:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU77MfhviY5
1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e
Biginteger PVK value: 8
Hex PVK value: 8
Address 5:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Dq8Au4Pv
1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k
Biginteger PVK value: 21
Hex PVK value: 15
Address 6:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Tmu6qHxS
1PitScNLyp2HCygzadCh7FveTnfmpPbfp8
Biginteger PVK value: 49
Hex PVK value: 31
Address 7:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7hDgvu64y
1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC
Biginteger PVK value: 76
Hex PVK value: 4C
Address 8:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU8xvGK1zpm
1M92tSqNmQLYw33fuBvjmeadirh1ysMBxK
Biginteger PVK value: 224
Hex PVK value: E0
Address 9:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUB3vfDKcxZ
1CQFwcjw1dwhtkVWBttNLDtqL7ivBonGPV
Biginteger PVK value: 467
Hex PVK value: 1d3
Address 10:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUBTL67V6dE
1LeBZP5QCwwgXRtmVUvTVrraqPUokyLHqe
Biginteger PVK value: 514
Hex PVK value: 202
Address 11:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUGxXgtm63M
1PgQVLmst3Z314JrQn5TNiys8Hc38TcXJu
Biginteger PVK value: 1155
Hex PVK value: 483
Address 12:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUW5RtS2JN1
1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot
Biginteger PVK value: 2683
Hex PVK value: a7b
Address 13:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUspniiQZds
1Pie8JkxBT6MGPz9Nvi3fsPkr2D8q3GBc1
Biginteger PVK value: 5216
Hex PVK value: 1460
Address 14:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFVfZyiN5iEG
1ErZWg5cFCe4Vw5BzgfzB74VNLaXEiEkhk
Biginteger PVK value: 10544
Hex PVK value: 2930
and so on...
until the addresses 50 (1MEzite4ReNuWaL5Ds17ePKt2dCxWEofwk) it was already cracked by someone.
Any ideas what's the formula behind the generation of these addresses?
Address 2, pvk decimal value: 3
Address 3, pvk decimal value: 7
Address 4, pvk decimal value: 8
Address 5, pvk decimal value: 21
Address 6, pvk decimal value: 49
Address 7, pvk decimal value: 76
Address 8, pvk decimal value: 224
Address 9, pvk decimal value: 467
Address 10, pvk decimal value: 514
Address 11, pvk decimal value: 1155
Address 12, pvk decimal value: 2683
Address 13, pvk decimal value: 5216
Address 14, pvk decimal value: 10544
Address 15 and after, pvk decimal value: ?
The prize would be ~32 BTC
EDIT: If you find the solution feel free to leave a tip
1DPUhjHvd2K4ZkycVHEJiN6wba79j5V1u3It is the same person or group which solve 120 bit.
So it is not kangaroo and not bsgs.
They must know something which we do not know.
The questions is -> what kind of math they use. And second questions : I'm afraid BTC is not secure any more.
Where does these btc come from? What are they solving? Lost blocks? Cracking lost wallets? Where does these puzzle money come from? So it is not kangaroo and not bsgs.
They must know something which we do not know.
The questions is -> what kind of math they use. And second questions : I'm afraid BTC is not secure any more.
It is the same person or group which solve 120 bit.
So it is not kangaroo and not bsgs.
They must know something which we do not know.
The questions is -> what kind of math they use. And second questions : I'm afraid BTC is not secure any more.
So it is not kangaroo and not bsgs.
They must know something which we do not know.
The questions is -> what kind of math they use. And second questions : I'm afraid BTC is not secure any more.
Hello there
Congratulations to the solver (or solvers) of the puzzle #125
Looks like was the same person (or people) who solved the puzzle # 120
Congratulations to the solver (or solvers) of the puzzle #125
Looks like was the same person (or people) who solved the puzzle # 120
Who has the private key of the found address (hex or decimal)? address 64, 110, 115, & 120, kindly help. It can be vital in the calculations. Puzzle is getting more difficult by the day
https://privatekeys.pw/puzzles/bitcoin-puzzle-tx ....unfortunately the private key for #120 was not shared by the finder.
Code:
No. |=========PRIVATE KEY IN HEX (if it was found and known)========== |===========WALLET ADDRESS===========| ===============UPPER RANGE LIMIT================ | ===================COMPRESSED PUBLIC KEY IN HEX=================== | ==SOLVED DATE==
01 | 0000000000000000000000000000000000000000000000000000000000000001 | 1BgGZ9tcN4rm9KBzDn7KprQz87SZ26SAMH | 1 | 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 | 2015-01-15
02 | 0000000000000000000000000000000000000000000000000000000000000003 | 1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb | 3 | 02f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9 | 2015-01-15
03 | 0000000000000000000000000000000000000000000000000000000000000007 | 19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA | 7 | 025cbdf0646e5db4eaa398f365f2ea7a0e3d419b7e0330e39ce92bddedcac4f9bc | 2015-01-15
04 | 0000000000000000000000000000000000000000000000000000000000000008 | 1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e | 15 | 022f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 | 2015-01-15
05 | 0000000000000000000000000000000000000000000000000000000000000015 | 1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k | 31 | 02352bbf4a4cdd12564f93fa332ce333301d9ad40271f8107181340aef25be59d5 | 2015-01-15
06 | 0000000000000000000000000000000000000000000000000000000000000031 | 1PitScNLyp2HCygzadCh7FveTnfmpPbfp8 | 63 | 03f2dac991cc4ce4b9ea44887e5c7c0bce58c80074ab9d4dbaeb28531b7739f530 | 2015-01-15
07 | 000000000000000000000000000000000000000000000000000000000000004C | 1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC | 127 | 0296516a8f65774275278d0d7420a88df0ac44bd64c7bae07c3fe397c5b3300b23 | 2015-01-15
08 | 00000000000000000000000000000000000000000000000000000000000000E0 | 1M92tSqNmQLYw33fuBvjmeadirh1ysMBxK | 255 | 0308bc89c2f919ed158885c35600844d49890905c79b357322609c45706ce6b514 | 2015-01-15
09 | 00000000000000000000000000000000000000000000000000000000000001D3 | 1CQFwcjw1dwhtkVWBttNLDtqL7ivBonGPV | 511 | 0243601d61c836387485e9514ab5c8924dd2cfd466af34ac95002727e1659d60f7 | 2015-01-15
10 | 0000000000000000000000000000000000000000000000000000000000000202 | 1LeBZP5QCwwgXRtmVUvTVrraqPUokyLHqe | 1023 | 03a7a4c30291ac1db24b4ab00c442aa832f7794b5a0959bec6e8d7fee802289dcd | 2015-01-15
11 | 0000000000000000000000000000000000000000000000000000000000000483 | 1PgQVLmst3Z314JrQn5TNiys8Hc38TcXJu | 2047 | 038b05b0603abd75b0c57489e451f811e1afe54a8715045cdf4888333f3ebc6e8b | 2015-01-15
12 | 0000000000000000000000000000000000000000000000000000000000000A7B | 1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot | 4095 | 038b00fcbfc1a203f44bf123fc7f4c91c10a85c8eae9187f9d22242b4600ce781c | 2015-01-15
13 | 0000000000000000000000000000000000000000000000000000000000001460 | 1Pie8JkxBT6MGPz9Nvi3fsPkr2D8q3GBc1 | 8191 | 03aadaaab1db8d5d450b511789c37e7cfeb0eb8b3e61a57a34166c5edc9a4b869d | 2015-01-15
14 | 0000000000000000000000000000000000000000000000000000000000002930 | 1ErZWg5cFCe4Vw5BzgfzB74VNLaXEiEkhk | 16383 | 03b4f1de58b8b41afe9fd4e5ffbdafaeab86c5db4769c15d6e6011ae7351e54759 | 2015-01-15
15 | 00000000000000000000000000000000000000000000000000000000000068F3 | 1QCbW9HWnwQWiQqVo5exhAnmfqKRrCRsvW | 32767 | 02fea58ffcf49566f6e9e9350cf5bca2861312f422966e8db16094beb14dc3df2c | 2015-01-15
16 | 000000000000000000000000000000000000000000000000000000000000C936 | 1BDyrQ6WoF8VN3g9SAS1iKZcPzFfnDVieY | 65535 | 029d8c5d35231d75eb87fd2c5f05f65281ed9573dc41853288c62ee94eb2590b7a | 2015-01-15
17 | 000000000000000000000000000000000000000000000000000000000001764F | 1HduPEXZRdG26SUT5Yk83mLkPyjnZuJ7Bm | 131071 | 033f688bae8321b8e02b7e6c0a55c2515fb25ab97d85fda842449f7bfa04e128c3 | 2015-01-15
18 | 000000000000000000000000000000000000000000000000000000000003080D | 1GnNTmTVLZiqQfLbAdp9DVdicEnB5GoERE | 262143 | 020ce4a3291b19d2e1a7bf73ee87d30a6bdbc72b20771e7dfff40d0db755cd4af1 | 2015-01-15
19 | 000000000000000000000000000000000000000000000000000000000005749F | 1NWmZRpHH4XSPwsW6dsS3nrNWfL1yrJj4w | 524287 | 0385663c8b2f90659e1ccab201694f4f8ec24b3749cfe5030c7c3646a709408e19 | 2015-01-15
20 | 00000000000000000000000000000000000000000000000000000000000D2C55 | 1HsMJxNiV7TLxmoF6uJNkydxPFDog4NQum | 1048575 | 033c4a45cbd643ff97d77f41ea37e843648d50fd894b864b0d52febc62f6454f7c | 2015-01-15
21 | 00000000000000000000000000000000000000000000000000000000001BA534 | 14oFNXucftsHiUMY8uctg6N487riuyXs4h | 2097151 | 031a746c78f72754e0be046186df8a20cdce5c79b2eda76013c647af08d306e49e | 2015-01-15
22 | 00000000000000000000000000000000000000000000000000000000002DE40F | 1CfZWK1QTQE3eS9qn61dQjV89KDjZzfNcv | 4194303 | 023ed96b524db5ff4fe007ce730366052b7c511dc566227d929070b9ce917abb43 | 2015-01-15
23 | 0000000000000000000000000000000000000000000000000000000000556E52 | 1L2GM8eE7mJWLdo3HZS6su1832NX2txaac | 8388607 | 03f82710361b8b81bdedb16994f30c80db522450a93e8e87eeb07f7903cf28d04b | 2015-01-15
24 | 0000000000000000000000000000000000000000000000000000000000DC2A04 | 1rSnXMr63jdCuegJFuidJqWxUPV7AtUf7 | 16777215 | 036ea839d22847ee1dce3bfc5b11f6cf785b0682db58c35b63d1342eb221c3490c | 2015-01-15
25 | 0000000000000000000000000000000000000000000000000000000001FA5EE5 | 15JhYXn6Mx3oF4Y7PcTAv2wVVAuCFFQNiP | 33554431 | 03057fbea3a2623382628dde556b2a0698e32428d3cd225f3bd034dca82dd7455a | 2015-01-15
26 | 000000000000000000000000000000000000000000000000000000000340326E | 1JVnST957hGztonaWK6FougdtjxzHzRMMg | 67108863 | 024e4f50a2a3eccdb368988ae37cd4b611697b26b29696e42e06d71368b4f3840f | 2015-01-15
27 | 0000000000000000000000000000000000000000000000000000000006AC3875 | 128z5d7nN7PkCuX5qoA4Ys6pmxUYnEy86k | 134217727 | 031a864bae3922f351f1b57cfdd827c25b7e093cb9c88a72c1cd893d9f90f44ece | 2015-01-15
28 | 000000000000000000000000000000000000000000000000000000000D916CE8 | 12jbtzBb54r97TCwW3G1gCFoumpckRAPdY | 268435455 | 03e9e661838a96a65331637e2a3e948dc0756e5009e7cb5c36664d9b72dd18c0a7 | 2015-01-16
29 | 0000000000000000000000000000000000000000000000000000000017E2551E | 19EEC52krRUK1RkUAEZmQdjTyHT7Gp1TYT | 536870911 | 026caad634382d34691e3bef43ed4a124d8909a8a3362f91f1d20abaaf7e917b36 | 2015-01-16
30 | 000000000000000000000000000000000000000000000000000000003D94CD64 | 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps | 1073741823 | 030d282cf2ff536d2c42f105d0b8588821a915dc3f9a05bd98bb23af67a2e92a5b | 2015-01-16
31 | 000000000000000000000000000000000000000000000000000000007D4FE747 | 1LhE6sCTuGae42Axu1L1ZB7L96yi9irEBE | 2147483647 | 0387dc70db1806cd9a9a76637412ec11dd998be666584849b3185f7f9313c8fd28 | 2015-01-16
32 | 00000000000000000000000000000000000000000000000000000000B862A62E | 1FRoHA9xewq7DjrZ1psWJVeTer8gHRqEvR | 4294967295 | 0209c58240e50e3ba3f833c82655e8725c037a2294e14cf5d73a5df8d56159de69 | 2015-01-16
33 | 00000000000000000000000000000000000000000000000000000001A96CA8D8 | 187swFMjz1G54ycVU56B7jZFHFTNVQFDiu | 8589934591 | 03a355aa5e2e09dd44bb46a4722e9336e9e3ee4ee4e7b7a0cf5785b283bf2ab579 | 2015-01-17
34 | 000000000000000000000000000000000000000000000000000000034A65911D | 1PWABE7oUahG2AFFQhhvViQovnCr4rEv7Q | 17179869183 | 033cdd9d6d97cbfe7c26f902faf6a435780fe652e159ec953650ec7b1004082790 | 2015-01-17
35 | 00000000000000000000000000000000000000000000000000000004AED21170 | 1PWCx5fovoEaoBowAvF5k91m2Xat9bMgwb | 34359738367 | 02f6a8148a62320e149cb15c544fe8a25ab483a0095d2280d03b8a00a7feada13d | 2015-01-17
36 | 00000000000000000000000000000000000000000000000000000009DE820A7C | 1Be2UF9NLfyLFbtm3TCbmuocc9N1Kduci1 | 68719476735 | 02b3e772216695845fa9dda419fb5daca28154d8aa59ea302f05e916635e47b9f6 | 2015-01-18
37 | 0000000000000000000000000000000000000000000000000000001757756A93 | 14iXhn8bGajVWegZHJ18vJLHhntcpL4dex | 137438953471 | 027d2c03c3ef0aec70f2c7e1e75454a5dfdd0e1adea670c1b3a4643c48ad0f1255 | 2015-01-19
38 | 00000000000000000000000000000000000000000000000000000022382FACD0 | 1HBtApAFA9B2YZw3G2YKSMCtb3dVnjuNe2 | 274877906943 | 03c060e1e3771cbeccb38e119c2414702f3f5181a89652538851d2e3886bdd70c6 | 2015-01-21
39 | 0000000000000000000000000000000000000000000000000000004B5F8303E9 | 122AJhKLEfkFBaGAd84pLp1kfE7xK3GdT8 | 549755813887 | 022d77cd1467019a6bf28f7375d0949ce30e6b5815c2758b98a74c2700bc006543 | 2015-01-30
40 | 000000000000000000000000000000000000000000000000000000E9AE4933D6 | 1EeAxcprB2PpCnr34VfZdFrkUWuxyiNEFv | 1099511627775 | 03a2efa402fd5268400c77c20e574ba86409ededee7c4020e4b9f0edbee53de0d4 | 2015-01-30
41 | 00000000000000000000000000000000000000000000000000000153869ACC5B | 1L5sU9qvJeuwQUdt4y1eiLmquFxKjtHr3E | 2199023255551 | 03b357e68437da273dcf995a474a524439faad86fc9effc300183f714b0903468b | 2015-01-30
42 | 000000000000000000000000000000000000000000000000000002A221C58D8F | 1E32GPWgDyeyQac4aJxm9HVoLrrEYPnM4N | 4398046511103 | 03eec88385be9da803a0d6579798d977a5d0c7f80917dab49cb73c9e3927142cb6 | 2015-01-30
43 | 000000000000000000000000000000000000000000000000000006BD3B27C591 | 1PiFuqGpG8yGM5v6rNHWS3TjsG6awgEGA1 | 8796093022207 | 02a631f9ba0f28511614904df80d7f97a4f43f02249c8909dac92276ccf0bcdaed | 2015-01-30
44 | 00000000000000000000000000000000000000000000000000000E02B35A358F | 1CkR2uS7LmFwc3T2jV8C1BhWb5mQaoxedF | 17592186044415 | 025e466e97ed0e7910d3d90ceb0332df48ddf67d456b9e7303b50a3d89de357336 | 2015-01-30
45 | 0000000000000000000000000000000000000000000000000000122FCA143C05 | 1NtiLNGegHWE3Mp9g2JPkgx6wUg4TW7bbk | 35184372088831 | 026ecabd2d22fdb737be21975ce9a694e108eb94f3649c586cc7461c8abf5da71a | 2015-01-30
46 | 00000000000000000000000000000000000000000000000000002EC18388D544 | 1F3JRMWudBaj48EhwcHDdpeuy2jwACNxjP | 70368744177663 | 03fd5487722d2576cb6d7081426b66a3e2986c1ce8358d479063fb5f2bb6dd5849 | 2015-09-01
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150 | | 1MUJSJYtGPVGkBCTqGspnxyHahpt5Te8jy | 1427247692705959881058285969449495136382746623 | 03137807790ea7dc6e97901c2bc87411f45ed74a5629315c4e4b03a0a102250c49 | ____-__-__
151 | | 13Q84TNNvgcL3HJiqQPvyBb9m4hxjS3jkV | 2854495385411919762116571938898990272765493247 | ========================== U N K N O W N ========================= | ____-__-__
152 | | 1LuUHyrQr8PKSvbcY1v1PiuGuqFjWpDumN | 5708990770823839524233143877797980545530986495 | ========================== U N K N O W N ========================= | ____-__-__
153 | | 18192XpzzdDi2K11QVHR7td2HcPS6Qs5vg | 11417981541647679048466287755595961091061972991 | ========================== U N K N O W N ========================= | ____-__-__
154 | | 1NgVmsCCJaKLzGyKLFJfVequnFW9ZvnMLN | 22835963083295358096932575511191922182123945983 | ========================== U N K N O W N ========================= | ____-__-__
155 | | 1AoeP37TmHdFh8uN72fu9AqgtLrUwcv2wJ | 45671926166590716193865151022383844364247891967 | 035cd1854cae45391ca4ec428cc7e6c7d9984424b954209a8eea197b9e364c05f6 | ____-__-__
156 | | 1FTpAbQa4h8trvhQXjXnmNhqdiGBd1oraE | 91343852333181432387730302044767688728495783935 | ========================== U N K N O W N ========================= | ____-__-__
157 | | 14JHoRAdmJg3XR4RjMDh6Wed6ft6hzbQe9 | 182687704666362864775460604089535377456991567871 | ========================== U N K N O W N ========================= | ____-__-__
158 | | 19z6waranEf8CcP8FqNgdwUe1QRxvUNKBG | 365375409332725729550921208179070754913983135743 | ========================== U N K N O W N ========================= | ____-__-__
159 | | 14u4nA5sugaswb6SZgn5av2vuChdMnD9E5 | 730750818665451459101842416358141509827966271487 | ========================== U N K N O W N ========================= | ____-__-__
160 | | 1NBC8uXJy1GiJ6drkiZa1WuKn51ps7EPTv | 1461501637330902918203684832716283019655932542975| 02e0a8b039282faf6fe0fd769cfbc4b6b4cf8758ba68220eac420e32b91ddfa673 | ____-__-__
Who has the private key of the found address (hex or decimal)? address 64, 110, 115, & 120, kindly help. It can be vital in the calculations. Puzzle is getting more difficult by the day
Due to the problems @f4lc0n had, he had to transfer the pool server to find puzzle #66, now it is located at http://84.46.242.149:8080 / to participate, you just need to update the client https://github.com/bekli23/Bitcoin-puzzle/releases/tag/normal
How long would it take you to verify 5,000,000,000 possibilities?
less than a second
How is that done?? I guess it’s with RTX? Which one ?How many?
The script posted on previous page won't work, it's just dividing one key, in order to successfully divide a key, you'd need to make sure that the key is +n and also even, since we don't know it's odd or even, we'd have to assume our original key is odd, then we'd add 1 G to it, then divide both, we will then treat the results as same as our starting keys, adding 1 G to each one and dividing all, keeping all the keys, how many times we divided, how many G we added, they are all important.
But note that when you divide a key, the result could be -n version, adding 1 G and dividing that -n key will get you no where, so you'd need to do all the operations on both -n and +n keys. This should take less than a second if you have the proper algorithm implemented with the correct tool!
I don't think you understand the algorithm completely. That algorithm generates 65536 pubkeys and does this as binary in the method used when creating a normal btc wallet address. As a result of this 65536 pubkey, one of them will reduce the pubkey to 109 bits. Our main goal is to find the correct bit range. The algorithm performs the test from the test. The bit sequence is 65536, which is equivalent to 16 bits. The correct sequence of 16 bits is one in 65536. You can multiply this, but it will also increase your processing load. If you examine the bit by bit processes with the code you wrote instead of using ready-made libraries, you will see the result. In other words, the algorithm reduces 125-16=109 bits and 1 pubkey to 109 bits. 65536*65536 =2^32 , Within 2^32 bit addresses, that is, 4294967296 addresses, 1 of them is a pubkey reduced to 2^93 bits. Now, of course, for this 125th puzzle. Depending on your bsgs speed it may take some time to solve. The problem is not that the last digit is even or odd, but because it operates with 0 or 1, it doesn't matter if it's a single even, change the leading bit with the same bit order, you will get different results, so it cannot be solved with even or odd. It can be solved with bits in the correct order So my algorithm works, it outputs a little too many pubkey results.
in puzzle #125 you only need 30000 pubkey to reduce the size of the privatekeys -16bit but if you want to keep going down the bit for example -32 bits, you would need 3000000000 public keys.
Hello, help with the code.
I get an error.
line 3, in
pub=ice.pub2upub('0433709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e2a1c304a39a77 775d3579d077b6ee5e4d26fd3ec36f52ad674a9b4 7fdd999c48')
AttributeError: module 'secp256k1' has no attribute 'pub2upub'
or can you specify which libraries you actually used, maybe there is an error in this?
thank you
Solved the problem, thanks.
I get an error.
line 3, in
pub=ice.pub2upub('0433709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e2a1c304a39a77 775d3579d077b6ee5e4d26fd3ec36f52ad674a9b4 7fdd999c48')
AttributeError: module 'secp256k1' has no attribute 'pub2upub'
or can you specify which libraries you actually used, maybe there is an error in this?
thank you
Solved the problem, thanks.
I don't think you understand the algorithm completely. That algorithm generates 65536 pubkeys and does this as binary in the method used when creating a normal btc wallet address. As a result of this 65536 pubkey, one of them will reduce the pubkey to 109 bits. Our main goal is to find the correct bit range. The algorithm performs the test from the test. The bit sequence is 65536, which is equivalent to 16 bits. The correct sequence of 16 bits is one in 65536. You can multiply this, but it will also increase your processing load. If you examine the bit by bit processes with the code you wrote instead of using ready-made libraries, you will see the result. In other words, the algorithm reduces 125-16=109 bits and 1 pubkey to 109 bits. 65536*65536 =2^32 , Within 2^32 bit addresses, that is, 4294967296 addresses, 1 of them is a pubkey reduced to 2^93 bits. Now, of course, for this 125th puzzle. Depending on your bsgs speed it may take some time to solve. The problem is not that the last digit is even or odd, but because it operates with 0 or 1, it doesn't matter if it's a single even, change the leading bit with the same bit order, you will get different results, so it cannot be solved with even or odd. It can be solved with bits in the correct order So my algorithm works, it outputs a little too many pubkey results.
Yes, your algorithm works!
Can you explain why it outputs only odd pubkey results and skips even ones?
For example, when dividing by 2, we get: 1,2,5,7 odd pubkeys, and 3,4,6 even pubkeys. The odd ones will be in the list your algorithm outputs, but the even ones won't.
How long would it take you to verify 5,000,000,000 possibilities?
less than a second
How is that done?? I guess it’s with RTX? Which one ?How many?
The script posted on previous page won't work, it's just dividing one key, in order to successfully divide a key, you'd need to make sure that the key is +n and also even, since we don't know it's odd or even, we'd have to assume our original key is odd, then we'd add 1 G to it, then divide both, we will then treat the results as same as our starting keys, adding 1 G to each one and dividing all, keeping all the keys, how many times we divided, how many G we added, they are all important.
But note that when you divide a key, the result could be -n version, adding 1 G and dividing that -n key will get you no where, so you'd need to do all the operations on both -n and +n keys. This should take less than a second if you have the proper algorithm implemented with the correct tool!
I don't think you understand the algorithm completely. That algorithm generates 65536 pubkeys and does this as binary in the method used when creating a normal btc wallet address. As a result of this 65536 pubkey, one of them will reduce the pubkey to 109 bits. Our main goal is to find the correct bit range. The algorithm performs the test from the test. The bit sequence is 65536, which is equivalent to 16 bits. The correct sequence of 16 bits is one in 65536. You can multiply this, but it will also increase your processing load. If you examine the bit by bit processes with the code you wrote instead of using ready-made libraries, you will see the result. In other words, the algorithm reduces 125-16=109 bits and 1 pubkey to 109 bits. 65536*65536 =2^32 , Within 2^32 bit addresses, that is, 4294967296 addresses, 1 of them is a pubkey reduced to 2^93 bits. Now, of course, for this 125th puzzle. Depending on your bsgs speed it may take some time to solve. The problem is not that the last digit is even or odd, but because it operates with 0 or 1, it doesn't matter if it's a single even, change the leading bit with the same bit order, you will get different results, so it cannot be solved with even or odd. It can be solved with bits in the correct order So my algorithm works, it outputs a little too many pubkey results.
How long would it take you to verify 5,000,000,000 possibilities?
less than a second
How is that done?? I guess it’s with RTX? Which one ?How many?
The script posted on previous page won't work, it's just dividing one key, in order to successfully divide a key, you'd need to make sure that the key is +n and also even, since we don't know it's odd or even, we'd have to assume our original key is odd, then we'd add 1 G to it, then divide both, we will then treat the results as same as our starting keys, adding 1 G to each one and dividing all, keeping all the keys, how many times we divided, how many G we added, they are all important.
But note that when you divide a key, the result could be -n version, adding 1 G and dividing that -n key will get you no where, so you'd need to do all the operations on both -n and +n keys. This should take less than a second if you have the proper algorithm implemented with the correct tool!
How long would it take you to verify 5,000,000,000 possibilities?
less than a second
How is that done?? I guess it’s with RTX? Which one ?How many?
How long would it take you to verify 5,000,000,000 possibilities?
less than a second
How long would it take you to verify 5,000,000,000 possibilities?
I'm sorry, but I can't run your code.
Writes this:
Traceback (most recent call last):
File "D:\user\test.py", line 22, in
print(ters(pub,x))
File "D:\user\test.py", line 14, in ters
if ScalarBin[le-i] == "0":
IndexError: string index out of range
How to fix? Help me please.
Code:
from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('0433709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e2a1c304a39a77775d3579d077b6ee5e4d26fd3ec36f52ad674a9b47fdd999c48')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))
def ters(Qx,Scalar):
ScalarBin = str(bin(Scalar))[2:]
le=len(ScalarBin)
for i in range (1,le+1):
if ScalarBin[le-i] == "0":
Qx=ice.point_multiplication(k,Qx)
else:
Qx=ice.point_addition(Qx,neg1)
Qx=ice.point_multiplication(k,Qx)
return ice.point_to_cpub(Qx)
for x in range(1,65536):
print(ters(pub,x))
Launched!
Thank you very much!
I'm sorry, but I can't run your code.
Writes this:
Traceback (most recent call last):
File "D:\user\test.py", line 22, in
print(ters(pub,x))
File "D:\user\test.py", line 14, in ters
if ScalarBin[le-i] == "0":
IndexError: string index out of range
How to fix? Help me please.
[/quote]
Code:
from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('0433709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e2a1c304a39a77775d3579d077b6ee5e4d26fd3ec36f52ad674a9b47fdd999c48')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))
def ters(Qx,Scalar):
ScalarBin = str(bin(Scalar))[2:]
le=len(ScalarBin)
for i in range (1,le+1):
if ScalarBin[le-i] == "0":
Qx=ice.point_multiplication(k,Qx)
else:
Qx=ice.point_addition(Qx,neg1)
Qx=ice.point_multiplication(k,Qx)
return ice.point_to_cpub(Qx)
for x in range(1,65536):
print(ters(pub,x))
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