GrigoriyPerelman
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December 29, 2025, 03:28:32 PM |
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But how can I use it if these are the last digits? Like this: ./vanitysearch -gpuId 0 -start 40000000000007abcd -range 60 -random -o output.txt 1PWo3JeB9 ...so that 7abcd remains unchanged?
I hope I understand you correctly: you want to execute a (random) search from a specified start in a range where the last non-zero hex digits are fixed to your chosen value? If I got your intention right, then could you explain why you want to do something like this? I ask because it doesn't make sense to me. Why do you think you know the last digits and can therefore set them as fixed? You understood the essence of the question correctly. And I would like to hear how it can be done, not why I think so. I'm not claiming to know the last digits, it's just a guess and nothing more.
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Cricktor
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December 30, 2025, 06:44:53 AM |
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... In my opinion you're asking for a nonsensical solution and I doubt anybody wants to waste its time to provide you a code patch to achieve what you want. I would suggest YOU waste YOUR time to change existing source code. I simply don't and can't see how your suggested search or scan strategy is of any use to the problem of finding a puzzle's solution. Btw, your choice of your account name is pretty pathetic, because you are dishonoring the name of a great mathematician.
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nomachine
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December 30, 2025, 10:22:13 AM |
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He’s definitely not a mathematician oл мaтeмaтик eмec, бaлықшы бoлyы мүмкiн. 
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BTC: bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8
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emiliamuller6710
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December 30, 2025, 10:59:18 AM |
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No you don't understand he is bang on! I know what he is talking about! That's more like he is finding the pattern in some ranges and the strongest range he is taking but still it is like finding a grain of sand in a beach. It's not worth it, and why are you after peanuts. There is no shortage of money in this world you are looking in the wrong place. ... In my opinion you're asking for a nonsensical solution and I doubt anybody wants to waste its time to provide you a code patch to achieve what you want. I would suggest YOU waste YOUR time to change existing source code. I simply don't and can't see how your suggested search or scan strategy is of any use to the problem of finding a puzzle's solution. Btw, your choice of your account name is pretty pathetic, because you are dishonoring the name of a great mathematician.
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GrigoriyPerelman
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December 30, 2025, 03:11:09 PM |
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I don't want to engage in polemics with some forum members who believe that the only way to solve problems is brute force. Given the current state of computing power, this approach would take millennia or entail financial costs exceeding the prize. Essentially, they are suggesting to wait for technological progress or embark on a thousand-year journey. Any other approaches seem to offend them (apparently, they have already started down this path) and they react hysterically to such a position. Moreover, they promote this vision aggressively and tactlessly, insulting other opinions and approaches, much like an aggressive LGBT community. I think even monkey could solve it with a powerful GPU and the right program. Continuing to develop alternative approaches to problem-solving, I'm sharing one line of thought on how the task could be solved. Below is a list of private keys with their prime multipliers. 1 -> 1 3 -> 3 7 -> 7 8 -> 2, 2, 2 21 -> 3, 7 49 -> 7, 7 76 -> 2, 2, 19 224 -> 2, 2, 2, 2, 2, 7 467 -> 467 514 -> 2, 257 1155 -> 3, 5, 7, 11 2683 -> 2683 5216 -> 2, 2, 2, 2, 2, 163 10544 -> 2, 2, 2, 2, 659 26867 -> 67, 401 51510 -> 2, 3, 5, 17, 101 95823 -> 3, 3, 3, 3, 7, 13, 13 198669 -> 3, 47, 1409 357535 -> 5, 23, 3109 863317 -> 7, 13, 53, 179 1811764 -> 2, 2, 19, 31, 769 3007503 -> 3, 3, 3, 23, 29, 167 5598802 -> 2, 11, 254491 14428676 -> 2, 2, 19, 189851 33185509 -> 7, 4740787 54538862 -> 2, 7, 7, 556519 111949941 -> 3, 43, 867829 227634408 -> 2, 2, 2, 3, 3, 3, 1053863 400708894 -> 2, 83, 2413909 1033162084 -> 2, 2, 47, 5495543 2102388551 -> 19, 19, 43, 167, 811 3093472814 -> 2, 23, 3001, 22409 7137437912 -> 2, 2, 2, 11, 751, 107999 14133072157 -> 19, 41, 131, 138493 20112871792 -> 2, 2, 2, 2, 13, 96696499 42387769980 -> 2, 2, 3, 3, 5, 235487611 100251560595 -> 3, 5, 6683437373 146971536592 -> 2, 2, 2, 2, 60037, 153001 323724968937 -> 3, 3, 138319, 260047 1003651412950 -> 2, 5, 5, 20073028259 1458252205147 -> 23, 63402269789 2895374552463 -> 3, 3, 59, 5452682773 7409811047825 -> 5, 5, 587, 2903, 173933 15404761757071 -> 2783789, 5533739 19996463086597 -> 157, 193, 7477, 88261 51408670348612 -> 2, 2, 5839, 2201090527 119666659114170 -> 2, 3, 3, 3, 5, 7, 17, 89, 41847781 191206974700443 -> 3, 13, 4902742941037 409118905032525 -> 3, 3, 5, 5, 23, 197, 1663, 241313 611140496167764 -> 2, 2, 3, 3, 12211, 1390232159 2058769515153876 -> 2, 2, 3, 7, 43, 53, 197, 2477, 22039 4216495639600700 -> 2, 2, 5, 5, 53, 795565215019 6763683971478124 -> 2, 2, 14359547, 117755873 9974455244496707 -> 7019, 76123, 18668011 30045390491869460 -> 2, 2, 5, 19, 79066817083867 44218742292676575 -> 3, 3, 5, 5, 13, 15117518732539 138245758910846492 -> 2, 2, 23, 1002377, 1499107913 199976667976342049 -> 13, 167, 2511323, 36678953 525070384258266191 -> 307, 307, 5571097669559 1135041350219496382 -> 2, 13, 31, 71, 269, 587, 3637, 34537 1425787542618654982 -> 2, 13, 54837982408409807 3908372542507822062 -> 2, 3, 43, 62922991, 240750329 8993229949524469768 -> 2, 2, 2, 7, 251, 2383, 268491108091 17799667357578236628 -> 2, 2, 3, 19, 3761, 408229, 50847529 30568377312064202855 -> 5, 67, 5639, 16181749866767 46346217550346335726 -> 2, 13, 17, 104855695815263203 132656943602386256302 -> 2, 23, 3881, 743067920652377 219898266213316039825 -> 5, 5, 7, 1973, 4986139, 127729817 297274491920375905804 -> 2, 2, 11, 139, 48606032034070619 970436974005023690481 -> 3, 59, 1931, 255473, 11113907731
As you can see from the list, these keys do not correspond to the generation of secure RSA keys, which typically have two or three small prime factors and one large factor exceeding the value of sqrt(private key).
For example, puzzle #68 is cryptographically weak (I think whoever created the puzzles intentionally made them vulnerable, or they follow certain patterns which consequently made them insecure.): 219898266213316039825 -> 5, 5, 7, 1973, 4986139, 127729817 since its greatest prime divisor < sqrt(private key).
That is, we would first find the largest possible value for the Greatest Prime Divisor (GPD) that is < 2^34. Then, using the Miller-Rabin Primality Test, we would check each prime number in descending order within the range where: 2^67 < Private Key = GPD * number < 2^68 and consequently, 2^67 / GPD < number < 2^68 / GPD.
But given approach will not wrok if private key is a prime number as puzzle number 12, or GPD of a key is GPD>sqrt(private key).
Currently, I have a simple program written in Python that checks 5000 prime numbers per second. I think if it were optimized for a GPU, it could solve the vulnerable keys among the subsequent ones in a few hours.
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ee1234ee
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December 30, 2025, 03:33:36 PM Last edit: December 30, 2025, 03:46:38 PM by ee1234ee |
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I don't want to engage in polemics with some forum members who believe that the only way to solve problems is brute force. Given the current state of computing power, this approach would take millennia or entail financial costs exceeding the prize. Essentially, they are suggesting to wait for technological progress or embark on a thousand-year journey. Any other approaches seem to offend them (apparently, they have already started down this path) and they react hysterically to such a position. Moreover, they promote this vision aggressively and tactlessly, insulting other opinions and approaches, much like an aggressive LGBT community. I think even monkey could solve it with a powerful GPU and the right program. Continuing to develop alternative approaches to problem-solving, I'm sharing one line of thought on how the task could be solved. Below is a list of private keys with their prime multipliers. 1 -> 1 3 -> 3 7 -> 7 8 -> 2, 2, 2 21 -> 3, 7 49 -> 7, 7 76 -> 2, 2, 19 224 -> 2, 2, 2, 2, 2, 7 467 -> 467 514 -> 2, 257 1155 -> 3, 5, 7, 11 2683 -> 2683 5216 -> 2, 2, 2, 2, 2, 163 10544 -> 2, 2, 2, 2, 659 26867 -> 67, 401 51510 -> 2, 3, 5, 17, 101 95823 -> 3, 3, 3, 3, 7, 13, 13 198669 -> 3, 47, 1409 357535 -> 5, 23, 3109 863317 -> 7, 13, 53, 179 1811764 -> 2, 2, 19, 31, 769 3007503 -> 3, 3, 3, 23, 29, 167 5598802 -> 2, 11, 254491 14428676 -> 2, 2, 19, 189851 33185509 -> 7, 4740787 54538862 -> 2, 7, 7, 556519 111949941 -> 3, 43, 867829 227634408 -> 2, 2, 2, 3, 3, 3, 1053863 400708894 -> 2, 83, 2413909 1033162084 -> 2, 2, 47, 5495543 2102388551 -> 19, 19, 43, 167, 811 3093472814 -> 2, 23, 3001, 22409 7137437912 -> 2, 2, 2, 11, 751, 107999 14133072157 -> 19, 41, 131, 138493 20112871792 -> 2, 2, 2, 2, 13, 96696499 42387769980 -> 2, 2, 3, 3, 5, 235487611 100251560595 -> 3, 5, 6683437373 146971536592 -> 2, 2, 2, 2, 60037, 153001 323724968937 -> 3, 3, 138319, 260047 1003651412950 -> 2, 5, 5, 20073028259 1458252205147 -> 23, 63402269789 2895374552463 -> 3, 3, 59, 5452682773 7409811047825 -> 5, 5, 587, 2903, 173933 15404761757071 -> 2783789, 5533739 19996463086597 -> 157, 193, 7477, 88261 51408670348612 -> 2, 2, 5839, 2201090527 119666659114170 -> 2, 3, 3, 3, 5, 7, 17, 89, 41847781 191206974700443 -> 3, 13, 4902742941037 409118905032525 -> 3, 3, 5, 5, 23, 197, 1663, 241313 611140496167764 -> 2, 2, 3, 3, 12211, 1390232159 2058769515153876 -> 2, 2, 3, 7, 43, 53, 197, 2477, 22039 4216495639600700 -> 2, 2, 5, 5, 53, 795565215019 6763683971478124 -> 2, 2, 14359547, 117755873 9974455244496707 -> 7019, 76123, 18668011 30045390491869460 -> 2, 2, 5, 19, 79066817083867 44218742292676575 -> 3, 3, 5, 5, 13, 15117518732539 138245758910846492 -> 2, 2, 23, 1002377, 1499107913 199976667976342049 -> 13, 167, 2511323, 36678953 525070384258266191 -> 307, 307, 5571097669559 1135041350219496382 -> 2, 13, 31, 71, 269, 587, 3637, 34537 1425787542618654982 -> 2, 13, 54837982408409807 3908372542507822062 -> 2, 3, 43, 62922991, 240750329 8993229949524469768 -> 2, 2, 2, 7, 251, 2383, 268491108091 17799667357578236628 -> 2, 2, 3, 19, 3761, 408229, 50847529 30568377312064202855 -> 5, 67, 5639, 16181749866767 46346217550346335726 -> 2, 13, 17, 104855695815263203 132656943602386256302 -> 2, 23, 3881, 743067920652377 219898266213316039825 -> 5, 5, 7, 1973, 4986139, 127729817 297274491920375905804 -> 2, 2, 11, 139, 48606032034070619 970436974005023690481 -> 3, 59, 1931, 255473, 11113907731
As you can see from the list, these keys do not correspond to the generation of secure RSA keys, which typically have two or three small prime factors and one large factor exceeding the value of sqrt(private key).
For example, puzzle #68 is cryptographically weak (I think whoever created the puzzles intentionally made them vulnerable, or they follow certain patterns which consequently made them insecure.): 219898266213316039825 -> 5, 5, 7, 1973, 4986139, 127729817 since its greatest prime divisor < sqrt(private key).
That is, we would first find the largest possible value for the Greatest Prime Divisor (GPD) that is < 2^34. Then, using the Miller-Rabin Primality Test, we would check each prime number in descending order within the range where: 2^67 < Private Key = GPD * number < 2^68 and consequently, 2^67 / GPD < number < 2^68 / GPD.
But given approach will not wrok if private key is a prime number as puzzle number 12, or GPD of a key is GPD>sqrt(private key).
Currently, I have a simple program written in Python that checks 5000 prime numbers per second. I think if it were optimized for a GPU, it could solve the vulnerable keys among the subsequent ones in a few hours.
I feel that your research is meaningless, as the private key of Bitcoin has nothing to do with the RSA encryption you mentioned. The private key of Bitcoin does not need to meet the requirements of RSA keys at all.
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WanderingPhilospher
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Shooters Shoot...
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December 30, 2025, 03:56:50 PM |
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Greetings, everyone. I recently learned about the BitPuzzle and it's pretty cool. On the forum, I explored several approaches to solving level 71—from statistical methods to mathematical computations that look for specific patterns. They are very diverse and all deserve respect. But I have a different question.
I'm currently using Fixed Paul. It works great and without errors. Special thanks to all the developers. With Fixed Paul, you can set the starting digits and a range, and search. For example: ./vanitysearch -gpuId 0 -start 7abcd0000000000000 -range 60 -o output.txt 1PWo3JeB9
But how can I use it if these are the last digits? Like this: ./vanitysearch -gpuId 0 -start 40000000000007abcd -range 60 -random -o output.txt 1PWo3JeB9 ...so that 7abcd remains unchanged?
ok, so this is your start, -start 40000000000007abcd, then what would be the end range to search in or are you wanting it to search in areas like: -start 40000000000007abcd -start 40000000000017abcd -start 40000000000027abcd -start 40000000000037abcd ... -start 7ffffade34c1f7abcd etc? Or, Would all private keys found that match the prefix end with 7abcd? I'd need more explanation.
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GrigoriyPerelman
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December 30, 2025, 04:38:15 PM |
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Greetings, everyone. I recently learned about the BitPuzzle and it's pretty cool. On the forum, I explored several approaches to solving level 71—from statistical methods to mathematical computations that look for specific patterns. They are very diverse and all deserve respect. But I have a different question.
I'm currently using Fixed Paul. It works great and without errors. Special thanks to all the developers. With Fixed Paul, you can set the starting digits and a range, and search. For example: ./vanitysearch -gpuId 0 -start 7abcd0000000000000 -range 60 -o output.txt 1PWo3JeB9
But how can I use it if these are the last digits? Like this: ./vanitysearch -gpuId 0 -start 40000000000007abcd -range 60 -random -o output.txt 1PWo3JeB9 ...so that 7abcd remains unchanged?
ok, so this is your start, -start 40000000000007abcd, then what would be the end range to search in or are you wanting it to search in areas like: -start 40000000000007abcd -start 40000000000017abcd -start 40000000000027abcd -start 40000000000037abcd ... -start 7ffffade34c1f7abcd etc? Or, Would all private keys found that match the prefix end with 7abcd? I'd need more explanation. simply -start 4000000000000 -range 49 + as last digits 7abcd (not mathemtical addition) is added.
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Bram24732
Member

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December 30, 2025, 04:39:48 PM |
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I don't want to engage in polemics with some forum members who believe that the only way to solve problems is brute force. Given the current state of computing power, this approach would take millennia or entail financial costs exceeding the prize. Essentially, they are suggesting to wait for technological progress or embark on a thousand-year journey. Any other approaches seem to offend them (apparently, they have already started down this path) and they react hysterically to such a position. Moreover, they promote this vision aggressively and tactlessly, insulting other opinions and approaches, much like an aggressive LGBT community. I think even monkey could solve it with a powerful GPU and the right program. Continuing to develop alternative approaches to problem-solving, I'm sharing one line of thought on how the task could be solved. Below is a list of private keys with their prime multipliers. 1 -> 1 3 -> 3 7 -> 7 8 -> 2, 2, 2 21 -> 3, 7 49 -> 7, 7 76 -> 2, 2, 19 224 -> 2, 2, 2, 2, 2, 7 467 -> 467 514 -> 2, 257 1155 -> 3, 5, 7, 11 2683 -> 2683 5216 -> 2, 2, 2, 2, 2, 163 10544 -> 2, 2, 2, 2, 659 26867 -> 67, 401 51510 -> 2, 3, 5, 17, 101 95823 -> 3, 3, 3, 3, 7, 13, 13 198669 -> 3, 47, 1409 357535 -> 5, 23, 3109 863317 -> 7, 13, 53, 179 1811764 -> 2, 2, 19, 31, 769 3007503 -> 3, 3, 3, 23, 29, 167 5598802 -> 2, 11, 254491 14428676 -> 2, 2, 19, 189851 33185509 -> 7, 4740787 54538862 -> 2, 7, 7, 556519 111949941 -> 3, 43, 867829 227634408 -> 2, 2, 2, 3, 3, 3, 1053863 400708894 -> 2, 83, 2413909 1033162084 -> 2, 2, 47, 5495543 2102388551 -> 19, 19, 43, 167, 811 3093472814 -> 2, 23, 3001, 22409 7137437912 -> 2, 2, 2, 11, 751, 107999 14133072157 -> 19, 41, 131, 138493 20112871792 -> 2, 2, 2, 2, 13, 96696499 42387769980 -> 2, 2, 3, 3, 5, 235487611 100251560595 -> 3, 5, 6683437373 146971536592 -> 2, 2, 2, 2, 60037, 153001 323724968937 -> 3, 3, 138319, 260047 1003651412950 -> 2, 5, 5, 20073028259 1458252205147 -> 23, 63402269789 2895374552463 -> 3, 3, 59, 5452682773 7409811047825 -> 5, 5, 587, 2903, 173933 15404761757071 -> 2783789, 5533739 19996463086597 -> 157, 193, 7477, 88261 51408670348612 -> 2, 2, 5839, 2201090527 119666659114170 -> 2, 3, 3, 3, 5, 7, 17, 89, 41847781 191206974700443 -> 3, 13, 4902742941037 409118905032525 -> 3, 3, 5, 5, 23, 197, 1663, 241313 611140496167764 -> 2, 2, 3, 3, 12211, 1390232159 2058769515153876 -> 2, 2, 3, 7, 43, 53, 197, 2477, 22039 4216495639600700 -> 2, 2, 5, 5, 53, 795565215019 6763683971478124 -> 2, 2, 14359547, 117755873 9974455244496707 -> 7019, 76123, 18668011 30045390491869460 -> 2, 2, 5, 19, 79066817083867 44218742292676575 -> 3, 3, 5, 5, 13, 15117518732539 138245758910846492 -> 2, 2, 23, 1002377, 1499107913 199976667976342049 -> 13, 167, 2511323, 36678953 525070384258266191 -> 307, 307, 5571097669559 1135041350219496382 -> 2, 13, 31, 71, 269, 587, 3637, 34537 1425787542618654982 -> 2, 13, 54837982408409807 3908372542507822062 -> 2, 3, 43, 62922991, 240750329 8993229949524469768 -> 2, 2, 2, 7, 251, 2383, 268491108091 17799667357578236628 -> 2, 2, 3, 19, 3761, 408229, 50847529 30568377312064202855 -> 5, 67, 5639, 16181749866767 46346217550346335726 -> 2, 13, 17, 104855695815263203 132656943602386256302 -> 2, 23, 3881, 743067920652377 219898266213316039825 -> 5, 5, 7, 1973, 4986139, 127729817 297274491920375905804 -> 2, 2, 11, 139, 48606032034070619 970436974005023690481 -> 3, 59, 1931, 255473, 11113907731
As you can see from the list, these keys do not correspond to the generation of secure RSA keys, which typically have two or three small prime factors and one large factor exceeding the value of sqrt(private key).
For example, puzzle #68 is cryptographically weak (I think whoever created the puzzles intentionally made them vulnerable, or they follow certain patterns which consequently made them insecure.): 219898266213316039825 -> 5, 5, 7, 1973, 4986139, 127729817 since its greatest prime divisor < sqrt(private key).
That is, we would first find the largest possible value for the Greatest Prime Divisor (GPD) that is < 2^34. Then, using the Miller-Rabin Primality Test, we would check each prime number in descending order within the range where: 2^67 < Private Key = GPD * number < 2^68 and consequently, 2^67 / GPD < number < 2^68 / GPD.
But given approach will not wrok if private key is a prime number as puzzle number 12, or GPD of a key is GPD>sqrt(private key).
Currently, I have a simple program written in Python that checks 5000 prime numbers per second. I think if it were optimized for a GPU, it could solve the vulnerable keys among the subsequent ones in a few hours.
Let’s put aside the fact that this is all trash. What the f does LGBT has to do with any of this ?
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WanderingPhilospher
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Shooters Shoot...
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December 30, 2025, 04:52:24 PM |
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Greetings, everyone. I recently learned about the BitPuzzle and it's pretty cool. On the forum, I explored several approaches to solving level 71—from statistical methods to mathematical computations that look for specific patterns. They are very diverse and all deserve respect. But I have a different question.
I'm currently using Fixed Paul. It works great and without errors. Special thanks to all the developers. With Fixed Paul, you can set the starting digits and a range, and search. For example: ./vanitysearch -gpuId 0 -start 7abcd0000000000000 -range 60 -o output.txt 1PWo3JeB9
But how can I use it if these are the last digits? Like this: ./vanitysearch -gpuId 0 -start 40000000000007abcd -range 60 -random -o output.txt 1PWo3JeB9 ...so that 7abcd remains unchanged?
ok, so this is your start, -start 40000000000007abcd, then what would be the end range to search in or are you wanting it to search in areas like: -start 40000000000007abcd -start 40000000000017abcd -start 40000000000027abcd -start 40000000000037abcd ... -start 7ffffade34c1f7abcd etc? Or, Would all private keys found that match the prefix end with 7abcd? I'd need more explanation. simply -start 4000000000000 -range 49 + as last digits 7abcd (not mathemtical addition) is added. I am taking what you said as: range = 4000000000000:500000007abcd So program would only search in that range, correct? Or am I missing something?
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GrigoriyPerelman
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December 30, 2025, 05:18:09 PM |
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I don't want to engage in polemics with some forum members who believe that the only way to solve problems is brute force. Given the current state of computing power, this approach would take millennia or entail financial costs exceeding the prize. Essentially, they are suggesting to wait for technological progress or embark on a thousand-year journey. Any other approaches seem to offend them (apparently, they have already started down this path) and they react hysterically to such a position. Moreover, they promote this vision aggressively and tactlessly, insulting other opinions and approaches, much like an aggressive LGBT community. I think even monkey could solve it with a powerful GPU and the right program. Continuing to develop alternative approaches to problem-solving, I'm sharing one line of thought on how the task could be solved. Below is a list of private keys with their prime multipliers. 1 -> 1 3 -> 3 7 -> 7 8 -> 2, 2, 2 21 -> 3, 7 49 -> 7, 7 76 -> 2, 2, 19 224 -> 2, 2, 2, 2, 2, 7 467 -> 467 514 -> 2, 257 1155 -> 3, 5, 7, 11 2683 -> 2683 5216 -> 2, 2, 2, 2, 2, 163 10544 -> 2, 2, 2, 2, 659 26867 -> 67, 401 51510 -> 2, 3, 5, 17, 101 95823 -> 3, 3, 3, 3, 7, 13, 13 198669 -> 3, 47, 1409 357535 -> 5, 23, 3109 863317 -> 7, 13, 53, 179 1811764 -> 2, 2, 19, 31, 769 3007503 -> 3, 3, 3, 23, 29, 167 5598802 -> 2, 11, 254491 14428676 -> 2, 2, 19, 189851 33185509 -> 7, 4740787 54538862 -> 2, 7, 7, 556519 111949941 -> 3, 43, 867829 227634408 -> 2, 2, 2, 3, 3, 3, 1053863 400708894 -> 2, 83, 2413909 1033162084 -> 2, 2, 47, 5495543 2102388551 -> 19, 19, 43, 167, 811 3093472814 -> 2, 23, 3001, 22409 7137437912 -> 2, 2, 2, 11, 751, 107999 14133072157 -> 19, 41, 131, 138493 20112871792 -> 2, 2, 2, 2, 13, 96696499 42387769980 -> 2, 2, 3, 3, 5, 235487611 100251560595 -> 3, 5, 6683437373 146971536592 -> 2, 2, 2, 2, 60037, 153001 323724968937 -> 3, 3, 138319, 260047 1003651412950 -> 2, 5, 5, 20073028259 1458252205147 -> 23, 63402269789 2895374552463 -> 3, 3, 59, 5452682773 7409811047825 -> 5, 5, 587, 2903, 173933 15404761757071 -> 2783789, 5533739 19996463086597 -> 157, 193, 7477, 88261 51408670348612 -> 2, 2, 5839, 2201090527 119666659114170 -> 2, 3, 3, 3, 5, 7, 17, 89, 41847781 191206974700443 -> 3, 13, 4902742941037 409118905032525 -> 3, 3, 5, 5, 23, 197, 1663, 241313 611140496167764 -> 2, 2, 3, 3, 12211, 1390232159 2058769515153876 -> 2, 2, 3, 7, 43, 53, 197, 2477, 22039 4216495639600700 -> 2, 2, 5, 5, 53, 795565215019 6763683971478124 -> 2, 2, 14359547, 117755873 9974455244496707 -> 7019, 76123, 18668011 30045390491869460 -> 2, 2, 5, 19, 79066817083867 44218742292676575 -> 3, 3, 5, 5, 13, 15117518732539 138245758910846492 -> 2, 2, 23, 1002377, 1499107913 199976667976342049 -> 13, 167, 2511323, 36678953 525070384258266191 -> 307, 307, 5571097669559 1135041350219496382 -> 2, 13, 31, 71, 269, 587, 3637, 34537 1425787542618654982 -> 2, 13, 54837982408409807 3908372542507822062 -> 2, 3, 43, 62922991, 240750329 8993229949524469768 -> 2, 2, 2, 7, 251, 2383, 268491108091 17799667357578236628 -> 2, 2, 3, 19, 3761, 408229, 50847529 30568377312064202855 -> 5, 67, 5639, 16181749866767 46346217550346335726 -> 2, 13, 17, 104855695815263203 132656943602386256302 -> 2, 23, 3881, 743067920652377 219898266213316039825 -> 5, 5, 7, 1973, 4986139, 127729817 297274491920375905804 -> 2, 2, 11, 139, 48606032034070619 970436974005023690481 -> 3, 59, 1931, 255473, 11113907731
As you can see from the list, these keys do not correspond to the generation of secure RSA keys, which typically have two or three small prime factors and one large factor exceeding the value of sqrt(private key).
For example, puzzle #68 is cryptographically weak (I think whoever created the puzzles intentionally made them vulnerable, or they follow certain patterns which consequently made them insecure.): 219898266213316039825 -> 5, 5, 7, 1973, 4986139, 127729817 since its greatest prime divisor < sqrt(private key).
That is, we would first find the largest possible value for the Greatest Prime Divisor (GPD) that is < 2^34. Then, using the Miller-Rabin Primality Test, we would check each prime number in descending order within the range where: 2^67 < Private Key = GPD * number < 2^68 and consequently, 2^67 / GPD < number < 2^68 / GPD.
But given approach will not wrok if private key is a prime number as puzzle number 12, or GPD of a key is GPD>sqrt(private key).
Currently, I have a simple program written in Python that checks 5000 prime numbers per second. I think if it were optimized for a GPU, it could solve the vulnerable keys among the subsequent ones in a few hours.
I feel that your research is meaningless, as the private key of Bitcoin has nothing to do with the RSA encryption you mentioned. The private key of Bitcoin does not need to meet the requirements of RSA keys at all. Seems like you are right. But knowing that Greatest Prime divisor <sqrt(private key) will give some advantages?
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Cricktor
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December 30, 2025, 07:30:38 PM |
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How does this make an advantage when the fundamental theorem of arithmetic states that every integer number greater than 1 is either a prime number itself or can be factorized as a product of prime numbers? This is applicable to every private key in the whole range of a particular puzzle. How do you want to exploit this to your advantage? Be more specific!
And the creator didn't make specific puzzles weaker than others, because if we take his words as true, all puzzles private keys were taken from a deterministic wallet (if taken from a specific sequence of keys is unknown and there was no specific detail stating it was a BIP39 HD wallet; we can assume this but have no proof).
The private keys appeared all as random numbers in the allowed range of Bitcoin private keys, little smaller than 2256. Those basically random numbers where then masked down to the ranges for each puzzle. I don't think you can distill some special properties of the masked private keys and have something to grasp on to exploit some magical procedure.
I mean, it would surprise me if it were possible. I don't say, I know all. Far from it.
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GrigoriyPerelman
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December 30, 2025, 09:25:25 PM |
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How does this make an advantage when the fundamental theorem of arithmetic states that every integer number greater than 1 is either a prime number itself or can be factorized as a product of prime numbers? This is applicable to every private key in the whole range of a particular puzzle. How do you want to exploit this to your advantage? Be more specific!
And the creator didn't make specific puzzles weaker than others, because if we take his words as true, all puzzles private keys were taken from a deterministic wallet (if taken from a specific sequence of keys is unknown and there was no specific detail stating it was a BIP39 HD wallet; we can assume this but have no proof).
The private keys appeared all as random numbers in the allowed range of Bitcoin private keys, little smaller than 2256. Those basically random numbers where then masked down to the ranges for each puzzle. I don't think you can distill some special properties of the masked private keys and have something to grasp on to exploit some magical procedure.
I mean, it would surprise me if it were possible. I don't say, I know all. Far from it.
Trial division principle: if a number has no prime divisor up to its square root, it is prime. For a composite key, there must be a prime factor ≤ sqrt(key). If we know the key is even, the search is twice easier — you check only even numbers (step +2). Here, instead of even/odd, I’m pointing out we should start searching from the largest possible prime divisor (GPD) just below sqrt(key) and go down. If the real GPD is far below sqrt(key), the search jumps quickly — like starting with the biggest possible step down, not checking small primes. It’s just trying to figure out the biggest jump.
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brainless
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Today at 07:48:43 AM |
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How does this make an advantage when the fundamental theorem of arithmetic states that every integer number greater than 1 is either a prime number itself or can be factorized as a product of prime numbers? This is applicable to every private key in the whole range of a particular puzzle. How do you want to exploit this to your advantage? Be more specific!
And the creator didn't make specific puzzles weaker than others, because if we take his words as true, all puzzles private keys were taken from a deterministic wallet (if taken from a specific sequence of keys is unknown and there was no specific detail stating it was a BIP39 HD wallet; we can assume this but have no proof).
The private keys appeared all as random numbers in the allowed range of Bitcoin private keys, little smaller than 2256. Those basically random numbers where then masked down to the ranges for each puzzle. I don't think you can distill some special properties of the masked private keys and have something to grasp on to exploit some magical procedure.
I mean, it would surprise me if it were possible. I don't say, I know all. Far from it.
Trial division principle: if a number has no prime divisor up to its square root, it is prime. For a composite key, there must be a prime factor ≤ sqrt(key). If we know the key is even, the search is twice easier — you check only even numbers (step +2). Here, instead of even/odd, I’m pointing out we should start searching from the largest possible prime divisor (GPD) just below sqrt(key) and go down. If the real GPD is far below sqrt(key), the search jumps quickly — like starting with the biggest possible step down, not checking small primes. It’s just trying to figure out the biggest jump. Puzzle creator already said in his message factor of 10 in range 0.1 to 0.001, I already said focus these lines, complete solution inside it
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13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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nomachine
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Today at 08:46:14 AM |
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Currently, I have a simple program written in Python that checks 5000 prime numbers per second. I think if it were optimized for a GPU, it could solve the vulnerable keys among the subsequent ones in a few hours.
Imagine trying to find a single, specific grain of sand across all the beaches on Earth. There are roughly 2⁶² to 2⁶³ grains of sand on all beaches combined. Puzzle 71 is like trying to search not just one Earth’s worth of sand, but 157.4 Earths’ worth (2⁷⁰). Even if you could check a billion grains per second, it would still take thousands of years to go through them all.  And that’s before you factor in the fact that GPU brute force doesn’t magically flatten an exponential search space. A single GPU doesn’t beat math. It just loses faster. 1 GPU = hopeless 10 GPUs = still hopeless 100 GPUs = noise 1,000 GPUs = maybe your great-great-grandkids hear back 10,000+ GPUs, tightly coordinated, executing a highly optimized, distributed Pollard’s kangaroo implementation.....
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BTC: bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8
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Wanderingaran
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Today at 09:09:22 AM |
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10,000+ GPUs, tightly coordinated
Who has 10k+ GPUs? Elon ? And even he’d be like: “Nice cluster. Still no.” 
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brainless
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Today at 09:10:53 AM |
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Currently, I have a simple program written in Python that checks 5000 prime numbers per second. I think if it were optimized for a GPU, it could solve the vulnerable keys among the subsequent ones in a few hours.
Imagine trying to find a single, specific grain of sand across all the beaches on Earth. There are roughly 2⁶² to 2⁶³ grains of sand on all beaches combined. Puzzle 71 is like trying to search not just one Earth’s worth of sand, but 157.4 Earths’ worth (2⁷⁰). Even if you could check a billion grains per second, it would still take thousands of years to go through them all.  And that’s before you factor in the fact that GPU brute force doesn’t magically flatten an exponential search space. A single GPU doesn’t beat math. It just loses faster. 1 GPU = hopeless 10 GPUs = still hopeless 100 GPUs = noise 1,000 GPUs = maybe your great-great-grandkids hear back 10,000+ GPUs, tightly coordinated, executing a highly optimized, distributed Pollard’s kangaroo implementation..... Sorry bro, could you tell us how much thousands year taken for solve 69 and 68 puzzle ?
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13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
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Wanderingaran
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Today at 09:14:03 AM |
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how much thousands year taken for solve 69 and 68 puzzle ?
I don't even believe that they were resolved in a normal way. That there is a conspiracy.
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nomachine
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Today at 09:31:15 AM |
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how much thousands year taken for solve 69 and 68 puzzle ?
I don't even believe that they were resolved in a normal way. That there is a conspiracy. Nobody can verify how a puzzle was solved. The blockchain only proves that a valid private key was found, nothing more. There’s no way anyone can verify how many GPUs or how long it took. Proving it would require a full dissertation, and even then it wouldn’t be conclusive. 
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BTC: bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8
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GrigoriyPerelman
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Today at 09:53:14 AM |
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How does this make an advantage when the fundamental theorem of arithmetic states that every integer number greater than 1 is either a prime number itself or can be factorized as a product of prime numbers? This is applicable to every private key in the whole range of a particular puzzle. How do you want to exploit this to your advantage? Be more specific!
And the creator didn't make specific puzzles weaker than others, because if we take his words as true, all puzzles private keys were taken from a deterministic wallet (if taken from a specific sequence of keys is unknown and there was no specific detail stating it was a BIP39 HD wallet; we can assume this but have no proof).
The private keys appeared all as random numbers in the allowed range of Bitcoin private keys, little smaller than 2256. Those basically random numbers where then masked down to the ranges for each puzzle. I don't think you can distill some special properties of the masked private keys and have something to grasp on to exploit some magical procedure.
I mean, it would surprise me if it were possible. I don't say, I know all. Far from it.
Trial division principle: if a number has no prime divisor up to its square root, it is prime. For a composite key, there must be a prime factor ≤ sqrt(key). If we know the key is even, the search is twice easier — you check only even numbers (step +2). Here, instead of even/odd, I’m pointing out we should start searching from the largest possible prime divisor (GPD) just below sqrt(key) and go down. If the real GPD is far below sqrt(key), the search jumps quickly — like starting with the biggest possible step down, not checking small primes. It’s just trying to figure out the biggest jump. Puzzle creator already said in his message factor of 10 in range 0.1 to 0.001, I already said focus these lines, complete solution inside it can you explain more detailed, please? who is creator? where he said? and how it can be factor of 10?
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