delta[trix]
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August 10, 2023, 03:47:28 PM |
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I have a small curiosity about these occasional small incoming transactions in wallet 66. Could they be some clue about the key's location, or is it just something random?
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Minase
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August 10, 2023, 03:58:17 PM |
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Not at all, just dust or some people sending small amounts. Have looked at the past addresses and no such hint was given
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delta[trix]
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August 10, 2023, 04:07:11 PM |
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Not at all, just dust or some people sending small amounts. Have looked at the past addresses and no such hint was given
But it seems to be something automated by the puzzle's creator itself, I looked at some transactions.
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citb0in
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August 10, 2023, 04:26:09 PM |
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But it seems to be something automated by the puzzle's creator itself, I looked at some transactions.
Be more specific, please
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_ _ _ __ _ _ _ __ |_) | / \ / |/ (_ / \ | \ / |_ |_) (_ |_) |_ \_/ \_ |\ __) \_/ |_ \/ |_ | \ __) --> citb0in Solo-Mining Group <--- low stake of only 0.001 BTC. We regularly rent about 5 PH/s hash power and direct it to SoloCK pool. Wanna know more? Read through the link and JOIN NOW
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james5000
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August 10, 2023, 05:16:41 PM |
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To mathematicians, there is a way to divide one point by another point on the elliptic curve.
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delta[trix]
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August 10, 2023, 05:17:39 PM |
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But it seems to be something automated by the puzzle's creator itself, I looked at some transactions.
Be more specific, please If you look at the transactions in wallet 66, and a few others, you will see that it appears to be something automated, sending small transactions with numbers that I believe might provide some clues in decimal/hexadecimal or binary code as to where the private key could be located. However, this is just an assumption.
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citb0in
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August 10, 2023, 07:17:14 PM |
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...you will see that it appears to be something automated, sending small transactions with numbers that I believe...
what makes you think that it's automated and not manually entered as a transaction? what do you mean by 'small transactions with numbers' in detail?
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_ _ _ __ _ _ _ __ |_) | / \ / |/ (_ / \ | \ / |_ |_) (_ |_) |_ \_/ \_ |\ __) \_/ |_ \/ |_ | \ __) --> citb0in Solo-Mining Group <--- low stake of only 0.001 BTC. We regularly rent about 5 PH/s hash power and direct it to SoloCK pool. Wanna know more? Read through the link and JOIN NOW
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digaran
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August 11, 2023, 12:24:23 AM |
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But it seems to be something automated by the puzzle's creator itself, I looked at some transactions.
Be more specific, please If you look at the transactions in wallet 66, and a few others, you will see that it appears to be something automated, sending small transactions with numbers that I believe might provide some clues in decimal/hexadecimal or binary code as to where the private key could be located. However, this is just an assumption. If you mean numbers 66 and 99, they are mirrors of each other, also converting 66 and 99 to hex twice you will get 2a for 66 and 3f for 99. Converting 66 to decimal is 102, and 99 is 153. Note that 6 and 9 are represented by each other in hex format, since there are only 6 alphabet letters, the remaining 9 digits have no alphabet representatives. So here is how it works : 0= f 1= e 2= d 3= c 4= b 5= a 6= 9 7= 8 8= 7 9= 6 It's a world of wonders this hexadecimal world!
To mathematicians, there is a way to divide one point by another point on the elliptic curve.
Are you asking or telling? Of course there is a way, look at my personal text, e.g. dividing n by 6 will give you this : 2aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa74727a26728c1ab49ff8651778090ae0 You can multiply any point by above key, and it will divide your key by 6 if it's divisible by 6 you will get a correct answer. Now if you divide a point by the key above, you will actually be multiplying your point by 6. Now if you divide a point by 4 and multiply the result by 2 you should naturally get half of your original point, if your point is divisible by 4 then your results are correct, otherwise you will have a much much bigger point than your original point. Essentially dividing a point by a number other than 2 will get you no where close if your key is not divisible by that number, you can try by dividing a number by 3, 4, 5, etc and if the fraction is something other than .5, then the result of point division will not be any where close, however as long as the fraction is .5, then you can always add n/your divisor to the result to get the correct result. For example, divide 23 by 2, you will get -n/2-12 as a result, and if you add n/2 to your result, you'll have -12. 101 on how to break cryptography by ~dig.😉
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james5000
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August 11, 2023, 12:48:45 AM |
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But it seems to be something automated by the puzzle's creator itself, I looked at some transactions.
Be more specific, please If you look at the transactions in wallet 66, and a few others, you will see that it appears to be something automated, sending small transactions with numbers that I believe might provide some clues in decimal/hexadecimal or binary code as to where the private key could be located. However, this is just an assumption. If you mean numbers 66 and 99, they are mirrors of each other, also converting 66 and 99 to hex twice you will get 2a for 66 and 3f for 99. Converting 66 to decimal is 102, and 99 is 153. Note that 6 and 9 are represented by each other in hex format, since there are only 6 alphabet letters, the remaining 9 digits have no alphabet representatives. So here is how it works : 0= f 1= e 2= d 3= c 4= b 5= a 6= 9 7= 8 8= 7 9= 6 It's a world of wonders this hexadecimal world!
To mathematicians, there is a way to divide one point by another point on the elliptic curve.
Are you asking or telling? Of course there is a way, look at my personal text, e.g. dividing n by 6 will give you this : 2aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa74727a26728c1ab49ff8651778090ae0 You can multiply any point by above key, and it will divide your key by 6 if it's divisible by 6 you will get a correct answer. Now if you divide a point by the key above, you will actually be multiplying your point by 6. Now if you divide a point by 4 and multiply the result by 2 you should naturally get half of your original point, if your point is divisible by 4 then your results are correct, otherwise you will have a much much bigger point than your original point. Essentially dividing a point by a number other than 2 will get you no where close if your key is not divisible by that number, you can try by dividing a number by 3, 4, 5, etc and if the fraction is something other than .5, then the result of point division will not be any where close, however as long as the fraction is .5, then you can always add n/your divisor to the result to get the correct result. For example, divide 23 by 2, you will get -n/2-12 as a result, and if you add n/2 to your result, you'll have -12. 101 on how to break cryptography by ~dig. for n = fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 print(toHex(round(n / 6)))
my result is 2aaaaaaaaaaaaa00000000000000000000000000000000000000000000000000  but what I meant is whether there is a way to divide a point on the curve x, y by another point on the curve x, y?
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digaran
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August 11, 2023, 01:55:58 AM Last edit: August 11, 2023, 06:51:16 AM by digaran |
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for n = fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 print(toHex(round(n / 6)))
my result is 2aaaaaaaaaaaaa00000000000000000000000000000000000000000000000000  but what I meant is whether there is a way to divide a point on the curve x, y by another point on the curve x, y? Your result is not correct because you should do the math mod n to have the correct result. Anyways, point by point multiplication and division without knowing at least one point's private key is impossible, why else they call it crypto-graphy for? If it was possible directly, we all could break ECC easily. Ps, I'm not a mathematician, but I'll find a way to divide point by point or die trying! 🤣
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james5000
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August 11, 2023, 02:02:52 AM |
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for n = fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 print(toHex(round(n / 6)))
my result is 2aaaaaaaaaaaaa00000000000000000000000000000000000000000000000000  but what I meant is whether there is a way to divide a point on the curve x, y by another point on the curve x, y? Your result is not correct because n is not divisible by 5, you should do the math mod n to have the correct result. Anyways, point by point multiplication and division without knowing at least one point's private key is impossible, why else they call it crypto-graphy for? If it was possible directly, we all could break ECC easily. Ps, I'm not a mathematician, but I'll find a way to divide point by point or die trying! 🤣 work with me 🤣🤣
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lordfrs
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August 11, 2023, 07:59:37 AM |
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for n = fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 print(toHex(round(n / 6)))
my result is 2aaaaaaaaaaaaa00000000000000000000000000000000000000000000000000  but what I meant is whether there is a way to divide a point on the curve x, y by another point on the curve x, y? Your result is not correct because n is not divisible by 5, you should do the math mod n to have the correct result. Anyways, point by point multiplication and division without knowing at least one point's private key is impossible, why else they call it crypto-graphy for? If it was possible directly, we all could break ECC easily. Ps, I'm not a mathematician, but I'll find a way to divide point by point or die trying! 🤣 work with me 🤣🤣 If the private key of a point is known, you can divide the other point, for this you take the inverse of the private key of the known point and multiply by the point, multiplying with the inverse is equal to dividing the point. Not: My native language is not english i translate from google sorry for the translation errors
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If you want to buy me a coffee
Btc = 3246y1G9YjnQQNRUrVMnaeCFrymZRgJAP7
Doge = DGNd8UTi8jVTVZ2twhKydyqicynbsERMjs
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zahid888
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August 11, 2023, 09:57:41 AM |
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Puzzle: 30 Zeros: 14 Ones: 16 Percent 0: 46.67% Percent 1: 53.33% Decimal: 1033162084 | Binary: 111101100101001100110101100100 Puzzle: 60 Zeros: 28 Ones: 32 Percent 0: 46.67% Percent 1: 53.33% Decimal: 1135041350219496382 | Binary: 111111000000011110100001100000100101001101100111101110111110
Puzzle: 32 Zeros: 17 Ones: 15 Percent 0: 53.12% Percent 1: 46.88% Decimal: 3093472814 | Binary: 10111000011000101010011000101110 Puzzle: 64 Zeros: 34 Ones: 30 Percent 0: 53.12% Percent 1: 46.88% Decimal: 17799667357578236628 | Binary: 1111011100000101000111110010011110110000100100010001001011010100
Puzzle: 33 Zeros: 17 Ones: 16 Percent 0: 51.52% Percent 1: 48.48% Decimal: 7137437912 | Binary: 110101001011011001010100011011000 Puzzle: 66 Zeros: 34 Ones: 32 Percent 0: 51.52% Percent 1: 48.48% Decimal: ? ? ? ? ? ? ? | Binary: ? ? ? ? ? ? ? ☝☝ ☝☝ ☝☝ ☝☝☝☝ ☝☝☝☝ Little Prediction according to previous binaries
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1BGvwggxfCaHGykKrVXX7fk8GYaLQpeixA
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bestie1549
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August 11, 2023, 10:55:31 AM |
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Puzzle: 30 Zeros: 14 Ones: 16 Percent 0: 46.67% Percent 1: 53.33% Decimal: 1033162084 | Binary: 111101100101001100110101100100 Puzzle: 60 Zeros: 28 Ones: 32 Percent 0: 46.67% Percent 1: 53.33% Decimal: 1135041350219496382 | Binary: 111111000000011110100001100000100101001101100111101110111110
Puzzle: 32 Zeros: 17 Ones: 15 Percent 0: 53.12% Percent 1: 46.88% Decimal: 3093472814 | Binary: 10111000011000101010011000101110 Puzzle: 64 Zeros: 34 Ones: 30 Percent 0: 53.12% Percent 1: 46.88% Decimal: 17799667357578236628 | Binary: 1111011100000101000111110010011110110000100100010001001011010100
Puzzle: 33 Zeros: 17 Ones: 16 Percent 0: 51.52% Percent 1: 48.48% Decimal: 7137437912 | Binary: 110101001011011001010100011011000 Puzzle: 66 Zeros: 34 Ones: 32 Percent 0: 51.52% Percent 1: 48.48% Decimal: ? ? ? ? ? ? ? | Binary: ? ? ? ? ? ? ? ☝☝ ☝☝ ☝☝ ☝☝☝☝ ☝☝☝☝ Little Prediction according to previous binaries
you skipped this part Puzzle: 31 Zeros: 10 Ones: 21 Percent 0: 32.26% Percent 1: 67.74% Decimal: 2102388551 | Binary: 1111101010011111110011101000111 Puzzle: 62 Zeros: 28 Ones: 34 Percent 0: 45.16% Percent 1: 54.84% Decimal: 3908372542507822062 | Binary: 11011000111101010101000001111010110110000100011010101111101110
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zahid888
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August 11, 2023, 11:38:46 AM |
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Puzzle: 30 Zeros: 14 Ones: 16 Percent 0: 46.67% Percent 1: 53.33% Decimal: 1033162084 | Binary: 111101100101001100110101100100 Puzzle: 60 Zeros: 28 Ones: 32 Percent 0: 46.67% Percent 1: 53.33% Decimal: 1135041350219496382 | Binary: 111111000000011110100001100000100101001101100111101110111110
Puzzle: 32 Zeros: 17 Ones: 15 Percent 0: 53.12% Percent 1: 46.88% Decimal: 3093472814 | Binary: 10111000011000101010011000101110 Puzzle: 64 Zeros: 34 Ones: 30 Percent 0: 53.12% Percent 1: 46.88% Decimal: 17799667357578236628 | Binary: 1111011100000101000111110010011110110000100100010001001011010100
Puzzle: 34 Zeros: 18 Ones: 16 Percent 0: 52.94% Percent 1: 47.06% Decimal: 14133072157 | Binary: 1101001010011001011001000100011101 Puzzle: 68 Zeros: 36 Ones: 32 Percent 0: 52.94% Percent 1: 47.06% Decimal: ? ? ? ? ? ? ? | Binary: ? ? ? ? ? ? ? ☝☝ ☝☝ ☝☝ ☝☝☝☝ ☝☝☝☝ including skip pattern.. lets skip 33 also
you skipped this part Puzzle: 31 Zeros: 10 Ones: 21 Percent 0: 32.26% Percent 1: 67.74% Decimal: 2102388551 | Binary: 1111101010011111110011101000111 Puzzle: 62 Zeros: 28 Ones: 34 Percent 0: 45.16% Percent 1: 54.84% Decimal: 3908372542507822062 | Binary: 11011000111101010101000001111010110110000100011010101111101110
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1BGvwggxfCaHGykKrVXX7fk8GYaLQpeixA
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frozenen
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August 11, 2023, 02:10:07 PM |
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zahid888 before you posted that my guess was the same : my guess: #63 0=27 correct 1=36 #64 0=34 correct 1=30 #65 0=36 correct 1=29 #66 0=34 guess 1=32 #67 0=40 guess 1=27 #68 0=33 guess 1=35 #69 0=32 guess 1=37 #70 0=36 correct 1=34
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bestie1549
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August 11, 2023, 03:36:11 PM |
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zahid888 before you posted that my guess was the same : my guess: #63 0=27 correct 1=36 #64 0=34 correct 1=30 #65 0=36 correct 1=29 #66 0=34 guess 1=32 #67 0=40 guess 1=27 #68 0=33 guess 1=35 #69 0=32 guess 1=37 #70 0=36 correct 1=34
Maybe James should publish the GPU code for the 1s and 0s searching and also make it possible for us to be able to arrange the positioning of the second 1s where we don't need to start all over again when the program stops working due to power loss or something and also make it in such a way that it program can continue from where it stopped so it doesn't have to keep doing the same thing over and over again. the code has so much potentials but we are going to give it a try too
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lordfrs
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August 11, 2023, 05:43:42 PM |
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zahid888 before you posted that my guess was the same : my guess: #63 0=27 correct 1=36 #64 0=34 correct 1=30 #65 0=36 correct 1=29 #66 0=34 guess 1=32 #67 0=40 guess 1=27 #68 0=33 guess 1=35 #69 0=32 guess 1=37 #70 0=36 correct 1=34
Maybe James should publish the GPU code for the 1s and 0s searching and also make it possible for us to be able to arrange the positioning of the second 1s where we don't need to start all over again when the program stops working due to power loss or something and also make it in such a way that it program can continue from where it stopped so it doesn't have to keep doing the same thing over and over again. the code has so much potentials but we are going to give it a try too import math n = 66 k = 32 combination = math.factorial(n) / (math.factorial(k) * math.factorial(n - k)) print(combination) 7007092303604022272 combination 2^62.6.....
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If you want to buy me a coffee
Btc = 3246y1G9YjnQQNRUrVMnaeCFrymZRgJAP7
Doge = DGNd8UTi8jVTVZ2twhKydyqicynbsERMjs
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bestie1549
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August 11, 2023, 06:33:06 PM |
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zahid888 before you posted that my guess was the same : my guess: #63 0=27 correct 1=36 #64 0=34 correct 1=30 #65 0=36 correct 1=29 #66 0=34 guess 1=32 #67 0=40 guess 1=27 #68 0=33 guess 1=35 #69 0=32 guess 1=37 #70 0=36 correct 1=34
Maybe James should publish the GPU code for the 1s and 0s searching and also make it possible for us to be able to arrange the positioning of the second 1s where we don't need to start all over again when the program stops working due to power loss or something and also make it in such a way that it program can continue from where it stopped so it doesn't have to keep doing the same thing over and over again. the code has so much potentials but we are going to give it a try too import math n = 66 k = 32 combination = math.factorial(n) / (math.factorial(k) * math.factorial(n - k)) print(combination) 7007092303604022272 combination 2^62.6..... 7,007,092,303,604,022,272/50,000,000,000,000/86,400=1.62 days with just 50 Trillion keys/s We go faster than that on keyhunt BSGS if you have sufficient amount of RAM and a good number of CPU in your machine you can even go as fast as 500 Trillion
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james5000
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August 11, 2023, 06:54:04 PM |
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zahid888 before you posted that my guess was the same : my guess: #63 0=27 correct 1=36 #64 0=34 correct 1=30 #65 0=36 correct 1=29 #66 0=34 guess 1=32 #67 0=40 guess 1=27 #68 0=33 guess 1=35 #69 0=32 guess 1=37 #70 0=36 correct 1=34
Maybe James should publish the GPU code for the 1s and 0s searching and also make it possible for us to be able to arrange the positioning of the second 1s where we don't need to start all over again when the program stops working due to power loss or something and also make it in such a way that it program can continue from where it stopped so it doesn't have to keep doing the same thing over and over again. the code has so much potentials but we are going to give it a try too import math n = 66 k = 32 combination = math.factorial(n) / (math.factorial(k) * math.factorial(n - k)) print(combination) 7007092303604022272 combination 2^62.6..... 7,007,092,303,604,022,272/50,000,000,000,000/86,400=1.62 days with just 50 Trillion keys/s We go faster than that on keyhunt BSGS if you have sufficient amount of RAM and a good number of CPU in your machine you can even go as fast as 500 Trillion This calculation is incorrect, all possible combinations are larger when the ratio of 1s and 0s is 50%, if 1 or 0 is smaller, the total decreases.
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