Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 200. (Read 215272 times)

id love to help in any way i have 3 3070ti gpus i would constintly run 24/7 for someone if you can push me in the right direction if youd want to split the price if the chance came one of hit. in conclusion what im saying is im a script kiddie who cant code worth shit but this is too apealing not to learn how or help atleast. under the correct cicomstances id be willing to basically give 24\7 acess to my computing power "probably not much" to help in this if anyone wants to help let me know.P.s ive tried kangaroo its too confusing i cant even get it to work thats how "not hip " i am lol... best of luck to all though ive been reading this thread and multipule others about it for a few months now with the right teacher i feel like i could be somewhat of help.

Hello, I've been following this thread for a year. I stocked all of them, including many archives that were deleted from the internet, along with their source code.

I just came up with a new idea and I need a little help with it.

Do you have fast point extraction algorithm for python ?

Thanks..

Not: sorry for my bad english, i don't know english i use google translate

Using tools will take much more time, this needs to be solved by math and by hands.😉

uncomprewssed key for 66 1Pir2jKv35rpYgUh8dUxNL6JTTm6d7HX6U

Therefore, the uncompressed public key for the P2PKH address "1Pir2jKv35rpYgUh8dUxNL6JTTm6d7HX6U" is:  04ce42244bfa7731d7088ee2f20a72b0dd834a723d2b8f563bc58cfa5d5dc4911a2f8aae5c5b

could be wrong could be wright do what you want with the info im just tossing ideas.



does this look like itd do anything am i in the right path?

RIGHT!!!!!!! the man the myth the legend "the creator" would you kindly please give us a hint of some sort a tool tool to look into something anything

OMG !!! 😱😱😱😱😱 Yesterday, I was thinking that I should give up now, but this has turned into a great motivation. need only one message from the puzzle creator in this situation, it would be the icing on the cake 👍🏼 👍🏼

WOW!
Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳

There's the math comes again,

66 Bit puzzle  


the total number of keys in this range can be calculated by taking 16 to the power of 17 (i.e., 16^17), which is equal to 295,147,905,179,352,825,856.

So, there are a total of 295,147,905,179,352,825,856 keys in the given range.

If you have a speed of 343 million keys per second (Mkeys/s), the result's,

Total number of keys = 295,147,905,179,352,825,856
Time = Total number of keys / Speed

Substituting the values, we get:

Time = 295,147,905,179,352,825,856 / 343,000,000
Time = 859,488,566,568 seconds

Therefore, it would take approximately 859,488,566,568 seconds (or about 27,247 years) to generate all the possible keys in the given range at a speed of 343 million keys per second. This is a very long time and may not be practically feasible.

what if i have 5000 mkeys to scan those number of key ?
it would take approximately 59,029,581,036 seconds (or about 1,872 years) to generate all the possible keys in the given range at a speed of 5000 million keys per second, still lame haha  

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and there's other scenario to win the ticket haha

For a 66-bit puzzle, the key range would be from 0 to 2^66 - 1, which is:

0 to 7,922,816,251,426,433

The number of keys in this range is:

2^66 = 73,786,976,294,838,206,464

At a scanning speed of 343 Mkey/s, the time it would take to scan this entire range would be:

Time = Number of Keys / Keys per Second

Time = (2^66) / 343000000 = 215625224 seconds
= 3593754 minutes
= 59896 hours
= 2495 days
≈ 6.84 years

So, it would take approximately 6.84 years to scan the entire key range of a 66-bit puzzle at the given scanning speed.

If we were to randomly hit the private key in the middle of the range, then we would only need to scan half of the key range, which is:

2^65 = 36,893,488,147,419,103,232

Using the same formula as above, the time it would take to scan half of the key range would be:

Time = (2^65) / 343000000 = 107812612 seconds
= 1796877 minutes
= 29948 hours
= 1248 days
≈ 3.42 years

So, if we randomly hit the private key in the middle of the key range, it would take approximately 3.42 years to scan half of the key range at the given speed of 343 Mkey/s.


btw, my close friend is neat, he's building something on ASICs L3+ to give a proof to scan the puzzle for mean time haha.
let's see..

Everyone thinks he will be the lucky one and prefers to avoid having to share the X10 is made to encourage people to pool their power to demonstrate that even with a collective effort the 66 67 68 and 69 will be difficult to obtain otherwise it won't find away
Also give a good boost to this puzzle which is now 8 years old

Let's do it one more time. With that much money at stake, we should grow a huge community.

What domain do you own/have?

I had a pool running over 2 years ago, for #64. People say they will run their GPUs but we only had around 2-4 people running consistently.

What would be nice is to know everyone’s fire power, and then decide what tools to use, and best setup. But most get excited in the beginning and then fizzle out a week or so later when the key hasn’t been found lol.

ive also seen alot of your work sir you sir are a legend  indeed. im going based off things ive done myself as long with the math and ideas throught the thousands of posts ive read through multiple threads. if you do think anything is off please let me know i get what you mean about my first part im too braindead to explain all of it rn that i put into it. id love to hear any output that you or anyone else has to say. please stay open minded in the factt that everyone is still learning something at this point its about brining all of our knowlge together and seeing what we get like the other guy said i think its time for something new we cannot let history repeat itself.

Yeah your math is way off.

We already knew/know the starting point of 61 and 65 and their applicable ranges and private keys.

61 is in the 61 bit range.
A 1 followed by 15 zeros thru a 1 followed by 15 f’s
65 is in the 65 bit range.
A 1 followed by 16 zeros thru a 1 followed by 16 f’s.

For 66:
The starting range is a 2 followed by 16 zeros
The ending range is a 3 followed by 16 f’s
20000000000000000-3ffffffffffffffff

Yeah your math is way off.

We already knew/know the starting point of 61 and 65 and their applicable ranges and private keys.

61 is in the 61 bit range.
A 1 followed by 15 zeros thru a 1 followed by 15 f’s
65 is in the 65 bit range.
A 1 followed by 16 zeros thru a 1 followed by 16 f’s.

For 66:
The starting range is a 2 followed by 16 zeros
The ending range is a 3 followed by 16 f’s
20000000000000000-3ffffffffffffffff

I do not understand your logic (not knocking it, I just don't get it)
#66 should have a starting range of 20000000000000000 and an ending range of 3ffffffffffffffff; that's the way every puzzle/challenge bit has worked up to this point.

i have no idea if this helps if it does contribute at all think of me when your doing a line off my mothers butt when ur sitting with 200k


Based on the information given, we can derive the starting point for Puzzle 49 and use it to calculate the starting point for Puzzle 66.

Puzzle 49 would be (1/16.532318326723157) times Puzzle 53, which gives us a factor of 0.06056415335786299. Multiplying this factor with the starting point for Puzzle 53 (0x9c8c956992e7aaf45d) gives us a starting point for Puzzle 49 of 0x5e2c3f6f1a2c80486f.

To calculate the starting point for Puzzle 66, we need to multiply the starting point for Puzzle 65 times 40, which gives us 0x4248dbb048e3e05018. We can then use the starting point for Puzzle 49 and the factor between Puzzles 49 and 65 to calculate the starting point for Puzzle 66:

Puzzle 65 is 21.439643985050658 times Puzzle 61.
Puzzle 61 is 10.313427000231762 times Puzzle 57.
Puzzle 57 is 20.439417260448156 times Puzzle 53.
Puzzle 53 is 16.532318326723157 times Puzzle 49.
Puzzle 49 is 0.06056415335786299 times Puzzle 66.

Therefore, the starting point for Puzzle 66 would be (0x4248dbb048e3e05018)/(21.43964398505065810.31342700023176220.4394172604481560.06056415335786299)(0x5e2c3f6f1a2c80486f) which evaluates to 0x1a274a956a7d53d977. edit--- with a ending point of 0x3500000000000000000 "based off my guess" or logically 0x3600000000000000000