There are only three ways to divide any number by 3.
I'll call them A B C.

A= the normal division of the number.
B= applying my script
C= adding + 1 and dividing.
Since you don't know what the pk is, you must apply A, B, C and one of the three will always be correct.
So you want to apply another division of 3 to the result.
you will get 9 pub one of them will be correct.
and so on..

3**X is the final amount of pubkeys where X is the number of times to divide the result (and of that result only one will be correct)

I took the number 150 just as an example.

For example Puzzle #65:

target = 30568377312064202855
target_2 = 30568377312064202855+1 #= 30568377312064202856

t1= target/3 #= 10189459104021400951.666666666666667
t2= target_2/3 #= 10189459104021400952

r= t1+t2 # = 20378918208042801903.666666666666667

30568377312064202855 − 20378918208042801903.666666666666667 = 10189459104021400951.333333333333333


I cannot understand how this method will help in solving the puzzle if the "target" is unknown to us.

in Python the // operator stands for floor division, that means it divides the first number by the second number and rounds the result down to the nearest integer.

If you're going to divide a point by 2, you multiply it by the inverse of 2 by N.

I'm bored, nothing interesting is happening around these woods no more.  Nobody has new and exciting ideas no more.

Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?

And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.

I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?

Good luck to all.😴

Speaking of spamming the topic with unimportant content!

When you guys realize that in order to find a key based on address alone, first you'd need to break rmd160, then sha256 to get the public key and then find the private key based on the public key, since only public key and private key are directly related mathematically.

What you should do instead now that you are using your time, you should find any relation between similar addresses and their public/ private keys with  their sha256 hashes, at least this way you are one step closer to solving 66 etc, because you won't have to face rmd160.

However, trying to do that will require more power and time than solving the puzzle #160.

So don't waste your time on something that difficult, try finding ways to go around the EC rules and break/ bend them either completely or partially.


For example, this is a method to partially break EC :

Divide a point by e.g, 5, then divide n by 5, and then add the result of p/5 to n/5, if your point is not divisible by 5 depending on the last digit of the private key, the division result differs.

Demonstration :
823 / 5 = 164.6
Now all you need is to extract 0.6 of n in secp256k1 and keep it for next operation, then after dividing 823 by 5, you just divide n by 5 and add the result of 823/5 to n/5 + 6/10 of n = 164

But what happens if we divide 822 / 5 = 164.4, just follow the previous steps this time extract 4/10 of n to add to your second result.

We just broke EC by simple math operations, though when your key is large and unknown, the fraction could be any percentage of n, but if you divide by 2 up to 9, since any key can only have 0 up to 9 on it's last digit, 1 out of 10 division + n/th + n/remainder will work.

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I could never find these things out if I was still working on addresses and hashes.😉

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You can use key subtracter by alberto, https://bitcointalk.org/index.php?topic=5360656.0

You could use a known public key and use -a - s with -r 0:f  -n 30 to have 15 + and -15 of your known public key, however use -z and -x to have only x coordinates, then you can convert them all to decimal.

Write a separate section of the function for generating the list and call it in your main function.

This doesn't call for a debate though...
Let us collectively achieve the goal we're all hoping to achieve in this forum.
By the way, does anyone know how to write codes for AntMiners?
I suggest the AntMiners are way faster than the GPUs in computational tasks.
We can solve 66, 67, 68 and 69 in one night if someone can write something to work on them
these machines are not profitable for now, so we need to gear them towards something that's worth it... like solving these puzzles
Let me know if anyone here knows how we can achieve that

I don't understand newbie preposition here! If you are referring me as new BIE than, I am happy to be it or even born bie if you like it! But the thing is we are all here making an effort to achieve a certain collective goal i guess. Someone posted logic, when I read it, I simply posted that Script that I developed probably 8 months ago, way before joining this forum! Well that didn't make me newbie there did it? Anyway, I shared my script, its simplest possible reasoning and logic behind,, the reason I said why digaran suggested formula won't work, the reason I commented on it because i tested it and tested it really hard not just spinning around! So basically I wanted to save someone's time pursuing something like that which has already proven to be waste of time, so that one can focus on other fresh ideas instead of wasting time on old ones!!! BTW how long one remains here a newBIE?