Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
subtract G 99 times from target, you will have 100 keys in total, then divide them all by 100, one of the results will definitely be the target/100. Then come here ask me what's next. But first I would need to test your script to see if it does what I said.😉
Hello, is this what you need?
Code:
import secp256k1 as ice
import bitcoin
target= "03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852"
print("Target:", target)
Target_upub= ice.pub2upub(target)
for i in range (100):
A= ice.scalar_multiplication(i)
B= ice.point_subtraction(Target_upub, A)
C= ice.to_cpub(B.hex())
D= bitcoin.divide(C, 100)
data = open("S-1.txt","a")
data.write(str(i)+" = "+str(C)+"\n")
data.close()
data = open("D-1.txt","a")
data.write(str(i)+" = "+str(D)+"\n")
data.close()
import bitcoin
target= "03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852"
print("Target:", target)
Target_upub= ice.pub2upub(target)
for i in range (100):
A= ice.scalar_multiplication(i)
B= ice.point_subtraction(Target_upub, A)
C= ice.to_cpub(B.hex())
D= bitcoin.divide(C, 100)
data = open("S-1.txt","a")
data.write(str(i)+" = "+str(C)+"\n")
data.close()
data = open("D-1.txt","a")
data.write(str(i)+" = "+str(D)+"\n")
data.close()
Hello.
Tell me how to remove this error?
No matter what I tried, it didn't work...
Code:
C:\BTCPazzle>test.py
Target: 0230210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b
Traceback (most recent call last):
File "C:\BTCPazzle>test.py", line 20, in <module>
data = open("D-1000.txt","a")
[color=red]PermissionError: [Errno 13] Permission denied:[/color] "D-1000.txt"
Target: 0230210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b
Traceback (most recent call last):
File "C:\BTCPazzle>test.py", line 20, in <module>
data = open("D-1000.txt","a")
[color=red]PermissionError: [Errno 13] Permission denied:[/color] "D-1000.txt"