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Same here, two days ago I sent from Freebitco FUN to my wallet and I dont know nothing about, only green banner showed, do you finally get your FUN Token?
Finally, Today I received an email confirmation and received FUN tokens ✅ |
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I am in the same situation as you, I have not received my FUN tokens and I do not have any type of information about the transaction, without a doubt there are problems in freebitco
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I made a withdraw yesterday of a big portion of my FUN Token to MEXC. It was said I would have it ~1hr. It’s now almost 12hr without receiving my FUN, there’s no way to find Tx of withdrawal, they just disappeared without a trace and I guess I just lost those 8300ish token I was purchasing everytime I had enough sats..
Yes, I double checked the address I entered into the withdrawal address zone and entered my 2FA (so definitely, they received my request, I even had the green rectangle box message on Freebitco that was confirming everything was ok….
I tried to get an answer on Twitter, but they never answer there…… nor on the message box section on the website…
Kinda stressed because I have still like.. 5days on MEXC before I’m probably gonna be kicked out and my assets will be frozen since I’m canadian and we can’t use MEXC anymore at around the end of the month (I think it’s the 29th)
Can they refund me the token I never received even though I followed everything as it should’ve been done..? I check on MEXC, they have a form for when transactions didn’t deposit, but since Freebitco didn’t gave the TX nor have updated the history of FUN assets, I can’t even use that form from MEXC…
Also, this week I received a message on Freebitco saying they were having trouble sending me my premium newsletter (with my daily free spin) on my email, then they sent me a email to confirm my address was still ok, in email there was no link to confirm, only saying that if I received it, everything was alright and nothing else was required….. but still, I don’t receive anything since almost a week.
Is Freebitco trying to rip me off? First time in all those years of fidelity that they have multiple issues at once and not seems to be working on them.
Same here, two days ago I sent from Freebitco FUN to my wallet and I dont know nothing about, only green banner showed, do you finally get your FUN Token? |
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Hello, I would like to know if it is possible to divide 2 public keys between them.
ex: 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 / 02c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 with "secp256k1 as ice" or even something else?
I can give you the result for your example, here it is : Pub = 0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63 Private = 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1 But if we don't know the private key of at least one of the points, we can never directly reach any meaningful result. If we don't know k for one of the points, how can an algorithm/library know it? There are ways to indirectly divide 2 points by using a k greater than 1 and smaller than the other. It would be nonsense/off topic if I try to explain it, so I just save everybody the headache and cut the nonsense.😉
He is learning maths and programming, so I think is not a waste of time..
Hello to everyone
Agreed, we should never stop learning, for example learning how to quote someone properly, lol I mean OMG, look the mess you posted. Jokes aside, welcome to the jungle.🤝 LOL I havn’t posted on forums in a long time, since AIO’s time I would say. Give me sometime 😌 Something I had made 😁  |
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By the way, and you know why their puzzle addresses were removed from the 161st?
My thought is that this is the limit of public addresses. That is the further search in the theory goes on a circle.
Because all the funds from 161 through 255 were redundant (Bitcoin addresses only use 160 bits) the author of the puzzle moved all the funds from the 160-255 ranged down into the 1-160 range. It is explained up thread. Is it? I thought it uses almost 256 bits (the last possible private key is FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140) OH, HIII, Good morning from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('Here Compressed Public Key')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))
def ters(Qx,Scalar):
ScalarBin = str(bin(Scalar))[2:]
le=len(ScalarBin)
for i in range (1,le+1):
if ScalarBin[le-i] == "0":
Qx=ice.point_multiplication(k,Qx)
else:
Qx=ice.point_addition(Qx,neg1)
Qx=ice.point_multiplication(k,Qx)
return ice.point_to_cpub(Qx)
for x in range(1,65536):
f= (ters(pub,x))
data= open(“halfpub.txt”,”a”)
data.write(f+”\n”)
data.close() Note this is where you decide how many bits should be reduced for x in range(1,65536): For example reducing 26 bits requires 67108864 to be generated, 1 of them is the correct 26 bit reduced key. Will you release for public to use? Yes this is my code, I can write bit reduction code in another way if you want. If you want, you can reduce bits by subtraction. I'm developing a new algorithm for downgrading from a known bit range to the bit range I want. When I complete it by giving only 1 correct pubkey, I will share it here. @lordfrs, I was thinking, why do we need to keep all the 65556 keys if we are reducing 16 bits? One other thing, if we multiply one of the keys from the end of the output file by 65556 we should see our target, correct? I haven't thought about on how to calculate, since we are subtracting one dividing by 2, one result if multiplied by 2^16 should reach the target or close to target, so why not discarding 90% of the generated keys and start multiplying the remaining 10% by 2^16? Could you add such operation to your script? I guess if we could operate on private keys instead of public keys using your script, we could reach some results, ( as a test of course ) To test with scalar only, I use this script : from sympy import mod_inverse
N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
def ters(scalar):
k = mod_inverse(2, N)
scalar_bin = bin(scalar)[2:
bro please stop replying with those comments copy past from chatgpt you are doing nothing except replying and talk too much, you are doing nothing for real men, stop it bro, :) thank you
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He is learning maths and programming, so I think is not a waste of time.. Hello to everyone |
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