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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 25. (Read 230658 times)

newbie
Activity: 9
Merit: 0
What about 135 puzzle? I have managed to reduce 135 bits down to 120 bits how long would it take?.

Why did you stop the reducing at 120 bits? I'd go full-blown to 1 bit. Let us know if it's a zero or not.

The possible public keys exponentially grow.. By the time i reduce 3 digits from the end if have 1 trillion plus possible public keys
member
Activity: 499
Merit: 38
I really want someone to work with..

Ask @kTimesG for that. He has the software, and you have the hardware. Good luck!
member
Activity: 165
Merit: 26
What about 135 puzzle? I have managed to reduce 135 bits down to 120 bits how long would it take?.

Why did you stop the reducing at 120 bits? I'd go full-blown to 1 bit. Let us know if it's a zero or not.
newbie
Activity: 9
Merit: 0
Haha, well, let's just say that 12 hours is like giving a Kangaroo a rocket boost but asking it to hop across the Sahara Desert! There's a chance we could hit the target, but the odds are a bit like finding a needle in the multiverse – still, we’ll have the universe’s fastest Kangaroo on our side.

Let's go for it, and if we solve it, it'll be legendary. And if not? We'll have an epic tale of trying to tame the impossible with 900 GPUs. Either way, it’s going to be one wild hop about 1.7% of the total search space of puzzle 135! Grin

Any chance people have tried using point division? I mean how effectively can it be used?
I have a way to reduce the bit size using point division I need a partner team work will definitely pay off and i have the resources so when we know we are at the right place we can use the resources.. I really want someone to work with..
member
Activity: 499
Merit: 38
Haha, well, let's just say that 12 hours is like giving a Kangaroo a rocket boost but asking it to hop across the Sahara Desert! There's a chance we could hit the target, but the odds are a bit like finding a needle in the multiverse – still, we’ll have the universe’s fastest Kangaroo on our side.

Let's go for it, and if we solve it, it'll be legendary. And if not? We'll have an epic tale of trying to tame the impossible with 900 GPUs. Either way, it’s going to be one wild hop about 1.7% of the total search space of puzzle 135! Grin
newbie
Activity: 9
Merit: 0
Hello guys.. My cousin leads a research team and they have a huge set up.. They have about 900 rtx4090 gpus for research.. I have been trying to convince him to grant me permission to use the set for 12 hours straight.. And finally did... Now I need a strategy to figure out the puzzle 135 or 67 which one would work fastest? I need someone to give me a plan and once it's solved I will give the person  who gave me a solid plan a reward.. I need someone who is just as passionate as I am so work with in solving atleast one puzzle...

1. This is steeling and your cousin can get into trouble.
2. With a single rtx4090 you can achieve about 4000MKey/sec. So with 900 GPUs in 12 hour you have one in 2**66/4_000_000_000/900/3600/12 = 474.45 chance to solve Puzzle 67. That is about 0.21% chance.

What about 135 puzzle? I have managed to reduce 135 bits down to 120 bits how long would it take?.
newbie
Activity: 20
Merit: 0
Hello guys.. My cousin leads a research team and they have a huge set up.. They have about 900 rtx4090 gpus for research.. I have been trying to convince him to grant me permission to use the set for 12 hours straight.. And finally did... Now I need a strategy to figure out the puzzle 135 or 67 which one would work fastest? I need someone to give me a plan and once it's solved I will give the person  who gave me a solid plan a reward.. I need someone who is just as passionate as I am so work with in solving atleast one puzzle...

1. This is steeling and your cousin can get into trouble.
2. With a single rtx4090 you can achieve about 4000MKey/sec. So with 900 GPUs in 12 hour you have one in 2**66/4_000_000_000/900/3600/12 = 474.45 chance to solve Puzzle 67. That is about 0.21% chance.
newbie
Activity: 9
Merit: 0
Hello guys.. My cousin leads a research team and they have a huge set up.. They have about 900 rtx4090 gpus for research.. I have been trying to convince him to grant me permission to use the set for 12 hours straight.. And finally did... Now I need a strategy to figure out the puzzle 135 or 67 which one would work fastest? I need someone to give me a plan and once it's solved I will give the person  who gave me a solid plan a reward.. I need someone who is just as passionate as I am so work with in solving atleast one puzzle...
member
Activity: 165
Merit: 26
So, you who criticize everything, what have you achieved?

It' anyone's right to criticize anything, from movies to food. But this isn't why we have junk movies and junk food. Otherwise we'd all be couch potatoes watching the Kardashians on repeat, in lack of anything better.

In your posts, you only refer to more computing power = more keys generated, but I don’t see anything we haven’t known for decades.

Because you didn't look.

2011 - https://eprint.iacr.org/2010/617.pdf
2012 - https://cr.yp.to/dlog/cuberoot-20120919.pdf
2021 - https://gcc.episciences.org/7180/pdf

Last one is ongoing and aims to break ECDLP. Where is YOUR paper?

Fermat’s Last Theorem...

Conjectures that were proven to be theorems 350 years later had a nice property attached to them: they also worked for the 350 years prior to them being proved. So this is the difference between a speculation that works and one that doesn't.

Without the pursuit of solving this “insignificant” problem for practical purposes over 300 years, technology would not be what it is today.

What do I mean by this? The beauty of math is that even when we fail, we make progress.

Sure. But all of these only make sense when it is under the umbrella of rational thinking.

Something that is proven to be irrational cannot be later proven to be rational. But a mad man will of course keep believing he is right, in his own rationality.

A true scientist is one who does not impose their beliefs on others as absolute truth, and an intelligent person knows this.

Beliefs != Facts

And I'm not a scientist, nor imposing MY facts. Facts are just facts. I'm on the very practical side of things. But I will never be able to add two numbers if a computer already tells me the result, would I? So it's not a matter of opinion if some code stinks, or if some algorithm fails to address the actual problem, as you do really seem to miss the point at its core.

I don't need to care how it works...

I stand 100% by my point - I DO NOT NEED TO KNOW HOW IT WORKS.
It is not an attack, it is a clear fact.

the other guy's database

Ah, and he is "@Digaran" the joke is told alone.

Well, since I'm "digaran", you're clearly "the other guy".
That is fair-play.
hero member
Activity: 630
Merit: 731
Bitcoin g33k
I never trust other people's code. Only the programs I write myself are the most stable.(linux) Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin

There are faster and more efficient tools out there. Python’s great for a lot of things, but for this? Not so much.

But he has the most stable (linux) programs he writes himself! ChatGPT must be envy of his code, searching for multiple matches between uncompressed and compressed addresses. Really stable stuff. If we combine this with COBRAS's 50000 public key division "how to get y" breakthroughs and the other guy's database, we might turn that needle into a giant Dune worm. But then it might be the one who finds us, we're digging into dangerous sands.

Meanwhile, 3Emi...YESs probably already spinning up his ASICs for 135.

So, you who criticize everything, what have you achieved? In your posts, you only refer to more computing power = more keys generated, but I don’t see anything we haven’t known for decades.

Fermat’s Last Theorem, although it may seem like an abstract problem with no direct practical applications, has had a significant impact on the development of mathematics. The quest for its proof led to advances in areas such as number theory, algebraic geometry, and the theory of automorphic forms.

Without the pursuit of solving this “insignificant” problem for practical purposes over 300 years, technology would not be what it is today.

What do I mean by this? The beauty of math is that even when we fail, we make progress.

A true scientist is one who does not impose their beliefs on others as absolute truth, and an intelligent person knows this.

Math lesson of the day: “humility.”

It’s unfortunate that instead of fostering a productive discussion your response seems to miss the mark and resorts to personal attacks. Let’s take a moment to set the record straight.

First, the person you’re criticizing has demonstrated a deep understanding of the topics he discuss, consistently backing up his points with logical reasoning, facts, and clear examples. His contributions have been informative and accessible which is the hallmark of someone who not only understands the subject but also knows how to communicate complex ideas effectively. This is invaluable in any discussion especially one involving technical topics like mathematics or computing power.

Your point about Fermat’s Last Theorem is certainly valid. Yes, theoretical pursuits often lead to unexpected practical advances ... this is one of the most beautiful aspects of mathematics. But it seems you’ve mischaracterized the person you’re addressing. Nowhere did he dismiss theoretical pursuits as "insignificant." Instead, he has focused on the practical application of increased computing power, which is a factual statement: more computing power does indeed generate more possibilities in key generation and encryption. This isn't a reductionist view of mathematics but an acknowledgment of how certain technological advancements operate.

The tone of your message suggests that you feel entitled to define what constitutes "real" scientific contribution. But a true scientist doesn’t rely on condescension or unfounded accusations. They engage with ideas based on merit and not by attempting to discredit others through personal attacks.

If you disagree with his perspective, by all means, engage with his arguments directly. However, resorting to accusations without offering your own valuable insights or knowledge weakens your position. Disagreement is welcome in any intellectual conversation but respect and humility... well ... values you mention, should go both ways.

Let’s elevate the conversation. Criticism can be constructive when it’s backed by evidence, but dismissing someone’s contributions without adding substance of your own is hardly the way forward.

I much appreciate the contribution of the person you're attacking and hope he'll keep on ...
newbie
Activity: 20
Merit: 0

I'm already provide good scrypt,in my previous post but no one interested. I was little modify and script was moree good, but no one interested I mot provide  fool modification code.

What a smart move. You noticed no one is interested. You were ask no to do so. So what are you doing? Spamming more sh*tty codes and lucky numbers.
member
Activity: 873
Merit: 22
$$P2P BTC BRUTE.JOIN NOW ! https://uclck.me/SQPJk
Meanwhile, 3Emi...YESs probably already spinning up his ASICs for 135.

Everybody can see that the same person has solved three puzzles in a row, but still there are people who think that they can solve #135 faster.
Well, good luck  Grin


no way  connect asick to pc with needed speed


you can do same with modify my previous scrypt, from my previous message:


priv 130:

0x367a476b147e734ceb0cabff07f69404e


I get number 0x99574916d09aa0


and get priv from this 0x99574916d09aa0 number :

0x367a476b147e733252d67bd6f96f96f97


DIFFERENCE between start priv and what I generate:

0x367a476b147e734ceb0cabff07f69404e - 0x367a476b147e733252d67bd6f96f96f97 = 0x1a983630280e86fd0b7  Grin Tongue

member
Activity: 873
Merit: 22
$$P2P BTC BRUTE.JOIN NOW ! https://uclck.me/SQPJk
I never trust other people's code. Only the programs I write myself are the most stable.(linux) Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin

There are faster and more efficient tools out there. Python’s great for a lot of things, but for this? Not so much.

But he has the most stable (linux) programs he writes himself! ChatGPT must be envy of his code, searching for multiple matches between uncompressed and compressed addresses. Really stable stuff. If we combine this with COBRAS's 50000 public key division "how to get y" breakthroughs and the other guy's database, we might turn that needle into a giant Dune worm. But then it might be the one who finds us, we're digging into dangerous sands.

Meanwhile, 3Emi...YESs probably already spinning up his ASICs for 135.

I'm already provide good scrypt,in my previous post but no one interested. I was little modify and script was moree good, but no one interested I mot provide  fool modification code.
member
Activity: 165
Merit: 26
I never trust other people's code. Only the programs I write myself are the most stable.(linux) Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin

There are faster and more efficient tools out there. Python’s great for a lot of things, but for this? Not so much.

But he has the most stable (linux) programs he writes himself! ChatGPT must be envy of his code, searching for multiple matches between uncompressed and compressed addresses. Really stable stuff. If we combine this with COBRAS's 50000 public key division "how to get y" breakthroughs and the other guy's database, we might turn that needle into a giant Dune worm. But then it might be the one who finds us, we're digging into dangerous sands.

Meanwhile, 3Emi...YESs probably already spinning up his ASICs for 135.
member
Activity: 499
Merit: 38
I never trust other people's code. Only the programs I write myself are the most stable.(linux) Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin

You're about to embark on a Python journey to solve BTC ranges from 67 to 160, right? Well, get ready for a thrilling race… at turtle speed. 🐢 Let’s be honest: this task is like trying to find a single needle in the Grand Canal of China using chopsticks, while blindfolded, during a sandstorm, with mittens on. Sure, Python is easy to learn—perfect for beginners, but this task? Your CPU might become a toaster before you even scratch the surface of that range.

Even if you add a single GPU, it's like giving roller skates to a sloth. Still slow, just now with wheels. 🔥 Yes, technically, there's a chance you'll find something, but that chance is about as likely as getting struck by lightning while holding a winning lottery ticket in one hand and a four-leaf clover in the other.

There are faster and more efficient tools out there. Python’s great for a lot of things, but for this? Not so much. You’d have better luck training a pigeon to pick numbers randomly than running a brute-force Python script. So unless you want your computer to burn through its lifespan in service of the RNG gods, maybe just grab some popcorn, sit back, and binge-watch a series on iQiyi.  Wink
hero member
Activity: 862
Merit: 662
I never trust other people's code.

Open source code It is written in that way for review.

Only the programs I write myself are the most stable.(linux) Grin Grin Grin Grin Grin

There is going to be always programs more stable than yours. Python maybe stable... but faster?Huh NO

And yes almost all users here can write some code in slow python, so please come here when you write something more useful than available options on github
copper member
Activity: 205
Merit: 1
I never trust other people's code. Only the programs I write myself are the most stable.(linux) Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin
 Reward address: 1JWM1RsRFuct6y9JCwhx3TibvhDJVB2Ldn

What is this code for??
First, it generates addresses sequentially, but the search space is so large that even with a GPU it would take years to sort through part of the range.
Second, this code generates uncompressed addresses when the puzzle uses compressed addresses.
That is, this code will never find anything.
Third, almost any user of this topic can write such code on their knee.
In extreme cases, using chatgpt, you can generate hundreds of variations of such code.
So stop posting meaningless posts.
newbie
Activity: 1
Merit: 0
I never trust other people's code. Only the programs I write myself are the most stable. Reward address: 1JWM1RsRFuct6y9JCwhx3TibvhDJVB2Ldn

import hashlib
import ecdsa
import base58
import threading
import logging

# 设置日志记录
logging.basicConfig(level=logging.INFO, format='%(asctime)s - %(threadName)s - %(message)s')

# 匹配的固定地址(写在代码中)
addresses_in_code = [
    "1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9",
    "1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ",
    "19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG",
    "19YZECXj3SxEZMoUeJ1yiPsw8xANe7M7QR",
    "1PWo3JeB9jrGwfHDNpdGK54CRas7fsVzXU",
    "1JTK7s9YVYywfm5XUH7RNhHJH1LshCaRFR",
    "12VVRNPi4SJqUTsp6FmqDqY5sGosDtysn4"
]

# 生成私钥对应的比特币地址
def private_key_to_address(private_key_hex):
    private_key_bytes = bytes.fromhex(private_key_hex)
    signing_key = ecdsa.SigningKey.from_string(private_key_bytes, curve=ecdsa.SECP256k1)
    verifying_key = signing_key.verifying_key
    pub_key = b'\x04' + verifying_key.to_string()  # 非压缩公钥格式

    # 计算公钥的哈希 (SHA256 + RIPEMD160)
    sha256_hash = hashlib.sha256(pub_key).digest()
    ripemd160_hash = hashlib.new('ripemd160', sha256_hash).digest()

    # 添加比特币网络前缀
    network_byte = b'\x00' + ripemd160_hash
    checksum = hashlib.sha256(hashlib.sha256(network_byte).digest()).digest()[:4]

    binary_address = network_byte + checksum
    return base58.b58encode(binary_address).decode()

# 按顺序递增生成私钥
def generate_private_keys_in_range(start_key_int, end_key_int):
    for key_int in range(start_key_int, end_key_int + 1):  # 递增 +1 生成私钥
        private_key_hex = format(key_int, '064x')
        yield private_key_hex

# 多线程生成私钥并匹配地址
def generate_and_match_private_keys(start_key_int, end_key_int):
    for private_key_hex in generate_private_keys_in_range(start_key_int, end_key_int):
        address = private_key_to_address(private_key_hex)

        # 匹配地址
        if address in addresses_in_code:
            logging.info(f"匹配成功: 私钥 {private_key_hex} 对应地址 {address}")
            # 匹配成功后立即写入 add.txt 文件
            with open('add.txt', 'a') as f_add:
                f_add.write(f"{private_key_hex},{address}\n")
        else:
            logging.info(f"未匹配: 私钥 {private_key_hex} 对应地址 {address}")

# 计算私钥区间的差值并划分成320份
def divide_key_range_into_parts(start_key, end_key, num_parts):
    start_int = int(start_key, 16)
    end_int = int(end_key, 16)
    total_range = end_int - start_int  # 修改为递增范围
    part_size = total_range // num_parts  # 每个线程的区间大小
    return [(start_int + i * part_size, start_int + (i + 1) * part_size - 1) for i in range(num_parts)]

# 启动320个线程,每个线程负责生成区间内的私钥
def start_threads_for_key_range(start_key, end_key, num_threads=320):
    key_ranges = divide_key_range_into_parts(start_key, end_key, num_threads)
    
    threads = []
    for i, (start_key_int, end_key_int) in enumerate(key_ranges):
        t = threading.Thread(target=generate_and_match_private_keys, args=(start_key_int, end_key_int), name=f"线程-{i+1}")
        threads.append(t)
        t.start()

    try:
        for t in threads:
            t.join()
    except KeyboardInterrupt:
        logging.info("捕获到 KeyboardInterrupt, 正在终止线程...")

# 主程序入口
if __name__ == "__main__":
    start_key = "00000000000000000000000000000000000000000000001b8f59c6c922e5426d"
    end_key = "00000000000000000000000000000000000000000000002c4221111a7294a017"
    
    try:
        # 启动320个线程,每个线程负责相应的私钥区间
        start_threads_for_key_range(start_key, end_key, num_threads=320)
    except KeyboardInterrupt:
        logging.info("程序手动终止")
        os._exit(0)



I never trust other people's code. Only the programs I write myself are the most stable.(linux) Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin Grin
 Reward address: 1JWM1RsRFuct6y9JCwhx3TibvhDJVB2Ldn

import hashlib
import binascii
import base58
import threading
import ecdsa

# 匹配的比特币地址列表
target_addresses = {
    "1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ",
    "19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG"
}

# 初始私钥列表(8个私钥)
initial_private_keys = [
    "000000000000000000000000000000000000000000000024351b116fe05a4ef1",
    "00000000000000000000000000000000000000000000001b0f680b1092d881e6",
    "00000000000000000000000000000000000000000000001c478360fcc5f7e4c3",
    "00000000000000000000000000000000000000000000001de29e256292c39398",
    "000000000000000000000000000000000000000000000024351b116fe05a426d",
    "0000000000000000000000000000000000000000000000288707c8ae1738f142",
    "000000000000000000000000000000000000000000000020cbae528b05b2a017",
    "00000000000000000000000000000000000000000000001ea2011cc8a10e4eec"
]

# 检查私钥是否符合规则(不允许三个连续相同的字符)
def is_valid_key_suffix(private_key_hex):
    for i in range(len(private_key_hex) - 2):
        if private_key_hex == private_key_hex[i + 1] == private_key_hex[i + 2]:
            return False
    return True

# 生成比特币地址(P2PKH Legacy)
def pubkey_to_address(public_key):
    sha256_hash = hashlib.sha256(public_key).digest()
    ripemd160_hash = hashlib.new('ripemd160', sha256_hash).digest()
    extended_ripemd160 = b'\x00' + ripemd160_hash  # 比特币地址前缀 0x00
    checksum = hashlib.sha256(hashlib.sha256(extended_ripemd160).digest()).digest()[:4]
    binary_address = extended_ripemd160 + checksum
    return base58.b58encode(binary_address).decode()

# 生成公钥(压缩)
def private_key_to_public_key(private_key_hex):
    private_key_bytes = binascii.unhexlify(private_key_hex)
    sk = ecdsa.SigningKey.from_string(private_key_bytes, curve=ecdsa.SECP256k1)
    vk = sk.verifying_key
    public_key = b'\x02' + vk.to_string()[:32] if int.from_bytes(vk.to_string()[32:], 'big') % 2 == 0 else b'\x03' + vk.to_string()[:32]
    return public_key

# 线程函数:从初始私钥开始递减,并生成对应的比特币地址
def worker_thread(start_private_key, thread_id):
    private_key_int = int(start_private_key, 16)
    total_generated = 0
    matched_count = 0
    
    while True:
        private_key_hex = format(private_key_int, '064x')

        # 检查私钥是否符合规则
        if is_valid_key_suffix(private_key_hex):
            public_key = private_key_to_public_key(private_key_hex)
            address = pubkey_to_address(public_key)

            # 检查生成的地址是否在目标地址列表中
            if address in target_addresses:
                with open("add.txt", "a") as f:
                    f.write(f"Private Key: {private_key_hex}, Address: {address}\n")
                matched_count += 1

        private_key_int -= 1
        total_generated += 1
        
        # 打印每个线程生成的私钥总数
        if total_generated % 100000 == 0:
            print(f"Thread {thread_id}: Total keys generated: {total_generated}, Matched: {matched_count}", end='\r')

# 启动8个线程,每个线程从不同的私钥开始递减
def start_threads():
    threads = []
    for idx, initial_key in enumerate(initial_private_keys):
        t = threading.Thread(target=worker_thread, args=(initial_key, idx + 1))
        threads.append(t)
        t.start()

    for t in threads:
        t.join()

# 启动程序
if __name__ == "__main__":
    start_threads()

member
Activity: 873
Merit: 22
$$P2P BTC BRUTE.JOIN NOW ! https://uclck.me/SQPJk
how to recover input from y ?


0x7404ea4a8c154c985f06f694467381d74885ba29676271cc1db16de2f71ab584
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x2678301d19f8bcc6328469b8ed8801bb 2500
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x2471dcbd410697e42fdbc2794ceece8a 5000
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x226b895d681473022d331b39ac68ae29 7500
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x206535fd8f224e202a8a73fa0bf5a098 10000
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x1e5ee29db630293e27e1ccba6b95a5d7 12500
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x1c588f3ddd3e045c2539257acb48bde6 15000
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x1a523bde044bdf7a22907e3b2b0ee8c5 17500
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x184be87e2b59ba981fe7d6fb8ae82674 20000
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x1645951e526795b61d3f2fbbead476f3 22500
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x143f41be797570d41a96887c4ad3da42 25000
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x1238ee5ea0834bf217ede13caae65061 27500
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x10329afec7912710154539fd0b0bd950 30000
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0xe2c479eee9f022e129c92bd6b44750f 32500
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0xc25f43f15acdd4c0ff3eb7dcb90239e 35000
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0xa1fa0df3cbab86a0d4b443e2beee4fd 37500
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x8194d7f63c893880aa29cfe8c60b92c 40000
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x612fa1f8ad66ea607f9f5beece5a02b 42500
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x40ca6bfb1e449c405514e7f4d7d99fa 45000
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0x206535fd8f224e202a8a73fae28a699 47500
input:
0x287e837cf2eae1a8352d10f88e3447bc
y 0xee6c608 50000


y = 0xee6c608


input=0x287e837cf2eae1a8352d10f88e3447bc


Code:

N =    115792089237316195423570985008687907852837564279074904382605163141518161494337

def inv(v): return pow(v, N-2, N)
def divnum(a, b): return ( (a * inv(b) ) % N )


x = 0x287e837cf2eae1a8352d10f88e3447bc #- 0x262794

y = 0x287e837cf2eae1a8352d10f88e3447bc  # - 0x262794

X = divnum(x ,50000)

print(hex(X))


i = 0
while y >=0:
   
    X = ((X - divnum(1,5) %N)%N)
   
    y = (y - X%N)%N
   
    i = i +1
   
    if y <= 2**190:
        print("input:")
        print(hex( 0x287e837cf2eae1a8352d10f88e3447bc))
       
        print("y",hex(y),i)
        #print(hex(X))



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Activity: 348
Merit: 34




Fishing offers a peaceful connection with nature, allowing the mind to relax and reset, while solving unsolvable BTC puzzles can lead to frustration and mental fatigue. Reduce stress, unlike the endless search for solutions in puzzles that may never yield results. Spending hours in front of a screen on futile tasks drains mental energy.
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