Private key for address with 5 BTC. | Bitcointalksearch.org

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legendary

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Quote from: Bill buffalo on July 27, 2024, 06:56:38 AM

The private key if found gives access to this BTC addres
s 1HqtKWKCLTs4eUuvTDPpJC4AaBYMiiXWok which contains 5btc if that isn't free money what is?

Here you can find all addresses that have balances https://bitcointalksearch.org/topic/list-of-all-bitcoin-addresses-with-a-balance-5254914
If you find private keys from all of them, you will have a lot of Bitcoins. A rare opportunity with so much free money, isn't it?

SamReomo

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Quote from: BlackHatCoiner on August 02, 2024, 11:01:55 AM

The most plausible scenario, given this information, is that the OP has no access to the 5 BTC.

I agree with you, I also believe that OP no longer has access to that wallet or has lost it somehow and those who still believe that they can solve the puzzle to earn 5 BTC should read your post at least. I suggest everyone to be careful as there's no chance to get anything from this puzzle anymore.

ABCbits

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Quote from: BlackHatCoiner on August 02, 2024, 11:01:55 AM

The private key is generated and partially displayed in PEM. When were private keys generated like that? It seems like this person manually generated an EC private key and generated its address with some custom script.

I would speculate 2013 or older. OpenSSL was popular library, before secp256k1 library created since late 2013. It's so popular that someone made bounty to convert PEM public key to Bitcoin address, https://bitcointalksearch.org/topic/50-btc-bounty-for-a-pem-public-key-to-bitcoin-address-convertor-2631.

Quote from: BlackHatCoiner on August 02, 2024, 11:01:55 AM

The 5 BTC have not moved. This is extremely odd, IMO, because when this "challenge" started, 5 BTC were worth about $70. Now, they are worth more than $300,000. It's even weirder, when you realize that the OP was last time active in 2018, discussing about altcoins. Usually, people who did that back in 2018, had likely sold their bitcoin for some shitcoin.

Or OP lost access to that address, since BTC in 2011 wasn't that valuable.

BlackHatCoiner

legendary

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Farewell, Leo

There are two things strange in this topic, and honestly, I don't know which one is stranger.

The private key is generated and partially displayed in PEM. When were private keys generated like that? It seems like this person manually generated an EC private key and generated its address with some custom script.
The 5 BTC have not moved. This is extremely odd, IMO, because when this "challenge" started, 5 BTC were worth about $70. Now, they are worth more than $300,000. It's even weirder, when you realize that the OP was last time active in 2018, discussing about altcoins. Usually, people who did that back in 2018, had likely sold their bitcoin for some shitcoin.

The most plausible scenario, given this information, is that the OP has no access to the 5 BTC. They funded an address which might have been created using a custom script, and which was later discovered it was flawed. There are many examples, similar to that, where people did testings in bitcoin mainnet and lost access to their coins forever.

apogio

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Activity: 560

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Warning!

Back in the day, yeah, 0.025 BTC was not a lot! But I will leave my post here, in case any new member thinks this post is still active!

DON'T engage with this puzzle game! Don't send coins to any address mentioned in the OP.

Basically OP is asking for money, so this is not acceptable!
All the puzzle games so far don't ask anything in return!
There are too many ways to get scammed with this.

1. OP doesn't have a sufficient amount of merit, nor a lot of trust. Even if they did, again, it would be very irresponsible to participate in a game like this.
2. The address 1HqtKWKCLTs4eUuvTDPpJC4AaBYMiiXWok has indeed 5 BTC, but the OP didn't provide a signed message from this address, so we are not sure they hold the keys to unlock them.
3. New members may get scammed, thinking this post is still active and sending coins to the address mentioned there.

@Mods: I suggest that you remove the address from the OP... You never know if anyone is silly enough to send coins to this address.

Bill buffalo

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The private key if found gives access to this BTC addres
s 1HqtKWKCLTs4eUuvTDPpJC4AaBYMiiXWok which contains 5btc if that isn't free money what is?

examplens

legendary

Activity: 3472

Merit: 3507

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Quote from: Bill buffalo on July 18, 2024, 01:09:37 PM

So you mean to tell me if you have the full private key of this string below MHQCAQEEI**7***************************e************oAcGBSuBBAAKoUQDQgAE********/********4*******4****f*********************************0*********************== if you can find it complete it will give access to this btc address 1HqtKWKCLTs4eUuvTDPpJC4AaBYMiiXWok that contains 5 BTC are you serious about this?

always check first a time when the thread is open. For example, this was launched in 2011, 13 years ago, it is unlikely that the OP is still current with this offer.
it seems that there is no free money here

Bill buffalo

newbie

Activity: 48

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So you mean to tell me if you have the full private key of this string below MHQCAQEEI**7***************************e************oAcGBSuBBAAKoUQDQgAE********/********4*******4****f*********************************0*********************== if you can find it complete it will give access to this btc address 1HqtKWKCLTs4eUuvTDPpJC4AaBYMiiXWok that contains 5 BTC are you serious about this?

paapa1268

newbie

Activity: 3

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Quote from: PawGo on April 23, 2022, 12:05:04 PM

ho ho ho
What a puzzle Wink

11 year old.
I see owner was not active during last 4 years - strange that coins are still there.

It gives known part:

Code:

MHQCAQEEI**7***************************e************oAcGBSuBBAAKoUQDQgAE********/********4*******4****f*********************************0*********************==

so it is rather not possible to bruteforce for a given hash...

the only solution is to use the same version script he used and repeate process until got hit the version used by him can be seen https://lapo.it/asn1js decoder see his other post he posted random pem key decode using above mention site and in option it will give all version info used

Seraphimjm

newbie

Activity: 12

Merit: 5

Would have probably taken the coins sent to reveal and up to the point where it was almost easily solvable just move the coins then declare found/solved.

classic raffle ticket scheme with a little slight of hand. The first 0.1btc sent was the bait, but there were no fish... Grin

PawGo

legendary

Activity: 952

Merit: 1385

ho ho ho
What a puzzle Wink

11 year old.
I see owner was not active during last 4 years - strange that coins are still there.

It gives known part:

Code:

MHQCAQEEI**7***************************e************oAcGBSuBBAAKoUQDQgAE********/********4*******4****f*********************************0*********************==

so it is rather not possible to bruteforce for a given hash...

mc_lovin

legendary

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www.bitcointrading.com

interesting little game.

anthony_

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Quote from: cuddlefish on July 20, 2011, 09:14:27 PM

Quote from: anthony_ on July 20, 2011, 03:49:17 PM

Quote from: cuddlefish on July 20, 2011, 09:19:23 AM

Ooops.

>>> random.seed(3)
>>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e-1]
...
39
e
89
4
60
B
98
4
>>>

it appears we get the first 8 digits then =P

No, those are the 0-indexed positions.

We already got the next 4 numbers for that seed as-well is what i'm saying Wink

cuddlefish

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Activity: 364

Merit: 250

Quote from: anthony_ on July 20, 2011, 03:49:17 PM

Quote from: cuddlefish on July 20, 2011, 09:19:23 AM

Ooops.

>>> random.seed(3)
>>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e-1]
...
39
e
89
4
60
B
98
4
>>>

it appears we get the first 8 digits then =P

No, those are the 0-indexed positions.

anthony_

member

Activity: 65

Merit: 10

Quote from: cuddlefish on July 20, 2011, 09:19:23 AM

Ooops.

>>> random.seed(3)
>>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e-1]
...
39
e
89
4
60
B
98
4
>>>

it appears we get the first 8 digits then =P

cuddlefish

sr. member

Activity: 364

Merit: 250

Ooops.

>>> random.seed(3)
>>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e-1]
...
39
e
89
4
60
B
98
4
>>>

Maged

legendary

Activity: 1204

Merit: 1015

Quote from: anthony_ on July 20, 2011, 01:49:59 AM

Quote from: cuddlefish on July 20, 2011, 12:22:09 AM

Quote from: anthony_ on July 19, 2011, 10:49:25 PM

Sent 0.1BTC for the first four characters to get things started.
http://blockexplorer.com/tx/5bc74811af3886fbb4ef551fa6e28cabe02ee8470559bcbbebf155c858aae361

>>> random.seed(3)
>>> for i in range(4): print random.randint(1, len(data))
...
39
89
60
98
>>> for i in range(4): print r[random.randint(1, len(data))]
...
f
7
C
0

from that output it doesn't show that you type random.seed the second time, so it would have continued from the last point in the PRNG and the f 7 C 0 would be wrong

If that's true, then this should be the correct assignment...

>>> random.seed(3)
>>> for i in range(4): print random.randint(1, 162)
...
39
89
60
98
>>> for i in range(4): print random.randint(1, 162)
...
102
11
3
136

anthony_

member

Activity: 65

Merit: 10

Quote from: cuddlefish on July 20, 2011, 12:22:09 AM

Quote from: anthony_ on July 19, 2011, 10:49:25 PM

Sent 0.1BTC for the first four characters to get things started.
http://blockexplorer.com/tx/5bc74811af3886fbb4ef551fa6e28cabe02ee8470559bcbbebf155c858aae361

>>> random.seed(3)
>>> for i in range(4): print random.randint(1, len(data))
...
39
89
60
98
>>> for i in range(4): print r[random.randint(1, len(data))]
...
f
7
C
0

from that output it doesn't show that you type random.seed the second time, so it would have continued from the last point in the PRNG and the f 7 C 0 would be wrong

cuddlefish

sr. member

Activity: 364

Merit: 250

Quote from: anthony_ on July 19, 2011, 10:49:25 PM

Sent 0.1BTC for the first four characters to get things started.
http://blockexplorer.com/tx/5bc74811af3886fbb4ef551fa6e28cabe02ee8470559bcbbebf155c858aae361

>>> random.seed(3)
>>> for i in range(4): print random.randint(1, len(data))
...
39
89
60
98
>>> for i in range(4): print r[random.randint(1, len(data))]
...
f
7
C
0

anthony_

member

Activity: 65

Merit: 10

Sent 0.1BTC for the first four characters to get things started.
http://blockexplorer.com/tx/5bc74811af3886fbb4ef551fa6e28cabe02ee8470559bcbbebf155c858aae361

Topic: Private key for address with 5 BTC. (Read 3763 times)