Bitcoin puzzle transaction ~32 BTC prize to who solves it

[Memmon3]

 no,I do not understand all this, If I'm smart enough to do this, it would be nice to do it. Because the gift is very large 32 bitcoin it is a very big gift, we can buy anything by having 32 bitcoin. but i cant o this ..

[1714027795]

1714027795

[1714027795]

1714027795

[1714027795]

1714027795

[1714027795]

1714027795

[FedorIzmailov]

we can’t solve this puzzle because the supercomputer invented it and only he can solve it

[zielar]

no,I do not understand all this, If I'm smart enough to do this, it would be nice to do it. Because the gift is very large 32 bitcoin it is a very big gift, we can buy anything by having 32 bitcoin. but i cant o this ..

You can not buy health or health for any currency. Such little wisdom at the end of the year: P

Hi Zielar
What brute force strategy do you adopt?
we could PM if you want.
A Very Healthy, Wealthy and Happy New Year to All

At the moment I am studying the operation of this script because I was interested:

import random
from bit import *
from PyRandLib import *
rand = FastRand63()
random.seed(rand())

c1 = str (random.choice("1"))
b1  = "01111111111111111111111111111111111111111111111111111111"
b2  = "00111111111111111111111111111111111111111111111111111111"
b3  = "00011111111111111111111111111111111111111111111111111111"
b4  = "00001111111111111111111111111111111111111111111111111111"
b5  = "00000111111111111111111111111111111111111111111111111111"
b6  = "00000011111111111111111111111111111111111111111111111111"
b7  = "00000001111111111111111111111111111111111111111111111111"
b8  = "00000000111111111111111111111111111111111111111111111111"
b9  = "00000000011111111111111111111111111111111111111111111111"
b10 = "00000000001111111111111111111111111111111111111111111111"
b11 = "00000000000111111111111111111111111111111111111111111111"
b12 = "00000000000011111111111111111111111111111111111111111111"
b13 = "00000000000001111111111111111111111111111111111111111111"
b14 = "00000000000000111111111111111111111111111111111111111111"
b15 = "00000000000000011111111111111111111111111111111111111111"
b16 = "00000000000000001111111111111111111111111111111111111111"
b17 = "00000000000000000111111111111111111111111111111111111111"
b18 = "00000000000000000011111111111111111111111111111111111111"
b19 = "00000000000000000001111111111111111111111111111111111111"
b20 = "00000000000000000000111111111111111111111111111111111111"
b21 = "00000000000000000000011111111111111111111111111111111111"
b22 = "00000000000000000000001111111111111111111111111111111111"
b23 = "00000000000000000000000111111111111111111111111111111111"
b24 = "00000000000000000000000011111111111111111111111111111111"
b25 = "00000000000000000000000001111111111111111111111111111111"
b26 = "00000000000000000000000000111111111111111111111111111111"
b27 = "00000000000000000000000000011111111111111111111111111111"
b28 = "00000000000000000000000000001111111111111111111111111111"
b29 = "00000000000000000000000000000111111111111111111111111111"
b30 = "00000000000000000000000000000011111111111111111111111111"
b31 = "00000000000000000000000000000001111111111111111111111111"
b32 = "00000000000000000000000000000000111111111111111111111111"
b33 = "00000000000000000000000000000000011111111111111111111111"
b34 = "00000000000000000000000000000000001111111111111111111111"
b35 = "00000000000000000000000000000000000111111111111111111111"
b36 = "00000000000000000000000000000000000011111111111111111111"
b37 = "00000000000000000000000000000000000001111111111111111111"
b38 = "00000000000000000000000000000000000000111111111111111111"
b39 = "00000000000000000000000000000000000000011111111111111111"
b40 = "00000000000000000000000000000000000000001111111111111111"
b41 = "00000000000000000000000000000000000000000111111111111111"
b42 = "00000000000000000000000000000000000000000011111111111111"
b43 = "00000000000000000000000000000000000000000001111111111111"
b44 = "00000000000000000000000000000000000000000000111111111111"
b45 = "00000000000000000000000000000000000000000000011111111111"
b46 = "00000000000000000000000000000000000000000000001111111111"
b47 = "00000000000000000000000000000000000000000000000111111111"
b48 = "00000000000000000000000000000000000000000000000011111111"
b49 = "00000000000000000000000000000000000000000000000001111111"
b50 = "00000000000000000000000000000000000000000000000000111111"
b51 = "00000000000000000000000000000000000000000000000000011111"
b52 = "00000000000000000000000000000000000000000000000000001111"
b53 = "00000000000000000000000000000000000000000000000000000111"
b54 = "00000000000000000000000000000000000000000000000000000011"
b55 = "00000000000000000000000000000000000000000000000000000001"
#xx = "00000000000000000000000000000000000000000000000000000001"

while True:
    spisok = [b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11,b12,b13,b14,b15,b16,b18,b19,b20,b21,b22,b23,b24,b25,b26,b27,b28,b30,b31,b32,b33,b34,b35,b36,b37,b38,b39,b40,b41,b42,b43,b44,b45,b46,b47,b48,b49,b50,b51,b52,b53,b54,b55]
    for element in (spisok):
        for spisok in range(1000):
            s = element
            d = ''.join(random.sample(s,len(s)))
            bina = (c1+d)
            b = int(c1+d,2)
            key = Key.from_int(b)
            addr = key.address
            if addr == "15c9mPGLku1HuW9LRtBf4jcHVpBUt8txKz":
                print ("found!!!",b,addr)
                s1 = str(b)
                s2 = addr
                f=open(u"C:/a.txt","a")
                f.write(s1)
                f.write(s2)      
                f.close()
                pass
            else:
                print (s,bina,b,addr)
    pass

I wonder how to change it so that he does not pick randomly only from the chosen threshold he flew up one by one. Does anyone have knowledge of how it could be remade?

BY USING OUR OPPORTUNITY, WE WOULD LIKE YOU TO SAY ALL OF YOUR SUCCESS AND JOY IN THE NEW YEAR!

[Andzhig]

there it is necessary little correct.

bina = (c1+d)
b = int(bina,2)

flew up one by one.

The meaning of this particular example is that it uses a fixed set, 01111111111111111111111111111111111111111111111111111111 and mixes it adding 1 in front so that it doesn’t appear there 0. So if you take 00000000000000000000000000000111111111111111111111111111 +1 = 100000000000000000000000000000111111111111111111111111111 72057594172145663 (can wrap and go in the opposite direction 111111111111111111111111111100000000000000000000000000000 144115187538984960) it will take as much time to find a number so that it satisfies the requests, in other words, it will be easy to go through step by step as without dancing with a tambourine.

i = 72057594037927936
while i <= 144115188075855872:
    ff = bin(i)[2:]
    ed = ff.count("0")
    if ed >= 10:
        if ed <= 35:
            print(i,ff,ed)
    i = i+1

try to change 10 and 35 to 30-31 and detect the time after which he will give you the result.

from bit import *
import random

while 1 == 1:
    a = 390000000000000000
    while a <= 554000000000000000:
        bina = bin(a)[2:]
        zeros = bina.count("0")
        if zeros >= 30:     # "00000000000000000000000000000011111111111111111111111111"
            if zeros <= 30: # "00000000000000000000000000000011111111111111111111111111"
                key = Key.from_int(a)
                addr = key.address
                if addr == "1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt":
                    print("found!!!",a,addr)
                    s1 = str(a)
                    s2 = addr
                    f=open(u"C:/a.txt","a")
                    f.write(s1)
                    f.write(s2)      
                    f.close()
                    break
                else:
                    print (a,bina,addr,zeros) #pass
        a = a +1
    pass

runs fast but still need gpu, the gain in speed is that he doesn’t compare unnecessary address spaces (using a bitcoin slowdown library), but nonetheless he goes through them.

Although, if we find crypto guru who will help us calculate to organize the program, we will be very grateful. for example if 2^256 this from 57896044618658097711785492504343953926634992332820282019728792003956564819968 to 115792089237316195423570985008687907853269984665640564039457584007913129639936 step by step and if we take numbers with only 127 zeros in the binary how much will it be in the end result, 2^?

***

Happy holiday https://www.youtube.com/watch?v=VVOXHmvRRh8  

[brainless]

What has this to do with this Puzzle transaction we discuss in this thread?

Dear Sir
i apologies if my question is not related to thread, i like to explain why i ask these question comes in my mind related to this thread,
1. every one posting here about random search scripts, some post about brute force techniques, some trying to explain logics for find 32 BTC PUZZLE
2. above my posted link, when explorer, you will find some services used by website, who offer protection for btc coins, what technique used is split btc into pieces and transfers to new series generated accounts, then more splits into new series generated accounts, all generated accounts are same time in seconds, (mean no manual transaction one by one),
3 same in 32 BTC PUZZLE series of satochi transferd to newly generated accounts same time with sequence of each satoshi plus in series of bit space
4 in 59 and 60 bit space, peoples need to explorer by brute force or random etc equal to total hardware used for btc minning in Ph/s,
5 for this reason people must think an other ways to definatly in formula by math, by brainwallet, logics will help
6 for path and ref , i post links for other to think and find solutions deeply by explorer links
7 for deep leaning inside for solve PUZZLE, IN MY VIEW , above and this link could change thinking, and aproch to solve puzzle
8 https://www.blockchain.com/btc/tx/1d6580dcd979951bd600252b741c22a3ea8e605e43168f8452c68915c3ea2bf3
   RETURN PUSHDATA(60)[424553544d495845522e494f207c204d495820594f555220424954434f494e5320544f2053544159205341464520414e442050524f54454354454421]
   (decoded) BESTMIXER.IO | MIX YOUR BITCOINS TO STAY SAFE AND PROTECTED!
9 bestmixer.io workstyle happen in 32 BTC PUZZLE and recenet transaction of https://www.blockchain.com/btc/address/1Ko8t47Mkvq8vyT2XgqLmdouVxM8tTryc
10 IN MY VIEW, brainwallet char series by formula \ calculation of series by formula,  could generate address in all Bit Space, etc, and work arund this may fruitfull instead of brute force by series or random

Sir if you feel Good about my refrence of research related to Solve PUZZLE, encourage me with thumb up, in other just update me, i will remove my post, like to say apologies again

[itod]

What has this to do with this Puzzle transaction we discuss in this thread?

...
Sir if you feel Good about my refrence of research related to Solve PUZZLE, encourage me with thumb up, in other just update me, i will remove my post, like to say apologies again
...

There is no puzzle to solve, and you do not have to delete anything. Here is the post from the creator of this "puzzle", it may get you to see things clearly:

I am the creator.

...

A few words about the puzzle.  There is no pattern.  It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty).  It is simply a crude measuring instrument, of the cracking strength of the community.

See? Nothing to solve, nothing to crack, from the man who created this. The "puzzle" is just a measurement of the brute forcing abilities of everyone together,  making us sleep peacefully showing that Bitcoin can not be bruteforced. The more of these private keys get discovered by this joined brute-forcing effort, the more (double exactly) difficult it gets to find next private key, and it continues to grow exponentially. There is nothing more to it, human brain is useless with this "puzzle", and any effort to use your brain to accomplish finding private keys for these 32 BTC is just big waist of your time.

[Elliptic23]

It takes brain power to come up with ways to check keys fast. Why do you think the BitCrack program can test a billion keys per second on a single video card while the Large Bitcoin Collider could test millions on the same card?

[PietCoin97]

Because you can run Bitcrack with a lot of cards without cpu.
At collider you need cpu power for every card so when you want to run lot of cards you need big cpu power.

[anythingtechpro]

Are we looking for specific wallets or just wallets in general that have a balance?

[itod]

Are we looking for specific wallets or just wallets in general that have a balance?

Neither, we are expecting that no private keys from a single transaction will ever being found faster then expected by current bruteforce difficulty. That's the best outcome - that everyone fails and Bitcoin is safe.

[anythingtechpro]

Are we looking for specific wallets or just wallets in general that have a balance?

Neither, we are expecting that no private keys from a single transaction will ever being found faster then expected by current bruteforce difficulty. That's the best outcome - that everyone fails and Bitcoin is safe.

Is it true that someone found the private key for this wallet? https://www.blockchain.com/btc/address/1AhTjUMztCihiTyA4K6E3QEpobjWLwKhkR

[racminer]

Are we looking for specific wallets or just wallets in general that have a balance?

Neither, we are expecting that no private keys from a single transaction will ever being found faster then expected by current bruteforce difficulty. That's the best outcome - that everyone fails and Bitcoin is safe.

Is it true that someone found the private key for this wallet? https://www.blockchain.com/btc/address/1AhTjUMztCihiTyA4K6E3QEpobjWLwKhkR

You may notice that many big wallets are receiving multiple very small deposits from https://www.blockchain.com/btc/address/1AK4LYE6PYwBmSYHQX3v2UsXXHTvCAsJeK  (Bitcoin buy or sell JUBTC.COM)
It seems to be a company which help people retrieve their private keys (they surely try to crack some for themselves ).
I think that if you narrow space search ( for example if you know part of mnemonic words ) , you can retrieve private key from transaction scripts ...

Any thoughts on this?
Does it seem possible?

[Lolo54]

arulbero Help!
what key can there be 1Dn8NF8qDyyfHMktmuoQLGyjWmZXgvosXf
2..............
3..............

Code:

Private key : 00000000000000000000000000000000000000000000000002c675b852189a21
Public key  : 11569442e870326ceec0de24eb5478c19e146ecd9d15e4666440f2f638875f42 524c08d882f868347f8b69d3330dc1913a159d8fb2b27864f197693a0eb39a23
 
PrKey WIF c.: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjv2kTamEYQT9BNC7o1
Address c.  : 8c2a6071f89c90c4dab5ab295d7729d1b54ea60f
Address c.  : 1Dn8NF8qDyyfHMktmuoQLGyjWmZXgvosXf

Bravo and thank you Arulbero you are impressive! If I have followed the topic well let’s say that from 59 to 65 addresses in just a few seconds you could get the private key?!
Could you get there keys for this address 1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt
1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt

[racminer]

arulbero Help!
what key can there be 1Dn8NF8qDyyfHMktmuoQLGyjWmZXgvosXf
2..............
3..............

Code:

Private key : 00000000000000000000000000000000000000000000000002c675b852189a21
Public key  : 11569442e870326ceec0de24eb5478c19e146ecd9d15e4666440f2f638875f42 524c08d882f868347f8b69d3330dc1913a159d8fb2b27864f197693a0eb39a23
 
PrKey WIF c.: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjv2kTamEYQT9BNC7o1
Address c.  : 8c2a6071f89c90c4dab5ab295d7729d1b54ea60f
Address c.  : 1Dn8NF8qDyyfHMktmuoQLGyjWmZXgvosXf

Bravo and thank you Arulbero you are impressive! If I have followed the topic well let’s say that from 59 to 65 addresses in just a few seconds you could get the private key?!
Could you get there keys for this address 1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt
1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt

No  one can. Unless whoever finds  the pvkey  spends partially or totally the  0.59 BTC.

 arulbero's program needs as input a spending script. And it is fast because the  pvkey range is small.

[umadbro]

I've modified bitcrack to work with hashtopolis (https://github.com/s3inlc/hashtopolis) in an attempt to better track and distribute keyspaces that i've already tried for TX 59. My current hashing power is around 6 GHS. If anyone is interested in running the client we could use this as a pool and combine our efforts. I realize the large bitcoin collider exists for this purpose but it is very slow in comparison.

If you're interested in joining, pm me and we can discuss details.

[zielar]

Hello, how have you "changed" bitcrack? As you can see from my posts on the last pages - I would gladly come in as a part of some mutual help. How can I check your revised version and check it on my 8x 1400mkeys / s?

[umadbro]

I'm unable to send or receive PMs here due to my newb status. I've set up a telegram group for anyone interested in learning about or joining the new pool.

t.me/puzzletxhunters

[zielar]

80 * 1400000000 = 112000000000
576460752303423488-288230376151711744 = 288230376151711744
288230376151711744 / 112000000000 = ~2573485 sec

~2573485 sec = 42891 min
42891 min = 714 hours
714 hours = 29 days

...it is full time what i need to scan full keyspace for actually address.
29 - 14 (actually searched) = max two weeks = to next puzzle :-)

I assume that the right key will be the last opportunity: P

[Andzhig]

max two weeks = to next puzzle :-)

Come on and then we already started to be bored. Moreover quantum computers will soon begin to roll up their demon eggs to the puzzle "IBM Q System One enables universal approximate superconducting quantum computers to operate outside the research lab for the first time. It's a major step forward in the commercialization of quantum computing, which could one day enable breakthroughs in such areas as materials and drug discovery, financial services, and artificial intelligence.". https://www.youtube.com/watch?v=LAA0-vjTaNY

Factoring large numbers here too especially not to pick out (without proper mathematical analysis). Any, exception method, come up. If take 2 9 digits of the number and multiply them, then at the exit 18 a significant number, in fact garbage, is not necessary in which with a 99% probability there will be no puzzle.

938831741 547147806 513679727291310246 1F3Dy1ZXMSLATum9JRrnHE4JK9DwS8e9k2 11100100000111101001111110110001001100001001010000010100110 32
632607579 761972272 482029434255049488 13fqgN2nDRh2wPiXb2ytESrTeM3p9ZhWCS 11010110000100000110011011101110000000011100000001100010000 37
930284466 462063150 429850170756027900 1BAg5YQ2xB4gKMCyXoRuWajM96NRwVjerz 10111110111001000100111001111101110100011011101010111111100 22
668510014 757844417 506626581818491838 19tBDCL6tKkaJkGAkqf5gGLTJbH3Y6UCR3 11100000111111001100011000100100011000011111100001110111110 28
597539789 891807805 532890647628253145 1LAK3HA8qKAbjjMs8Wek56JV5JTYzusSc7 11101100101001101010011100001111100000000101000011111011001 30
844591555 549728690 464296209115212950 15JBiekJW559eqrw7qe4J4hCTjm1iNjcdr 11001110001100000101111000100110110110111110110110010010110 27
732978027 653067936 478684447226242272 1GWXMr94nxk2EDf5h8CbzMfCRDdQgUCK9u 11010100100101000001111011111110100011011001010010011100000 30
763876069 543620238 415258490432284422 1EfDVsB2NnJMpiqA9PruhHRRVfqPTrbTQu 10111000011010010110110010110011100001011100110011100000110 30
629186867 638375952 401657765207022384 1CsPFVU64ucnGe6BQpdgmJ7d2NxWK465iU 10110010010111110011001110000000000011100010001101100110000 34
564723991 954488317 539022451739113147 1Kqn6vNXNUCjCPnaiir4Wan6Btf5V1YP4f 11101111010111111100001000001100001110111101011011010111011 23
875865707 475953115 416871011568327305 12yecuGZmuBxqsm8hCd6ue68A9Q7YmwvNh 10111001001000001011111100111110001101100000001101010001001 31
977325742 476974937 466159884218928254 19LpJpXqjrnUE4SZHvd7CVSdZEkRz9gGxw 11001111000001000011111000111100011011010000000110001111110 30
512674146 780048099 399910492993748454 1Br89YTpXawNHQr7z255pcHegiN412vTbt 10110001100110001000111100101111100110101111111110111100110 23
808812590 623132064 503997058595885760 1FPZMeN2nj67DU1kiZLFk2gBfu6pMnsf9j 11011111110100011101010011110001101011100110001001011000000 28
751799896 665682109 500459740315260664 1HMEeFbvcAdUWVm7gibGeyxk2KpNdJyXBh 11011110001111111010111101101110101111000000001011011111000 23
914920715 450766694 412415785972666210 1AeQpMLzSs9SH4AV3nd27Vf9TaKMFLMcyB 10110111001001100011111100011100110101010100000111101100010 28
541893875 730376406 395786500855913250 16uhZqPBHvMRXGhvuLANMtcV49zd4mRT6R 10101111110000111011011100110111110011010010101101100100010 25
742351746 661837156 491315968324274376 18KZfx7exdDzJ67NCa41DCVzrtVLTuidt2 11011010001100000010100010100100100011100100000100011001000 38
640479904 637703132 408436040763859328 1NVmf3kwES9oJLDXuLVEfTVS5g928T8eYk 10110101011000011100110101001000011001100100110000110000000 35
836538012 471895399 394758438951406788 18BjU9Ne4caWwABVBz17maQTHJ9JMPeVg8 10101111010011101101011010101101101110010010100110011000100 27
954216342 427817816 408230751425949072 1ESuMkPEMdRaxtgevkFhVcvdc9cqks7Vzn 10110101010010100111011010010011101001000100110110110010000 31
982503052 420048273 412698710209829196 16TAzPTHafofBHpk5i1jRt6sJ8BBFJnip8 10110111010001100110100101001010101010100011001100101001100 31
558546443 756797499 422706551137746057 16zvGSAEo8qu1ZYrq3usfDqpFa9u6rEUZc 10111011101110000010101111001100000110010111011110010001001 28
627917575 677787768 425594851647222600 1DFYQbDtYTYguyUjRL54ixqNmMXuTW9Dfw 10111101000000001000100001100111110101010101111001101001000 32
600424028 749967667 450298607489902676 1GGwTUSFrngjyBvRkRipwTN2gUzWJxscV8 11000111111110010000011001010011110010010011110100001010100 30
806914533 552014915 445428857346259695 19tSsZEiTnP5QivbB6i3t6jpKxdMFegDtN 11000101110011110110010111110001110110101001111101011101111 21
869130918 518297841 450468678345748038 132ZZFAWbbD1aFwuqCUBQMVsoaM8oM95BL 11001000000011000101110000001010011001111001110111001000110 33
468020592 860548543 402754438539597456 1JaxSTTVc3PhRroZvdAaKHTAXVTtKYbRvb 10110010110110111110000011100101000110111010101011010010000 29
538571208 902044595 485815247199020760 1NVL9WTxUHnBJJT8otKsfnP8NaXWfoetNB 11010111101111101100110010010110011110001000010011011011000 27
540854249 789587036 427051503375915964 1JMV8tviznVeMbBXTwLRVk5erzQvPi4cf 10111101101001100010001010001100111100001001001101110111100 29
576712145 676633259 390222618176230555 1BUskK9PMWkqHvD64CR2PCjj5Me74cpiEo 10101101010010110010110011101011011101101111001010010011011 25
841676700 589145791 495870285187769700 1D5tiBFp31uLEwM75XAjUgWUGRM8WvijSm 11011100001101011110110010110101011010010011110010101100100 27
818100651 599135207 490152902883719757 14AJhmbKC7CGxxQ5JAd6zuz6oDfAP42MW1 11011001101010111110111011111011110000000101011101001001101 24
662409474 620143188 410788722967763112 14bZUy6THC3tPLfL3Wawmq2CHi53MhYqvu 10110110011011010100010101011000110001110101101100010101000 30
600646677 702524396 421968943968832092 18ekDdyh94rt7Dm8ZgKqfRk3cSixMSY1Mg 10111011011001000101000010011010011100000100100101001011100 33
664622932 753645105 500889819372547860 19WqzLHRzDw1WaNJoGb2wtisY8fw6xKSvG 11011110011100001001010001100001011010011110110111100010100 29
427141223 949702206 405656961756637938 1KhDrSKjYHjCH4ANitkrgh39juEVVwSCdJ 10110100001001011101101101110000010001010110010101011110010 30

[/size]

Factorial decomposition of the last 8 puzzle addresses

2058769515153876   ~ 2, 2, 3, 7, 43, 53, 197, 2477, 22039
4216495639600700   ~ 2, 2, 5, 5, 53, 795565215019
6763683971478124   ~ 2, 2, 14359547, 117755873
9974455244496707   ~ 7019, 76123, 18668011
30045390491869460  ~ 2, 2, 5, 19, 79066817083867
44218742292676575  ~ 3, 5120419, 2878588796
138245758910846492 ~ 2, 2, 2, 2, 2, 3, 3, 7, 7, 13, 163, 3313, 1395439
199976667976342049 ~ 13, 31, 563, 881385470324

[umadbro]

80 * 1400000000 = 112000000000
576460752303423488-288230376151711744 = 288230376151711744
288230376151711744 / 112000000000 = ~2573485 sec

~2573485 sec = 42891 min
42891 min = 714 hours
714 hours = 29 days

...it is full time what i need to scan full keyspace for actually address.
29 - 14 (actually searched) = max two weeks = to next puzzle :-)

I assume that the right key will be the last opportunity: P

We might beat you to it