Bitcoin puzzle transaction ~32 BTC prize to who solves it

[iparktur]

I can't see the address on bitcoin blockchain.

Code:

19YZECXj3SxEZMoUeJ1yiPsw8xANe7M7GR 

and i want to ask, what equipment is needed to solve the puzzle and what software?

https://www.blockchain.com/ru/btc/address/19YZECXj3SxEZMoUeJ1yiPsw8xANe7M7QR

[1714735370]

1714735370

[1714735370]

1714735370

[1714735370]

1714735370

[PrivatePerson]

I think that beyond #85 it will be very difficult to recover the private key, even with 1 TB of RAM (with the Baby-Giant Step algorithm).

Nvme ssd as a swap?

[domozhirov]

I think that beyond #85 it will be very difficult to recover the private key, even with 1 TB of RAM (with the Baby-Giant Step algorithm).

Nvme ssd as a swap?

SSD is much slower than RAM. (DDR4 47GB/s)

[arulbero]

#80 found! Not by me ...

https://www.blockchain.com/btc/address/1BCf6rHUW6m3iH2ptsvnjgLruAiPQQepLe

[virus-cyber]

arulbero
well done now it is clear who takes Bitcoin

[iparktur]

It seems that this opens a new perspective. We do have now spending scripts for these puzzle wallet

#65 18ZMbwUFLMHoZBbfpCjUJQTCMCbktshgpe (0.00001 BTC - Sortie)
#70 19YZECXj3SxEZMoUeJ1yiPsw8xANe7M7QR (0.00001 BTC - Sortie)
#75 1J36UjUByGroXcCvmj13U6uwaVv9caEeAt (0.00001 BTC - Sortie)
#80 1BCf6rHUW6m3iH2ptsvnjgLruAiPQQepLe (0.00001 BTC - Sortie)
#85 1Kh22PvXERd2xpTQk3ur6pPEqFeckCJfAr (0.00001 BTC - Sortie)
#90 1L12FHH2FHjvTviyanuiFVfmzCy46RRATU (0.00001 BTC - Sortie)
#95 19eVSDuizydXxhohGh8Ki9WY9KsHdSwoQC (0.00001 BTC - Sortie)
#100  1KCgMv8fo2TPBpddVi9jqmMmcne9uSNJ5F (0.00001 BTC - Sortie)
#105 1CMjscKB3QW7SDyQ4c3C3DEUHiHRhiZVib (0.00001 BTC - Sortie)
#110 12JzYkkN76xkwvcPT6AWKZtGX6w2LAgsJg (0.00001 BTC - Sortie)
#115  1NLbHuJebVwUZ1XqDjsAyfTRUPwDQbemfv (0.00001 BTC - Sortie)
#120  17s2b9ksz5y7abUm92cHwG8jEPCzK3dLnT (0.00001 BTC - Sortie)
#125  1PXAyUB8ZoH3WD8n5zoAthYjN15yN5CVq5 (0.00001 BTC - Sortie)
#130  1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua (0.00001 BTC - Sortie)
#135  16RGFo6hjq9ym6Pj7N5H7L1NR1rVPJyw2v (0.00001 BTC - Sortie)
#140  1QKBaU6WAeycb3DbKbLBkX7vJiaS8r42Xo (0.00001 BTC - Sortie)
#145  19GpszRNUej5yYqxXoLnbZWKew3KdVLkXg (0.00001 BTC - Sortie)
#150  1MUJSJYtGPVGkBCTqGspnxyHahpt5Te8jy (0.00001 BTC - Sortie)
#155 1AoeP37TmHdFh8uN72fu9AqgtLrUwcv2wJ (0.00001 BTC - Sortie)
#160 1NBC8uXJy1GiJ6drkiZa1WuKn51ps7EPTv (0.00001 BTC - Sortie)

 a GPU version of this code https://gist.github.com/jhoenicke/2e39b3c6c49b1d7b216b8626197e4b89  or arulbero code is worthed .... 

It is possible that all these addresses will be reset to zero in the next few days.

[arulbero]

It is possible that all these addresses will be reset to zero in the next few days.

I don't think so,  the #85 and #90 for sure, maybe #95 and #100, but not the other addresses. And it will take many many days, not just a few.

[AndreuSmetanin]

My life would change abruptly if I found the key to one of the addresses with the cue ball here is my address just in case  39xoA35q27BZEvc5acyPmBwvZ3xVfqvnn5

[Thirdspace]

arulbero
It seems that someone is doing something very odd here. the above recent transaction https://www.blockchain.com/de/btc/tx/17e4e323cfbc68d7f0071cad09364e8193eedf8fefbcbd8a21b4b65717a4b3d3
~
Who else other than puzzle owner can spend from theses wallets??

I think that's the reason he can find #65 private key, exposed public key makes it easier  
@arulbero can you tell us how you found it? details on how you used baby-step giant-step algorithm

I think that beyond #85 it will be very difficult to recover the private key, even with 1 TB of RAM (with the Baby-Giant Step algorithm).

With my 32 GB finding the #70 is already a hard task.

But there are other algorithms more suitable that don't need so much ram.

I don't understand thoroughly the algorithm
can you explain in short why the size of RAM matters in this algorithm?
I thought it just generates privkey hex sequentially, finds corresponding pubkey and compares it to target pubkey

[Bajula]

Since I don't have the resources to hope to compete here and everyone else is likely frantically converting pubkeys into byte arrays and re-writing thier stuff...   So anyway the breakshort program uses the baby step giant step algo and the public key  ( https://en.wikipedia.org/wiki/Baby-step_giant-step ) to basically cut the searching down to the squareroot  but it is ram intensive (like build a hashtable that can hold 2^80 and it can then search 2^160 and that is VERY cool, BUT the sheer ram needed to do something like that doesn't exist at the moment. ) say you have 8gb ram you can do maybe* 2^27 in the hashtable which seems to search "about" 2^55 keyspace - in a freakishly short amount of time. Kinda awesome right? say you wanna do 2^28 hashtable.. now you just doubled memory requirements. 
* maybe is because of certain variables.. like the breakshort program as is, you can in theory do 2^29? (i think) with 8gb but with unint32_t you have potential for false collisions - and on my comp for some reason while 2^28 "should" work fine - 2^27 takes 52% mem so if I try 2^28 it starts using the swap file and SIGNIFICANTLY slowing it down. you figure that 4% wouldn't be THAT big a deal but it is the difference between driving a car a mile vs riding a skateboard a mile and a half.  All over this thread is all kinds of info that is far more informative than I'm being - look around, have fun with it. (seriously I started out as "What?!?! free bitcoin!!! and then got sucked into teaching myself C (With quite a bit of help - you know who you are and thank you again) At my age this is kind of a "thing" - and going from not having a clue what an elliptic curve is to being fascinated by cryptography (well okay that whole journey started in 2012? 13? when my son first said the word bitcoin and I was like "Huh?" ) -:) anyway go back as many pages as you need to and happy hunting.

arulbero
It seems that someone is doing something very odd here. the above recent transaction https://www.blockchain.com/de/btc/tx/17e4e323cfbc68d7f0071cad09364e8193eedf8fefbcbd8a21b4b65717a4b3d3
~
Who else other than puzzle owner can spend from theses wallets??

I think that's the reason he can find #65 private key, exposed public key makes it easier  
@arulbero can you tell us how you found it? details on how you used baby-step giant-step algorithm

I think that beyond #85 it will be very difficult to recover the private key, even with 1 TB of RAM (with the Baby-Giant Step algorithm).

With my 32 GB finding the #70 is already a hard task.

But there are other algorithms more suitable that don't need so much ram.

I don't understand thoroughly the algorithm
can you explain in short why the size of RAM matters in this algorithm?
I thought it just generates privkey hex sequentially, finds corresponding pubkey and compares it to target pubkey

[j2002ba2]

I don't understand thoroughly the algorithm
can you explain in short why the size of RAM matters in this algorithm?
I thought it just generates privkey hex sequentially, finds corresponding pubkey and compares it to target pubkey

You could look up the algorithms here: https://www.math.auckland.ac.nz/~sgal018/crypto-book/crypto-book.html
BSGS is in chapter 13.3
When memory is limited - Distributed Kangaroo in chapter 14.6

[mrxtraf]

I try description algorim. But I do not understand a few moments.

Code:

int main(int argc, char **argv) {
    secp256k1_context *ctx = secp256k1_context_create(SECP256K1_CONTEXT_NONE);

    int next = 0;//Initial varibale next for cicl search in publick kes
    
    //Convert publik keys from raw to eckey format. Nothing intresested. 
    for (int i = 0; i < NUMPUBKEYS; i++) {
        if (!secp256k1_eckey_pubkey_parse(&pubkeys[i], rawpubkeys[i], 33)) {
            printf("Unparsable pubkey %2d\n", i);
            return -1;
        }
    }

    printf("Build Hash\n");
    secp256k1_gej pt;//Init variable pt
    secp256k1_gej_set_ge(&pt, &secp256k1_ge_const_g);//??????
    //Start cicle from 1 to GSTEP (1<<25 or other count bits). With step one.
    for (size_t i = 1; i < GSTEP; i++) {
        /*if(i%1000000==0){
    	    printf("Generate %zu from %2d \n", i, GSTEP);
        }*/
        secp256k1_fe x,zinv;//Init variable z and zinv
        secp256k1_fe_storage xst;//Init variable xst
        secp256k1_fe_inv_var(&zinv, &pt.z);//????????? Maybe inverted variable. But zinv or pt.z?
        secp256k1_fe_sqr(&zinv, &zinv);//Sqr from who? Sqr zinv from zinv?
        secp256k1_fe_mul(&x, &pt.x, &zinv);//Multiple x pt.x and zinv. But who changes?
        secp256k1_fe_to_storage(&xst, &x);//Return hash data to xst from x . xst this array from 8 part of hashes.
        uint32_t entry = xst.n[0] & (HASH_SIZE-1);//In entry getted last (25-1 or other setted count bit) bit from xst.n[0] (first part from hash)
        while (table[entry].exponent != 0) {//Cicle run if in table with key entry already setted data.
            entry = (entry + (xst.n[1] | 1)) & (HASH_SIZE - 1);//Changed entry. In current xst.n[1] setted last bit to 1 plus add current entry. From this value get last 25-1 or other setted bits.
        }//This is algortim searned free row in table with changed key by algoritm in up.
        table[entry].exponent = i;//Set in table with key entry to subkey exponent varuibale i (currently pozition in main for)
        table[entry].x = xst.n[2];//Set in table with key entry to subkey x, xst.n[2] (part of hash from storage)
        //------------
        //I try inserte here searched this hash in curently public keys. Algoritm founded, but maksimal found 25 or other setted bit, no more! Logical is true.
        //------------
        secp256k1_gej_add_ge_var(&pt, &pt, &secp256k1_ge_const_g, NULL);//????????????
    }
    //End generated main table
    //But undestord. Variable i not used for generated hashes. Who is used aka privatkey?

    printf("Search Keys\n");
    secp256k1_ge ptgstep;//Init variable ptgstep
    secp256k1_gej_neg(&pt, &pt);//Negativation pt ????? pt from previos step?
    secp256k1_gej_double_var(&pt, &pt, NULL);//Double pt ???
    secp256k1_ge_set_gej(&ptgstep, &pt);//????
    secp256k1_gej_set_infinity(&pt);//????
    //In up init variable from main cicl.
    
    //Start cicl i from 0 (previos cilck start from 1). To 2*GSTEP (double gstep). With step 1.
    for (size_t i = 0; i < 2*GSTEP; i++) {
        //Start cicl j from next (dinamic variable for exclude founded keys on begin array). To NUMPUBKEYS (coun publick keys). With step 1.
        for (int j = next; j < NUMPUBKEYS; j++) {
            secp256k1_gej diff;//Init variable diff
            secp256k1_fe x,zinv;//Init variable x and zinv
            secp256k1_fe_storage xst;//Init variable xst
            secp256k1_gej_add_ge_var(&diff, &pt, &pubkeys[j],  NULL);//????? May be added variable diff, pt and pubkeys[j] (currently publick key)
            secp256k1_fe_inv_var(&zinv, &diff.z);//??????Maybe inverted variable. But zinv or pt.z?
            secp256k1_fe_sqr(&zinv, &zinv);//Sqr from who? Sqr zinv from zinv?
            secp256k1_fe_mul(&x, &diff.x, &zinv);//Multiple x pt.x and zinv. But who changes?
            secp256k1_fe_to_storage(&xst, &x);//Return hash data to xst from x . xst this array from 8 part of hashes.
            uint32_t entry = xst.n[0] & (HASH_SIZE-1);//In entry getted last (25-1 or other setted count bit) bit from xst.n[0] (first part from hash)
            //-----------------
            //I try showed this entry for each publcik addr. And with each new cycle, this value was different for the same public key. Why?
            //-----------------
            while (table[entry].exponent != 0) {//Cicl run if in table with key entry present data.
                if (table[entry].x == (uint32_t) xst.n[2]) {//If table[entry].x (hash part 2 generated in in prevos loop) equal xst.n[2] (hash part 2 from currently publik key) run block down
                    uint64_t key = (uint64_t) i *  (uint64_t) (2 * GSTEP);//Generate varibale key. i * (2*GSTEP) . !!!!!!!!!!! Do not understand the logic. !!!!!!!!!!
                    //show founded key
                    printf("Found private key %2d: %16lx or %16lx\n", j + 1,
                           key - table[entry].exponent,
                           key + table[entry].exponent);
                    next++;//Add in variable next +1; Exclude this publick key from next searched.
                    if (next == NUMPUBKEYS)//if next equal NUMPUBKEYS, found ded last key, programm is stop
                        return 0;
                }        
                entry = (entry + (xst.n[1] | 1)) & (HASH_SIZE - 1);//Changed entry. In current xst.n[1] setted last bit to 1 plus add current entry. From this value get last 25-1 or other setted bits.
            }
            if (j == next)///????????? In cycl each firt loop this is true, break? not searched other public keys.
                break;
        }
        secp256k1_gej_add_ge_var(&pt, &pt, &ptgstep, NULL);//?????????? Choto gdeto kakto
    }
    return 0;
}

[arulbero]

It is possible that all these addresses will be reset to zero in the next few days.

I don't think so,  the #85 and #90 for sure, maybe #95 and #100, but not the other addresses. And it will take many many days, not just a few.

From : https://en.wikipedia.org/wiki/Discrete_logarithm_records

In July 2009, Joppe W. Bos, Marcelo E. Kaihara, Thorsten Kleinjung, Arjen K. Lenstra and Peter L. Montgomery announced that they had carried out a discrete logarithm computation on an elliptic curve modulo a 112-bit prime. The computation was done on a cluster of over 200 PlayStation 3 game consoles over about 6 months. They used the common parallelized version of Pollard rho method.

112 bit is the current record for the ECDLP (Elliptic Curve Discrete Logarithm Problem = retrieve the private key from the public key)

[PrivatePerson]

112 bit is the current record for the ECDLP (Elliptic Curve Discrete Logarithm Problem = retrieve the private key from the public key)

I understood correctly: if you have 200 playstation3 and 6 months, then you can find a private key to any Bitcoin address that had outgoing transactions?

[Dr.Hash]

112 bit is the current record for the ECDLP (Elliptic Curve Discrete Logarithm Problem = retrieve the private key from the public key)

I understood correctly: if you have 200 playstation3 and 6 months, then you can find a private key to any Bitcoin address that had outgoing transactions?

Um no it doesnt work like that

[j2002ba2]

112 bit is the current record for the ECDLP (Elliptic Curve Discrete Logarithm Problem = retrieve the private key from the public key)

I understood correctly: if you have 200 playstation3 and 6 months, then you can find a private key to any Bitcoin address that had outgoing transactions?

If you have 2^143 * 200 = 2230074519853062314153571827264836150598041600 PS3 and 6 months. And the electricity to power them, let's say about 100W per PS3.

[PrivatePerson]

If you have 2^143 * 200 = 2230074519853062314153571827264836150598041600 PS3 and 6 months. And the electricity to power them, let's say about 100W per PS3.

Just? Count all bitcoins in my pocket 

[arulbero]

112 bit is the current record for the ECDLP (Elliptic Curve Discrete Logarithm Problem = retrieve the private key from the public key)

I understood correctly: if you have 200 playstation3 and 6 months, then you can find a private key to any Bitcoin address that had outgoing transactions?

No, you could find any private key in a 120 bit space, but Bitcoin uses a 256 bit space.

You could find only the private keys of the puzzle transaction below 120 bit: #85, #90, #95, #100, #105 ... #115 bit

[JDScreesh]

01 0000000000000000000000000000000000000000000000000000000000000001 1BgGZ9tcN4rm9KBzDn7KprQz87SZ26SAMH
02 0000000000000000000000000000000000000000000000000000000000000003 1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb
03 0000000000000000000000000000000000000000000000000000000000000007 19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA
04 0000000000000000000000000000000000000000000000000000000000000008 1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e
05 0000000000000000000000000000000000000000000000000000000000000015 1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k
06 0000000000000000000000000000000000000000000000000000000000000031 1PitScNLyp2HCygzadCh7FveTnfmpPbfp8
07 000000000000000000000000000000000000000000000000000000000000004c 1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC
08 00000000000000000000000000000000000000000000000000000000000000e0 1M92tSqNmQLYw33fuBvjmeadirh1ysMBxK
09 00000000000000000000000000000000000000000000000000000000000001d3 1CQFwcjw1dwhtkVWBttNLDtqL7ivBonGPV
10 0000000000000000000000000000000000000000000000000000000000000202 1LeBZP5QCwwgXRtmVUvTVrraqPUokyLHqe
11 0000000000000000000000000000000000000000000000000000000000000483 1PgQVLmst3Z314JrQn5TNiys8Hc38TcXJu
12 0000000000000000000000000000000000000000000000000000000000000a7b 1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot
13 0000000000000000000000000000000000000000000000000000000000001460 1Pie8JkxBT6MGPz9Nvi3fsPkr2D8q3GBc1
14 0000000000000000000000000000000000000000000000000000000000002930 1ErZWg5cFCe4Vw5BzgfzB74VNLaXEiEkhk
15 00000000000000000000000000000000000000000000000000000000000068f3 1QCbW9HWnwQWiQqVo5exhAnmfqKRrCRsvW
16 000000000000000000000000000000000000000000000000000000000000c936 1BDyrQ6WoF8VN3g9SAS1iKZcPzFfnDVieY
17 000000000000000000000000000000000000000000000000000000000001764f 1HduPEXZRdG26SUT5Yk83mLkPyjnZuJ7Bm
18 000000000000000000000000000000000000000000000000000000000003080d 1GnNTmTVLZiqQfLbAdp9DVdicEnB5GoERE
19 000000000000000000000000000000000000000000000000000000000005749f 1NWmZRpHH4XSPwsW6dsS3nrNWfL1yrJj4w
20 00000000000000000000000000000000000000000000000000000000000d2c55 1HsMJxNiV7TLxmoF6uJNkydxPFDog4NQum
21 00000000000000000000000000000000000000000000000000000000001ba534 14oFNXucftsHiUMY8uctg6N487riuyXs4h
22 00000000000000000000000000000000000000000000000000000000002de40f 1CfZWK1QTQE3eS9qn61dQjV89KDjZzfNcv
23 0000000000000000000000000000000000000000000000000000000000556e52 1L2GM8eE7mJWLdo3HZS6su1832NX2txaac
24 0000000000000000000000000000000000000000000000000000000000dc2a04 1rSnXMr63jdCuegJFuidJqWxUPV7AtUf7
25 0000000000000000000000000000000000000000000000000000000001fa5ee5 15JhYXn6Mx3oF4Y7PcTAv2wVVAuCFFQNiP
26 000000000000000000000000000000000000000000000000000000000340326e 1JVnST957hGztonaWK6FougdtjxzHzRMMg
27 0000000000000000000000000000000000000000000000000000000006ac3875 128z5d7nN7PkCuX5qoA4Ys6pmxUYnEy86k
28 000000000000000000000000000000000000000000000000000000000d916ce8 12jbtzBb54r97TCwW3G1gCFoumpckRAPdY
29 0000000000000000000000000000000000000000000000000000000017e2551e 19EEC52krRUK1RkUAEZmQdjTyHT7Gp1TYT
30 000000000000000000000000000000000000000000000000000000003d94cd64 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps
31 000000000000000000000000000000000000000000000000000000007d4fe747 1LhE6sCTuGae42Axu1L1ZB7L96yi9irEBE
32 00000000000000000000000000000000000000000000000000000000b862a62e 1FRoHA9xewq7DjrZ1psWJVeTer8gHRqEvR
33 00000000000000000000000000000000000000000000000000000001a96ca8d8 187swFMjz1G54ycVU56B7jZFHFTNVQFDiu
34 000000000000000000000000000000000000000000000000000000034a65911d 1PWABE7oUahG2AFFQhhvViQovnCr4rEv7Q
35 00000000000000000000000000000000000000000000000000000004aed21170 1PWCx5fovoEaoBowAvF5k91m2Xat9bMgwb
36 00000000000000000000000000000000000000000000000000000009de820a7c 1Be2UF9NLfyLFbtm3TCbmuocc9N1Kduci1
37 0000000000000000000000000000000000000000000000000000001757756a93 14iXhn8bGajVWegZHJ18vJLHhntcpL4dex
38 00000000000000000000000000000000000000000000000000000022382facd0 1HBtApAFA9B2YZw3G2YKSMCtb3dVnjuNe2
39 0000000000000000000000000000000000000000000000000000004b5f8303e9 122AJhKLEfkFBaGAd84pLp1kfE7xK3GdT8
40 000000000000000000000000000000000000000000000000000000e9ae4933d6 1EeAxcprB2PpCnr34VfZdFrkUWuxyiNEFv
41 00000000000000000000000000000000000000000000000000000153869acc5b 1L5sU9qvJeuwQUdt4y1eiLmquFxKjtHr3E
42 000000000000000000000000000000000000000000000000000002a221c58d8f 1E32GPWgDyeyQac4aJxm9HVoLrrEYPnM4N
43 000000000000000000000000000000000000000000000000000006bd3b27c591 1PiFuqGpG8yGM5v6rNHWS3TjsG6awgEGA1
44 00000000000000000000000000000000000000000000000000000e02b35a358f 1CkR2uS7LmFwc3T2jV8C1BhWb5mQaoxedF
45 0000000000000000000000000000000000000000000000000000122fca143c05  1NtiLNGegHWE3Mp9g2JPkgx6wUg4TW7bbk
46 00000000000000000000000000000000000000000000000000002ec18388d544 1F3JRMWudBaj48EhwcHDdpeuy2jwACNxjP
47 00000000000000000000000000000000000000000000000000006cd610b53cba 1Pd8VvT49sHKsmqrQiP61RsVwmXCZ6ay7Z
48 0000000000000000000000000000000000000000000000000000ade6d7ce3b9b 1DFYhaB2J9q1LLZJWKTnscPWos9VBqDHzv
49 000000000000000000000000000000000000000000000000000174176b015f4d 12CiUhYVTTH33w3SPUBqcpMoqnApAV4WCF
50 00000000000000000000000000000000000000000000000000022bd43c2e9354 1MEzite4ReNuWaL5Ds17ePKt2dCxWEofwk
51 00000000000000000000000000000000000000000000000000075070a1a009d4 1NpnQyZ7x24ud82b7WiRNvPm6N8bqGQnaS
52 000000000000000000000000000000000000000000000000000efae164cb9e3c  15z9c9sVpu6fwNiK7dMAFgMYSK4GqsGZim
53 00000000000000000000000000000000000000000000000000180788e47e326c 15K1YKJMiJ4fpesTVUcByoz334rHmknxmT
54 00000000000000000000000000000000000000000000000000236fb6d5ad1f43  1KYUv7nSvXx4642TKeuC2SNdTk326uUpFy
55 000000000000000000000000000000000000000000000000006abe1f9b67e114 1LzhS3k3e9Ub8i2W1V8xQFdB8n2MYCHPCa
56 000000000000000000000000000000000000000000000000009d18b63ac4ffdf   17aPYR1m6pVAacXg1PTDDU7XafvK1dxvhi
57 00000000000000000000000000000000000000000000000001eb25c90795d61c 15c9mPGLku1HuW9LRtBf4jcHVpBUt8txKz
58 00000000000000000000000000000000000000000000000002c675b852189a21 1Dn8NF8qDyyfHMktmuoQLGyjWmZXgvosXf
59 00000000000000000000000000000000000000000000000007496cbb87cab44f  1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt
60 0000000000000000000000000000000000000000000000000fc07a1825367bbe  1Kn5h2qpgw9mWE5jKpk8PP4qvvJ1QVy8su
61 00000000000000000000000000000000000000000000000013C96A3742F64906 1AVJKwzs9AskraJLGHAZPiaZcrpDr1U6AB
.....
65 000000000000000000000000000000000000000000000001a838b13505b26867 18ZMbwUFLMHoZBbfpCjUJQTCMCbktshgpe
.....
70 ----------------------------- Unknown Private Key spent ----------------------------- 19YZECXj3SxEZMoUeJ1yiPsw8xANe7M7QR
.....
75 ----------------------------- Unknown Private Key spent ----------------------------- 1J36UjUByGroXcCvmj13U6uwaVv9caEeAt
.....
80 ----------------------------- Unknown Private Key spent ----------------------------- 1BCf6rHUW6m3iH2ptsvnjgLruAiPQQepLe

    

[mrxtraf]

.....
80 ----------------------------- Unknown Private Key spent ----------------------------- 1BCf6rHUW6m3iH2ptsvnjgLruAiPQQepLe
.....
85 ---------- Unknown Private Key spent 0.00001 BTC from 0.85001 BTC ------------ 1Kh22PvXERd2xpTQk3ur6pPEqFeckCJfAr
.....
90 ---------- Unknown Private Key spent 0.00001 BTC from 0.90001 BTC ------------ 1L12FHH2FHjvTviyanuiFVfmzCy46RRATU
.....
95 ---------- Unknown Private Key spent 0.00001 BTC from 0.95001 BTC ------------ 19eVSDuizydXxhohGh8Ki9WY9KsHdSwoQC
.....
100 --------- Unknown Private Key spent 0.00001 BTC from 1.00001 BTC ------------ 1KCgMv8fo2TPBpddVi9jqmMmcne9uSNJ5F
.....
105 --------- Unknown Private Key spent 0.00001 BTC from 1.05001 BTC ------------ 1CMjscKB3QW7SDyQ4c3C3DEUHiHRhiZVib
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110 --------- Unknown Private Key spent 0.00001 BTC from 1.10001 BTC ------------ 12JzYkkN76xkwvcPT6AWKZtGX6w2LAgsJg
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115 --------- Unknown Private Key spent 0.00001 BTC from 1.15001 BTC ------------ 1NLbHuJebVwUZ1XqDjsAyfTRUPwDQbemfv
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120 --------- Unknown Private Key spent 0.00001 BTC from 1.20001 BTC ------------ 17s2b9ksz5y7abUm92cHwG8jEPCzK3dLnT
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125 --------- Unknown Private Key spent 0.00001 BTC from 1.25001 BTC ------------ 1PXAyUB8ZoH3WD8n5zoAthYjN15yN5CVq5
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130 --------- Unknown Private Key spent 0.00001 BTC from 1.30001 BTC ------------ 1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua
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135 --------- Unknown Private Key spent 0.00001 BTC from 1.35001 BTC ------------ 16RGFo6hjq9ym6Pj7N5H7L1NR1rVPJyw2v
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140 --------- Unknown Private Key spent 0.00001 BTC from 1.40001 BTC ------------ 1QKBaU6WAeycb3DbKbLBkX7vJiaS8r42Xo
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145 --------- Unknown Private Key spent 0.00001 BTC from 1.45001 BTC ------------ 19GpszRNUej5yYqxXoLnbZWKew3KdVLkXg
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150 --------- Unknown Private Key spent 0.00001 BTC from 1.50001 BTC ------------ 1MUJSJYtGPVGkBCTqGspnxyHahpt5Te8jy
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155 --------- Unknown Private Key spent 0.00001 BTC from 1.55001 BTC ------------ 1AoeP37TmHdFh8uN72fu9AqgtLrUwcv2wJ
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160 --------- Unknown Private Key spent 0.00001 BTC from 1.60001 BTC ------------ 1NBC8uXJy1GiJ6drkiZa1WuKn51ps7EPTv