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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 186268 times)
Andzhig
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October 17, 2020, 08:35:56 PM
Last edit: October 18, 2020, 08:39:39 PM by Andzhig
 #1441

Continuation "road to calvary".

looking for a 20 digit number, example 30568377312064202855 by 2 > 30 56 83 77 31 20 64 20 28 55 (10 pieces for mixing)

30568377312064202855 by 2 > 30 56 83 77 31 20 64 20 28 55

take the first 100 numbers power of 2 (4096-32768 or 4096-30000000)

these first 100 digits split by 2 (50 elements in list) and look at what positions our numbers (30 56 83 77 31 20 64 20 28 55) will be, then we sort them from smallest to largest,

some like this...

1   [0, 6, 7, 10, 16, 21, 24, 34, 35, 40]
2   [3, 4, 5, 11, 12, 23, 27, 40, 43, 48]
3   [8, 9, 10, 13, 17, 24, 27, 28, 35, 42]
4   [4, 7, 8, 18, 20, 22, 24, 27, 33, 36]
5   [4, 7, 8, 11, 13, 14, 21, 29, 38, 48]
6   [2, 5, 7, 12, 19, 23, 31, 33, 40, 44]
7   [0, 3, 7, 11, 13, 18, 20, 25, 35, 43]
8   [4, 13, 15, 17, 19, 25, 30, 33, 40, 46]
9   [1, 2, 4, 15, 18, 29, 35, 40, 43, 45]
10   [0, 3, 4, 5, 14, 17, 22, 32, 36, 38]

and see the difference between the numbers.

some like this...

95465   [0, 1, 10, 11, 14, 15, 22, 31, 40, 44, 47]     1 9 1 3 1 7 9 9 4 3
273880   [0, 2, 12, 15, 19, 21, 23, 31, 32, 37, 44]    2 10 3 4 2 2 8 1 5 7
310872   [0, 9, 10, 19, 23, 27, 28, 29, 41, 43, 47]    9 1 9 4 4 1 1 12 2 4
324831   [6, 10, 19, 20, 22, 23, 26, 33, 34, 35, 40]  4 9 1 2 1 3 7 1 1 5
327766   [1, 2, 5, 11, 14, 24, 27, 28, 45, 46, 49]     1 3 6 3 10 3 1 17 1 3
342364   [1, 5, 13, 15, 16, 20, 21, 26, 32, 38, 43]    4 8 2 1 4 1 5 6 6 5
484256   [3, 7, 10, 11, 15, 17, 18, 20, 41, 43, 49]    4 3 1 4 2 1 2 21 2 6
519343   [0, 2, 5, 6, 10, 24, 27, 28, 29, 40, 48]       2 3 1 4 14 3 1 1 11 8
540844   [2, 3, 6, 7, 8, 13, 29, 36, 37, 39, 45]         1 3 1 1 5 16 7 1 2 6
744223   [4, 8, 10, 13, 14, 16, 18, 26, 29, 44, 48]     4 2 3 1 2 2 8 3 15 4
773619   [2, 3, 11, 12, 13, 18, 20, 28, 31, 33, 49]     1 8 1 1 5 2 8 3 2 16
804181   [2, 6, 10, 12, 16, 33, 35, 37, 39, 45, 46]     4 4 2 4 17 2 2 2 6 1
816903   [3, 6, 11, 18, 20, 22, 23, 30, 31, 38, 45]     3 5 7 2 2 1 7 1 7 7

now we filter and look for the "formula")))

here we have several components in charge of search time, the size power of 2, first difference numbers to which to stick, the difference, and possible differences + ~10000000 for mixing 10 parts by 2. i.e. some like this    

60466176 6x10  60466176×2^262144 = 15850845241344
9765625   5x10  9765625×2^262144  = 2560000000000
blablabla

besides need to do an offset (take the first 100 numbers power of 2) 0-100 num, 100-200 num, 200-300 num...

exemple 1

2^4096-2^32768

filt <=6 <=6 <=6 <=6 <=6 <=6 <=6 <=6 <=6

0 100
2^15047   [8, 11, 12, 15, 17, 19, 24, 24, 28, 34]     3 1 3 2 2 5 0 4 6
100 200
200 300
300 400
400 500
500 600
600 700
700 800
800 900
900 1000
2^21946   [12, 15, 15, 18, 22, 25, 31, 37, 39, 44]     3 0 3 4 3 6 6 2 5
1000 1100

exemple 2

2^4096-2^30000000

filt <=1 <=1 <=1 <=1 <=1 <=30 <=30 <=30 <=30

2^7788201   [1, 1, 2, 3, 4, 5, 19, 33, 35, 48]          0 1 1 1 1 14 14 2 13
2^20398384   [0, 0, 1, 2, 3, 4, 16, 17, 33, 41]         0 1 1 1 1 12 1 16 8
0-100
2^371466   [5, 6, 7, 7, 8, 9, 10, 11, 19, 20]            1 1 0 1 1 1 1 8 1
2^21846679   [2, 3, 4, 5, 5, 6, 21, 25, 27, 31]         1 1 1 0 1 15 4 2 4
100-200
200-300
2^23749755   [7, 8, 9, 10, 10, 11, 16, 18, 32, 36]     1 1 1 0 1 5 2 14 4
2^29029857   [5, 5, 6, 7, 8, 9, 14, 19, 35, 36]          0 1 1 1 1 5 5 16 1
300-400
2^1061949   [0, 0, 1, 2, 3, 4, 20, 38, 45, 47]            0 1 1 1 1 16 18 7 2
...


Quote
#import collections
#import matplotlib.pyplot as plt
#import time



j=[]

count = 0

ii = 0
while ii <= 1000:
    
    i = 4096 # 2^ start
    while i <= 32768: # 2^ end #1024 2048 4096 8192 16384 32768 65536 131072
        

    #L = open(u"D:/prime/pow65206000/pow65.txt", "r")   #(u"C:/pow65_2.txt", "r")
    #for line in L:
        a = pow(2,i)
        s1=0+ii
        s2=100+ii
        n = str(a)[s1:s2]
        gg = ([n[i:i + 2] for i in range(0, len(n), 2)])
        ggg = gg
        
        
        nn1= "30"
        nn2= "56"
        nn3= "83"
        nn4= "77"
        nn5= "31"
        nn6= "20"
        nn7= "64"
        nn8= "20"
        nn9= "28"
        nn10= "55"
        
        count += 1
        if nn1 in ggg:
            v1 = ggg.index(nn1)
            gg1 = int(v1)
            if gg1 <= 100:
                jj = j.append(gg1)

                if nn2 in ggg:
                    v2 = ggg.index(nn2)
                    gg2 = int(v2)
                    if gg2 <= 100:
                        jj = j.append(gg2)

                        if nn3 in ggg:
                            v3 = ggg.index(nn3)
                            gg3 = int(v3)
                            if gg3 <= 100:
                                jj = j.append(gg3)

                                if nn4 in ggg:
                                    v4 = ggg.index(nn4)
                                    gg4 = int(v4)
                                    if gg4 <= 100:
                                        jj = j.append(gg4)

                                        if nn5 in ggg:
                                            v5 = ggg.index(nn5)
                                            gg5 = int(v5)
                                            if gg5 <= 100:
                                                jj = j.append(gg5)

                                                if nn6 in ggg:
                                                    v6 = ggg.index(nn6)
                                                    gg6 = int(v6)
                                                    if gg6 <= 100:
                                                        jj = j.append(gg6)

                                                        if nn7 in ggg:
                                                            v7 = ggg.index(nn7)
                                                            gg7 = int(v7)
                                                            if gg7 <= 100:
                                                                jj = j.append(gg7)

                                                                if nn8 in ggg:
                                                                    v8 = ggg.index(nn8)
                                                                    gg8 = int(v8)
                                                                    if gg8 <= 100:
                                                                        jj = j.append(gg8)

                                                                        if nn9 in ggg:
                                                                            v9 = ggg.index(nn9)
                                                                            gg9 = int(v9)
                                                                            if gg9 <= 100:
                                                                                jj = j.append(gg9)

                                                                                if nn10 in ggg:
                                                                                    v10 = ggg.index(nn10)
                                                                                    gg10 = int(v10)
                                                                                    if gg10 <= 100:
                                                                                        jj = j.append(gg10)

                                                                                        

                                                                                              
                                                                                                #count += 1
                                                                                        kkk = gg1,gg2,gg3,gg4,gg5,gg6,gg7,gg8,gg9,gg10
                                                                                        kkkk = sorted(kkk)
                                                                                        fa1 = kkkk[1]-kkkk[0]
                                                                                        if fa1 <= 6:
                                                                                            fa2 = kkkk[2]-kkkk[1]
                                                                                            if fa2 <=6:
                                                                                                fa3 = kkkk[3]-kkkk[2]
                                                                                                if fa3 <=6:
                                                                                                    fa4 = kkkk[4]-kkkk[3]
                                                                                                    if fa4 <=6:
                                                                                                        fa5 = kkkk[5]-kkkk[4]
                                                                                                        if fa5 <=6:
                                                                                                            fa6 = kkkk[6]-kkkk[5]
                                                                                                            if fa6 <=6:
                                                                                                                fa7 = kkkk[7]-kkkk[6]
                                                                                                                if fa7 <=6:
                                                                                                                    fa8 = kkkk[8]-kkkk[7]
                                                                                                                    if fa8 <=6:
                                                                                                                        fa9 = kkkk[9]-kkkk[8]
                                                                                                                        if fa9 <=6:
                                                                                                                                

                                                                                                                            print(count," ",kkkk,"   ",fa1,fa2,fa3,fa4,fa5,fa6,fa7,fa8,fa9)
                                                                                                
                                                                                                                                                                                                                        

                                                                                                                                                                                                                        

                                                                              
            else:
                pass #print(d1,d2)
        i=i+1
    count = 0        
    ii=ii+100
    print(s1,s2)
#print(j)









#l = j
#w = collections.Counter(l)
#plt.bar(w.keys(), w.values())
#plt.show()
#time.sleep(360.0)



all filters by 30

4923   [9, 10, 15, 23, 37, 43, 46, 48, 49, 49]     1 5 8 14 6 3 2 1 0
8124   [4, 6, 19, 19, 27, 30, 34, 38, 41, 48]     2 13 0 8 3 4 4 3 7
8817   [1, 2, 3, 6, 9, 11, 11, 15, 26, 43]     1 1 3 3 2 0 4 11 17
12047   [1, 9, 15, 32, 33, 33, 40, 45, 46, 49]     8 6 17 1 0 7 5 1 3
18834   [0, 0, 3, 6, 7, 10, 13, 22, 23, 26]     0 3 3 1 3 3 9 1 3
21377   [1, 9, 14, 25, 30, 33, 40, 40, 43, 46]     8 5 11 5 3 7 0 3 3
24186   [3, 4, 10, 14, 15, 20, 20, 29, 33, 48]     1 6 4 1 5 0 9 4 15
24901   [1, 3, 4, 4, 5, 11, 20, 21, 32, 47]     2 1 0 1 6 9 1 11 15
0 100
7632   [3, 5, 5, 7, 9, 13, 29, 38, 39, 46]     2 0 2 2 4 16 9 1 7
11284   [15, 19, 21, 24, 25, 26, 30, 30, 42, 45]     4 2 3 1 1 4 0 12 3
15047   [8, 11, 12, 15, 17, 19, 24, 24, 28, 34]     3 1 3 2 2 5 0 4 6
15604   [0, 2, 14, 17, 28, 29, 29, 32, 47, 49]     2 12 3 11 1 0 3 15 2
15979   [9, 11, 12, 17, 17, 27, 31, 32, 36, 43]     2 1 5 0 10 4 1 4 7
16582   [2, 7, 18, 22, 24, 40, 40, 41, 48, 49]     5 11 4 2 16 0 1 7 1
17250   [0, 3, 9, 12, 14, 14, 15, 21, 27, 39]     3 6 3 2 0 1 6 6 12
100 200
17445   [0, 4, 5, 7, 11, 17, 17, 26, 29, 33]     4 1 2 4 6 0 9 3 4
18118   [2, 3, 3, 5, 13, 22, 28, 43, 46, 49]     1 0 2 8 9 6 15 3 3
18680   [1, 9, 12, 20, 21, 23, 23, 26, 30, 44]     8 3 8 1 2 0 3 4 14
18954   [0, 3, 21, 22, 23, 42, 43, 43, 44, 48]     3 18 1 1 19 1 0 1 4
20267   [7, 7, 11, 14, 22, 25, 34, 38, 41, 46]     0 4 3 8 3 9 4 3 5
20327   [5, 16, 17, 35, 38, 39, 39, 42, 44, 48]     11 1 18 3 1 0 3 2 4
20470   [2, 13, 19, 22, 23, 25, 25, 27, 45, 46]     11 6 3 1 2 0 2 18 1
200 300
873   [0, 6, 6, 10, 11, 14, 18, 31, 34, 38]     6 0 4 1 3 4 13 3 4
11157   [5, 8, 10, 10, 17, 22, 27, 36, 39, 47]     3 2 0 7 5 5 9 3 8
11516   [1, 4, 9, 11, 19, 23, 32, 32, 39, 42]     3 5 2 8 4 9 0 7 3
28033   [0, 2, 4, 5, 9, 9, 19, 20, 28, 46]     2 2 1 4 0 10 1 8 18
300 400
3435   [1, 1, 2, 20, 22, 27, 28, 30, 36, 37]     0 1 18 2 5 1 2 6 1
8807   [7, 11, 14, 18, 24, 24, 32, 42, 43, 48]     4 3 4 6 0 8 10 1 5
9620   [0, 3, 11, 25, 26, 31, 33, 33, 37, 48]     3 8 14 1 5 2 0 4 11
10560   [0, 3, 8, 18, 19, 21, 24, 24, 26, 40]     3 5 10 1 2 3 0 2 14
15523   [1, 6, 13, 14, 23, 30, 33, 33, 34, 39]     5 7 1 9 7 3 0 1 5
16869   [2, 4, 12, 19, 20, 20, 37, 41, 43, 49]     2 8 7 1 0 17 4 2 6
19038   [4, 4, 5, 14, 18, 29, 35, 37, 41, 49]     0 1 9 4 11 6 2 4 8
23824   [1, 3, 12, 19, 29, 30, 31, 32, 32, 46]     2 9 7 10 1 1 1 0 14
27905   [3, 4, 5, 9, 26, 33, 33, 36, 37, 42]     1 1 4 17 7 0 3 1 5
400 500
1132   [1, 6, 7, 10, 10, 19, 26, 36, 45, 47]     5 1 3 0 9 7 10 9 2
1337   [1, 10, 15, 15, 20, 28, 30, 35, 45, 48]     9 5 0 5 8 2 5 10 3
9327   [3, 7, 20, 22, 27, 32, 38, 47, 47, 48]     4 13 2 5 5 6 9 0 1
14501   [1, 9, 31, 33, 41, 44, 44, 47, 48, 49]     8 22 2 8 3 0 3 1 1
500 600
9921   [0, 11, 15, 17, 19, 28, 39, 40, 40, 42]     11 4 2 2 9 11 1 0 2
14861   [1, 8, 15, 17, 18, 31, 32, 38, 38, 46]     7 7 2 1 13 1 6 0 8
15781   [1, 7, 10, 26, 26, 27, 32, 36, 37, 49]     6 3 16 0 1 5 4 1 12
17596   [5, 11, 26, 28, 34, 39, 40, 40, 41, 46]     6 15 2 6 5 1 0 1 5
600 700
8504   [3, 11, 14, 17, 19, 19, 25, 35, 39, 47]     8 3 3 2 0 6 10 4 8
13467   [0, 13, 19, 22, 22, 24, 25, 27, 39, 42]     13 6 3 0 2 1 2 12 3
20033   [9, 11, 12, 13, 15, 24, 28, 28, 36, 44]     2 1 1 2 9 4 0 8 8
700 800
18281   [2, 7, 10, 12, 21, 35, 43, 43, 44, 46]     5 3 2 9 14 8 0 1 2
18290   [7, 9, 20, 30, 30, 35, 41, 43, 44, 48]     2 11 10 0 5 6 2 1 4
19776   [7, 13, 17, 18, 18, 24, 25, 26, 32, 42]     6 4 1 0 6 1 1 6 10
22846   [1, 4, 12, 15, 21, 23, 23, 25, 28, 48]     3 8 3 6 2 0 2 3 20
800 900
405   [0, 1, 2, 11, 20, 21, 40, 40, 42, 44]     1 1 9 9 1 19 0 2 2
4441   [0, 6, 10, 23, 25, 28, 28, 33, 37, 43]     6 4 13 2 3 0 5 4 6
4487   [6, 11, 20, 20, 26, 27, 29, 35, 42, 47]     5 9 0 6 1 2 6 7 5
16432   [1, 9, 12, 15, 17, 17, 21, 35, 47, 48]     8 3 3 2 0 4 14 12 1
900 1000
17947   [2, 3, 3, 8, 19, 22, 34, 35, 38, 48]     1 0 5 11 3 12 1 3 10
21946   [12, 15, 15, 18, 22, 25, 31, 37, 39, 44]     3 0 3 4 3 6 6 2 5
28025   [0, 3, 9, 10, 10, 12, 14, 17, 18, 34]     3 6 1 0 2 2 3 1 16
1000 1100

if take just a large 2 ^ and move to the side along it, the same will be... "filter" question...

i.e. if we take  (2^371466  [5, 6, 7, 7, 8, 9, 10, 11, 19, 20]       1 1 0 1 1 1 1 8 1) 5+1=6, 6+1=7, 7+0=7, e.t.c...  5, 6, 7, 7, 8, 9, 10, 11, 19, 20 these positions in the first 100 digits 2^371466 (do not exactly take from the cropped file) and there will be ours 30 56 83 77 31 20 64 20 28 55 and they will need to be mixed to find 30568377312064202855...

search essence "take the first 100 numbers power of 2 (4096-32768 or 4096-30000000)" the first position is always either 0 or 1 (or from 0 to 20) and add to them our set (filt <=6 <=6 <=6 <=6 <=6 <=6 <=6 <=6 <=6) 0+6, 6+6, 12+6, e.t.c  these will be the positions for mixing.

***

looming here

2^4096-2^30000000

filt <=1 <=1 <=1 <=1 <=1 <=30 <=30 <=30 <=30

2^7788201    [1, 1, 2, 3, 4, 5, 19, 33, 35, 48]          0 1 1 1 1 14 14 2 13
2^20398384   [0, 0, 1, 2, 3, 4, 16, 17, 33, 41]          0 1 1 1 1 12 1 16 8
0-100                                                  
2^371466     [5, 6, 7, 7, 8, 9, 10, 11, 19, 20]           1 1 0 1 1 1 1 8 1
2^21846679   [2, 3, 4, 5, 5, 6, 21, 25, 27, 31]          1 1 1 0 1 15 4 2 4
100-200                                                  
200-300                                                  
2^23749755   [7, 8, 9, 10, 10, 11, 16, 18, 32, 36]     1 1 1 0 1 5 2 14 4
2^29029857   [5, 5, 6, 7, 8, 9, 14, 19, 35, 36]          0 1 1 1 1 5 5 16 1
300-400                                                  
2^1061949    [0, 0, 1, 2, 3, 4, 20, 38, 45, 47]          0 1 1 1 1 16 18 7 2
2^16811419   [1, 2, 2, 3, 4, 5, 12, 18, 19, 21]         1 0 1 1 1 7 6 1 2
2^19105979   [3, 4, 5, 6, 7, 7, 19, 20, 30, 48]         1 1 1 1 0 12 1 10 18
400-500                                                  
2^11871324   [4, 4, 5, 6, 7, 8, 10, 19, 29, 49]            0 1 1 1 1 2 9 10 20
2^16508824   [0, 0, 1, 2, 3, 4, 23, 29, 33, 35]            0 1 1 1 1 19 6 4 2
2^28301237   [10, 11, 12, 12, 13, 14, 24, 31, 41, 47]  1 1 0 1 1 10 7 10 6
500-600                                                  
2^4991975    [3, 4, 5, 6, 6, 7, 8, 20, 33, 43]              1 1 1 0 1 1 12 13 10
2^6765622    [3, 4, 4, 5, 6, 7, 11, 34, 41, 49]             1 0 1 1 1 4 23 7 8
2^13036039   [3, 4, 5, 6, 6, 7, 11, 14, 15, 31]             1 1 1 0 1 4 3 1 16
2^14362774   [2, 3, 4, 5, 6, 7, 11, 27, 27, 32]             1 1 1 1 1 4 16 0 5      
2^22753731   [5, 6, 6, 7, 8, 9, 17, 21, 29, 38]             1 0 1 1 1 8 4 8 9
600-700                                                  
2^11472284   [0, 1, 2, 3, 3, 4, 17, 34, 35, 47]             1 1 1 0 1 13 17 1 12
2^13954777   [15, 15, 16, 17, 18, 19, 28, 33, 37, 38]    0 1 1 1 1 9 5 4 1
2^24653194   [4, 5, 6, 6, 7, 8, 20, 21, 37, 46]             1 1 0 1 1 12 1 16 9
700-800
2^14201449   [2, 2, 3, 4, 5, 6, 14, 18, 21, 22]             0 1 1 1 1 8 4 3 1
2^14583083   [1, 2, 3, 4, 4, 5, 29, 39, 43, 47]             1 1 1 0 1 24 10 4 4
2^21591045   [12, 13, 14, 15, 15, 16, 20, 26, 33, 34]    1 1 1 0 1 4 6 7 1
800-900
2^2859506    [2, 3, 4, 5, 6, 7, 7, 8, 38, 40]                   1 1 1 1 1 0 1 30 2
2^7259060    [13, 14, 14, 15, 16, 17, 39, 44, 48, 49]       1 0 1 1 1 22 5 4 1
2^13594185   [0, 1, 2, 3, 3, 4, 6, 14, 19, 22]                 1 1 1 0 1 2 8 5 3
2^16188178   [11, 12, 12, 13, 14, 15, 20, 21, 27, 29]      1 0 1 1 1 5 1 6 2
...

2^7788201    [1, 1, 2, 3, 4, 5, 19, 33, 35, 48]            0 1 1 1 1 14 14 2 13
2^20398384   [0, 0, 1, 2, 3, 4, 16, 17, 33, 41]           0 1 1 1 1 12 1 16 8

2^29029857   [5, 5, 6, 7, 8, 9, 14, 19, 35, 36]           0 1 1 1 1 5 5 16 1

2^1061949    [0, 0, 1, 2, 3, 4, 20, 38, 45, 47]            0 1 1 1 1 16 18 7 2

2^11871324   [4, 4, 5, 6, 7, 8, 10, 19, 29, 49]            0 1 1 1 1 2 9 10 20
2^16508824   [0, 0, 1, 2, 3, 4, 23, 29, 33, 35]           0 1 1 1 1 19 6 4 2

2^13954777   [15, 15, 16, 17, 18, 19, 28, 33, 37, 38]   0 1 1 1 1 9 5 4 1

2^14201449   [2, 2, 3, 4, 5, 6, 14, 18, 21, 22]             0 1 1 1 1 8 4 3 1

consistency  0 1 1 1 1 X X X X

can start from 0, fixed 0 1 1 1 1 and brute force X X X X, 20×20×20×20=160000

by time for 1 pass, 2^20000000, 1 second per 10000000 mixing =.... 20000000/60 = 333333 minutes, 333333/60 = 5555 hours, 5555/24 = 231 days (again unattainable for cpu((()  231 days x (20×20×20×20=160000) = 36960000 days. but if run multiple copies of the program, 1000000 for example))),  36960000/1000000 = 36 days.

***
with filter 1-30-1-30-1-30-1-30-1

2^1212405   [8, 9, 12, 13, 16, 17, 20, 21, 42, 42]     1 3 1 3 1 3 1 21 0

***

it is clear that can reduce the time by reducing 2^... 2^20000000 > 2^32768 but "the filter" will grow... 3 passes per day by one program run on cpu, take 100 programs by 3 = 300 per day and for "the filter" 60466176 6x10, 60466176/300 = 201553 days? O_o  

***
So this means only a computer genius that eats programs who can solve it? Lucky for you guys to crack it as it has huge rewards waiting to be claimed. I wonder if there were puzzles out there that does not require any computer skills just to crack it. Someone like me will surely cannot solve puzzles like this. 32BTC is a lot of money maybe I will have to find ways to participate and this is fun because the odds of me cracking it is so high maybe thousand times. 😅

Another puzzle that doesn't require any computer skills.
 https://i.redd.it/n1x7g8ceaur51.png
https://www.blockchain.com/id/btc/address/1KfZGvwZxsvSmemoCmEV75uqcNzYBHjkHZ

Designing and creating a puzzle seems more expensive than a reward.
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October 25, 2020, 09:07:23 AM
 #1442

Hello everyone I decided to join you, but I have a weak hardware, so I won't be able to win by consistently, hope for a random. Did I set up the program correctly?   Smiley







Sale the first NFT of the first foto
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October 25, 2020, 01:51:25 PM
 #1443

Hello everyone I decided to join you, but I have a weak hardware, so I won't be able to win by consistently, hope for a random. Did I set up the program correctly?   Smiley








Yes that setup you're using is correct.
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October 25, 2020, 02:05:13 PM
Last edit: October 25, 2020, 05:25:23 PM by alevlaslo
 #1444

thanks


Sale the first NFT of the first foto
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December 03, 2020, 09:25:14 AM
 #1445

No discussion for a long time  Huh Huh Huh Why Roll Eyes

#66     2a183647529c49b3c     13zbaHbtmYwRPRSBMVwtQ3GSnrFXsPijso
#67     59265959791de58c7     1BY8Lw3n6TYv3ARN538RRu4VMFVYfkvTW9
#68     9e8b4c743400da30     1MVDUrdV1fbyu5N23qqmn5AHyVp9HW6tvZ

1BGvwggxfCaHGykKrVXX7fk8GYaLQpeixA
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December 06, 2020, 10:15:14 AM
 #1446

Anyone who has a max prefix of puzzle 64, please post here.. it will definitely help me with my new experiments soon I will share it..  Huh

1BGvwggxfCaHGykKrVXX7fk8GYaLQpeixA
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December 06, 2020, 10:58:56 PM
 #1447

No discussion for a long time  Huh Huh Huh Why Roll Eyes

#66     2a183647529c49b3c     13zbaHbtmYwRPRSBMVwtQ3GSnrFXsPijso
#67     59265959791de58c7     1BY8Lw3n6TYv3ARN538RRu4VMFVYfkvTW9
#68     9e8b4c743400da30     1MVDUrdV1fbyu5N23qqmn5AHyVp9HW6tvZ


Nobody can solve it at this moment, that's why no discussion for a long time.
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December 24, 2020, 02:40:52 PM
Last edit: December 24, 2020, 02:59:49 PM by cyptomania
 #1448

Puzzle 64 was cracked 6 days ago.

https://www.blockchain.com/btc/address/16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN





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December 24, 2020, 03:14:54 PM
 #1449

Huh?
Balance: 0.64016564 BTC
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December 24, 2020, 03:44:01 PM
 #1450

My fault guys sorry,just woke up and didn't even have my morning coffee at the time I posted that,had Blockchain.com up and saw the 0 and got confused as the Total was off-screen and I didn't scroll down,it's not solved,MY BAD.
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January 10, 2021, 12:52:59 AM
Merited by Symmetrick (1)
 #1451

All space the private key in hexadecimal is located  - keyspace:
FROM - 0000000000000000000000000000000000000000000000000000000000000001:
TO - FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

No! This is a fallacy, these addresses (out of range) are only mirrored addresses (X / Y axis) Example:

                                                                                          
fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd036418d                      -U                                                   -C

5Km2kuu7vtFDPpxywn4u3NLpbr5jKpTB3jsuDU2KYEqf3VKty2N - 1ESJVfV5UVERkWgVNfMjsLwJT88yMJHi8R - 1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC


000000000000000000000000000000000000000000000000000000000000004c

5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreKD614Nu - 1ESJVfV5UVERkWgVNfMjsLwJT88yMJHi8R - 1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC

an address with two keys .. the same applies to your determined address:

"ucompressed" - 12M4QznuNZH2BRVbLK8SKvNqGTPJpCpST7
"compressed" - 1PRWyFKTsQSJaUdX9VKgQNw8JERPw2kMFm

All other addresses are in space
FROM - 0000000000000000000000000000000000000000000000000000000000000001:
TO - (FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF) - 5Km2kuu7vtFDPpxywn4u3NLu8iSdrqhxWT8tUKjeEXs2f9yxoWz  No!  Grin -> you cannot import these key in to the wallet!

"ucompressed" - 12M4QznuNZH2BRVbLK8SKvNqGTPJpCpST7
"compressed" - 1PRWyFKTsQSJaUdX9VKgQNw8JERPw2kMFm

000000000000000000000000000000014551231950b75fc4402da1732fc9bebe
(5HpHagT65TZzG1PH3CSu63kCkUBpkA7sBXKmV2m4wAt1vyA9SA4)





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January 14, 2021, 12:41:24 AM
Last edit: January 14, 2021, 01:41:54 PM by BASE16
 #1452

All space the private key in hexadecimal is located  - keyspace:
FROM - 0000000000000000000000000000000000000000000000000000000000000001:
TO - FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

No! This is a fallacy, these addresses (out of range) are only mirrored addresses (X / Y axis) Example:

                                                                                         
fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd036418d                      -U                                                   -C

5Km2kuu7vtFDPpxywn4u3NLpbr5jKpTB3jsuDU2KYEqf3VKty2N - 1ESJVfV5UVERkWgVNfMjsLwJT88yMJHi8R - 1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC


000000000000000000000000000000000000000000000000000000000000004c

5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreKD614Nu - 1ESJVfV5UVERkWgVNfMjsLwJT88yMJHi8R - 1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC

an address with two keys .. the same applies to your determined address:

"ucompressed" - 12M4QznuNZH2BRVbLK8SKvNqGTPJpCpST7
"compressed" - 1PRWyFKTsQSJaUdX9VKgQNw8JERPw2kMFm

All other addresses are in space
FROM - 0000000000000000000000000000000000000000000000000000000000000001:
TO - (FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF) - 5Km2kuu7vtFDPpxywn4u3NLu8iSdrqhxWT8tUKjeEXs2f9yxoWz  No!  Grin -> you cannot import these key in to the wallet!

"ucompressed" - 12M4QznuNZH2BRVbLK8SKvNqGTPJpCpST7
"compressed" - 1PRWyFKTsQSJaUdX9VKgQNw8JERPw2kMFm

000000000000000000000000000000014551231950b75fc4402da1732fc9bebe
(5HpHagT65TZzG1PH3CSu63kCkUBpkA7sBXKmV2m4wAt1vyA9SA4)


Well if you look at the known private keys in BASE2 format then you can see that the first bit always starts with a 1 and the bit count is also fixed so that narrows down the search space because a 64 bits key will be found somewhere between 0x8000000000000000 and 0xFFFFFFFFFFFFFFFF in BASE16 or in BASE2 between 0b10000000000000000000000000000000000000000000000000
00000000000000 and 0b1111111111111111111111111111111111111111111111111111111111
111111 respectively and there is no need to search for it outside of those boundaries.

In plain BASE10 language that is a number somewhere between 9223372036854775808 and 18446744073709551615 so it is a large number.
But not merely as large as the 160 bit key which clocks in somewhere in between 730750818665451459101842416358141509827966271488 and 1461501637330902918203684832716283019655932542975.
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January 14, 2021, 02:47:29 AM
 #1453

I know all that, I only reacted to a post from "iparktur" in which the keyspace was expanded, which is nonsense and thus proved to him that the addresses only repeat themselves ... and nothing else ...  Wink
https: // bitcointalk.org/index.php?topic=1306983.msg51264673#msg51264673
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January 29, 2021, 02:25:20 PM
 #1454



Quote
The next one old address was opened 1EaXdZypzsWcTqEJQk2esJDck3oSLpra2T 13pd4UFwUk3HFpk4HkCFV8xvzBDC4aQTC1
Would you like to say that you found the keys to these addresses from 2010 with this program? Apparently she works with the kangaroo method with a public key. Or the authors decided to joke and they have access to the old addresses in order to lure a couple of hundred bucks from the simpletons. Show the video on YouTube how the program works. Smiley If it works and you really found the keys to these addresses, then this should excite the crypto community.

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January 29, 2021, 07:26:11 PM
 #1455



Quote
The next one old address was opened 1EaXdZypzsWcTqEJQk2esJDck3oSLpra2T 13pd4UFwUk3HFpk4HkCFV8xvzBDC4aQTC1
Would you like to say that you found the keys to these addresses from 2010 with this program? Apparently she works with the kangaroo method with a public key. Or the authors decided to joke and they have access to the old addresses in order to lure a couple of hundred bucks from the simpletons. Show the video on YouTube how the program works. Smiley If it works and you really found the keys to these addresses, then this should excite the crypto community.



Reason no one is excited is simple: It's a scam.

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February 06, 2021, 10:41:13 AM
 #1456

Quote
Reason no one is excited is simple: It's a scam.


yes scam 158LUYbF38KGu8rRRWPDL6gxe8GrozF862    -) Grin
update: 12bXbfa4rjh1uwbgs25JGPWjPCpaz6VK8e  Tongue
Who know privat?
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February 06, 2021, 06:45:29 PM
 #1457

hey so I decided to upgrade BitCrack to CUDA 11.2 to see if it would improve my performance but nope, originally I was getting 216Mkey/s but after compiling for CUDA 11.2 it was drastically lower 4.2Mkey/s lol

Here's the forked code upgraded to CUDA 11.2 if anyone wants it: https://github.com/GonzoTheDev/BitCrack

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February 09, 2021, 06:43:09 AM
 #1458

https://satoshidisk.com/pay/CBd7TZ
The public key of the Bitcoin wallet from Puzzle 64 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN. I can't crack the private key with Kangaroo Sad


The public key of this address has not been exposed, nobody but the owner of the puzzle can give it to you.
I guess you've been scammed.
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February 09, 2021, 07:24:13 PM
Last edit: February 09, 2021, 09:28:52 PM by MeBender
 #1459

This seems very fishy...
It doesn't even seem to be that complicated but the prize is very high. Huh

Yeah its only 21,575,147,362,539,733,087 addresses that need to be checked pfff couldn't be that hard Roll Eyes  Cheesy

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February 10, 2021, 02:10:37 AM
 #1460

https://satoshidisk.com/pay/CBd7TZ
The public key of the Bitcoin wallet from Puzzle 64 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN. I can't crack the private key with Kangaroo Sad


The public key of this address has not been exposed, nobody but the owner of the puzzle can give it to you.
I guess you've been scammed.


I used the site https://iancoleman.io/bitcoin-key-compression/, if you enter this public key, the compression address appears 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN. However, I am using Kangaroo and it doesn't find anything. Perhaps the hex range is incorrect?
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