Bitcoin puzzle transaction ~32 BTC prize to who solves it

[Niekko]

NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed
[2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4
[2026-05-13.11:37:57] [Info] Ending at:   0000000000000000000000000000000000000000000001EE119475F12CEE4EF4
[2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090
[2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB)
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[2026-05-13.11:38:09] [Info] Done
NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27]

Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0

it's not very much bad, think that BitCrack it's a real bruteforce, and you step at 0x01. With my 5090 i going at 4750 MKey/s, but I've modded it a little.

You will never going at speed like BSGS, because it's not a real speed, yours it's real.

[0xastraeus]

NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed
[2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4
[2026-05-13.11:37:57] [Info] Ending at:   0000000000000000000000000000000000000000000001EE119475F12CEE4EF4
[2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090
[2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB)
[2026-05-13.11:38:07] [Info] 10.0%
[2026-05-13.11:38:08] [Info] 20.0%
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[2026-05-13.11:38:09] [Info] 100.0%
[2026-05-13.11:38:09] [Info] Done
NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27]

Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0

Did you compile correctly for 5090? try lowering blocks and higher points like -b 256 -p 4096 or something see how that goes. Play around with it.

[brainless]

I use cuBitCrack, but getting 4GKeys/s from RTX5090
Could you please anyone help me about optimize for speed?

Below code provides auto random jumping, when a space scanning finished

Code:

cat > auto_random.sh << 'EOF'
#!/bin/bash
TARGET="12VVRNPi4SJqUTsp6FmqDqY5sGosDtysn4"
DEVICE=0
BLOCKS=1024
THREADS=256
POINTS=512

while true; do
    pkill cuBitCrack 2>/dev/null
    sleep 2

    RESULT=$(python3 << 'PYEOF'
import random
start = 0x1000000000000000000
end   = 0x1ffffffffffffffffff
chunk_size = 2**46
max_start = end - chunk_size
random_start = random.randint(start, max_start)
random_end = random_start + chunk_size
print(f"{random_start:020x} {random_end:020x}")
PYEOF
)

    RANDOM_START=$(echo $RESULT | awk '{print $1}')
    RANDOM_END=$(echo $RESULT | awk '{print $2}')

    echo "$(date): Yeni dilim $RANDOM_START - $RANDOM_END" >> auto_log.txt

    ./bin/cuBitCrack -d $DEVICE \
        -b $BLOCKS -t $THREADS -p $POINTS \
        -c -o Found.txt \
        --keyspace ${RANDOM_START}:${RANDOM_END} \
        ${TARGET} >> output.log 2>&1

    if [ -s Found.txt ]; then
        echo "$(date): KEY BULUNDU!" >> auto_log.txt
        cat Found.txt >> auto_log.txt
        break
    fi
done
EOF

chmod +x auto_random.sh

When running bitcrack, share that screen to guide u best optimal setting

NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed
[2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4
[2026-05-13.11:37:57] [Info] Ending at:   0000000000000000000000000000000000000000000001EE119475F12CEE4EF4
[2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090
[2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB)
[2026-05-13.11:38:07] [Info] 10.0%
[2026-05-13.11:38:08] [Info] 20.0%
[2026-05-13.11:38:08] [Info] 30.0%
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[2026-05-13.11:38:09] [Info] Done
NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27]

Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0

BLOCKS=170
THREADS=512
POINTS=8192
Now need to optimize -t -p, mostly point up and down with make set
When ever need to adjust t and p
Use 32 +- in above mention , report here screen shot for see what's going on

[OzBtcOz]

NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed
[2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4
[2026-05-13.11:37:57] [Info] Ending at:   0000000000000000000000000000000000000000000001EE119475F12CEE4EF4
[2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090
[2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB)
[2026-05-13.11:38:07] [Info] 10.0%
[2026-05-13.11:38:08] [Info] 20.0%
[2026-05-13.11:38:08] [Info] 30.0%
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[2026-05-13.11:38:09] [Info] Done
NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27]

Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0

it's not very much bad, think that BitCrack it's a real bruteforce, and you step at 0x01. With my 5090 i going at 4750 MKey/s, but I've modded it a little.

You will never going at speed like BSGS, because it's not a real speed, yours it's real.

Thanks your answer, but when I checked other forums, they presented their benchmarks up to 8 GKeys/s. So, I thouht that there is something wrong with my GPU or compiling or etc.

[Niekko]

it's not very much bad, think that BitCrack it's a real bruteforce, and you step at 0x01. With my 5090 i going at 4750 MKey/s, but I've modded it a little.

You will never going at speed like BSGS, because it's not a real speed, yours it's real.

Thanks your answer, but when I checked other forums, they presented their benchmarks up to 8 GKeys/s. So, I thouht that there is something wrong with my GPU or compiling or etc.

with BitCrack ? My 5090 was never going at 8GKeys/s.

[Grzegorz2022]

NVIDIA GeForce R 12799 / 32109MB | 1 target 3891.43 MKey/s (1,593,567,084,544 total) [00:06:47][2026-05-13.11:37:57] [Info] Compression: compressed
[2026-05-13.11:37:57] [Info] Starting at: 0000000000000000000000000000000000000000000001EE119455F12CEE4EF4
[2026-05-13.11:37:57] [Info] Ending at:   0000000000000000000000000000000000000000000001EE119475F12CEE4EF4
[2026-05-13.11:37:57] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2026-05-13.11:37:57] [Info] Initializing NVIDIA GeForce RTX 5090
[2026-05-13.11:37:57] [Info] Generating 134,217,728 starting points (5120.0MB)
[2026-05-13.11:38:07] [Info] 10.0%
[2026-05-13.11:38:08] [Info] 20.0%
[2026-05-13.11:38:08] [Info] 30.0%
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[2026-05-13.11:38:09] [Info] 90.0%
[2026-05-13.11:38:09] [Info] 100.0%
[2026-05-13.11:38:09] [Info] Done
NVIDIA GeForce R 12799 / 32109MB | 1 target 3864.52 MKey/s (15,749,779,292,160 total) [01:07:27]

Any Above Value of 134,217,728 starting points (5120.0MB), I get out of memory error. And CUDA version is 13.0

it's not very much bad, think that BitCrack it's a real bruteforce, and you step at 0x01. With my 5090 i going at 4750 MKey/s, but I've modded it a little.

You will never going at speed like BSGS, because it's not a real speed, yours it's real.

Thanks your answer, but when I checked other forums, they presented their benchmarks up to 8 GKeys/s. So, I thouht that there is something wrong with my GPU or compiling or etc.

how long does it take you to find the key to puzzle 70

[Divaytis]

Anyone here know how to code and want to help me solve the puzzles? I created a system from the ground up complete with axioms definitions and procedural operations i just designed my system so that it happens to use the exact same values as secp256k1 im not saying i solved the discrete log thats impossible instead i created a system where the discrete log doesnt even apply at all

If it's still relevant, I can help write the code for you.

Bro if you can code we can cook

I can help with writing the code without any problems, how can I contact you to discuss everything?

[speed_user_113]

As i see there is GPU crisys and you cannot find anymore on a good price 4090 or 5090 to rent.
I want to see the ones that had scripts and GPU farms rental to steal now the key (unless they have their own GPU farm)

[SecretAdmirere]

Thanks your answer, but when I checked other forums, they presented their benchmarks up to 8 GKeys/s. So, I thouht that there is something wrong with my GPU or compiling or etc.

Don't know what and from who, you are getting your BitCrack code, but 5090 is easily capable of 11b+ checked hash160 addresses per second, if not 12b+, but for those numbers you need a custom code beacuse none of those publicly avaliable can achive it (they are not that very well "optimised")..

You should get the https://github.com/FixedPaul/VanitySearch-Bitcrack, that one is the fastest publicly avaliable, and it would run at 8b+ at stock speeds and power draw, and if you want to go beyond that well, you need to drasticly change the hashing and FE functions.

how long does it take you to find the key to puzzle 70

You can't simply compare "how long it takes to find 2⁷⁰ private key", that is a mode where it hashes pubkeys and checks hash160 target to the computed hash160, totally different space where your bsgs is being used. You should compile and benchmark your FE mul and compare it against  kTimesG or RetiredCoder (115/140 Gmul/s on 4090) and get a rough estimate where your performance is at beacuse that one is by far the most impactful one.

[username666187]

Anyone here know how to code and want to help me solve the puzzles? I created a system from the ground up complete with axioms definitions and procedural operations i just designed my system so that it happens to use the exact same values as secp256k1 im not saying i solved the discrete log thats impossible instead i created a system where the discrete log doesnt even apply at all

If it's still relevant, I can help write the code for you.

Bro if you can code we can cook

I can help with writing the code without any problems, how can I contact you to discuss everything?

Dude allow new members to send you message im not trying to dox my phone number and email address

Anyone here know how to code and want to help me solve the puzzles? I created a system from the ground up complete with axioms definitions and procedural operations i just designed my system so that it happens to use the exact same values as secp256k1 im not saying i solved the discrete log thats impossible instead i created a system where the discrete log doesnt even apply at all

If it's still relevant, I can help write the code for you.

Bro if you can code we can cook

Allow messages from newbies on your profile so i can pm you

Same goes for you "And24u"

[cctv5go]

The interesting math behind this (private key binary):
Plz27
13,14
Plz31
10,21--->The two numbers differ by 11
Plz35
20,15
Plz39
19,20
Plz43
19,24
Plz47
23,24
Plz51
32,19--->The two numbers differ by 13
Plz55
24,31
Plz59
26,33
Plz63
27,36
Plz67
36,31
Plz71
 , --->The two numbers differ by 15？？？(43,28)?

[Divaytis]

Anyone here know how to code and want to help me solve the puzzles? I created a system from the ground up complete with axioms definitions and procedural operations i just designed my system so that it happens to use the exact same values as secp256k1 im not saying i solved the discrete log thats impossible instead i created a system where the discrete log doesnt even apply at all

If it's still relevant, I can help write the code for you.

Bro if you can code we can cook

I can help with writing the code without any problems, how can I contact you to discuss everything?

Dude allow new members to send you message im not trying to dox my phone number and email address

Allowed

[OzBtcOz]

As i see there is GPU crisys and you cannot find anymore on a good price 4090 or 5090 to rent.
I want to see the ones that had scripts and GPU farms rental to steal now the key (unless they have their own GPU farm)

Funny you! 
You think it is rental GPU ? Don't worry about it, my rig has enough RTX5090 on it 

[nomachine]

I am open to collaboration with anyone who has strong mathematical expertise.

Nah bruh, this ain’t “hidden math hacks unlock Bitcoin” territory, this straight up cap physics cosplay.

You talkin’ like you can reverse-engineer secp256k1 with some +7 denominator wizardry like the curve just gon’ fold under algebra pressure. That’s not cryptography, that’s wishful thinking with extra steps.

Bitcoin keys ain’t solved, they’re picked 256-bit entropy, pure randomness, no pattern to reverse. Public key is just scalar multiplication on an elliptic curve: priv × G. That’s it. One-way street with no U-turn, no cheat code, no “formula bridge.”

Trying to reverse it is like watching an explosion and thinking you can reconstruct the uranium atom by reading the smoke.

And this “gravity leaking from parallel dimensions” talk? That’s like you took Randall Sundrum brane world physics and started remixing it like it explains Bitcoin.

In real theory, extra dimensions could in principle let gravity spread through a higher dimensional bulk while the Standard Model forces stay locked on a 3 plus 1 dimensional brane. That is part of why these models exist in the first place, to try to explain why gravity is so weak compared to the other forces.

But even in that setup it still does not line up cleanly with real observations like gravitational lensing, the cosmic microwave background, and large scale structure without adding extra assumptions on top of extra assumptions.

So trying to connect that kind of physics to Bitcoin is just category error. Bitcoin does not sit in some hidden geometry you can probe. It sits on deterministic elliptic curve multiplication where the private key is random entropy and the public key is just a one way transformation of it.

Different universe of problems entirely. 

[And24r]

I am open to collaboration with anyone who has strong mathematical expertise.

Nah bruh, this ain’t “hidden math hacks unlock Bitcoin” territory, this straight up cap physics cosplay.

You talkin’ like you can reverse-engineer secp256k1 with some +7 denominator wizardry like the curve just gon’ fold under algebra pressure. That’s not cryptography, that’s wishful thinking with extra steps.

Bitcoin keys ain’t solved, they’re picked 256-bit entropy, pure randomness, no pattern to reverse. Public key is just scalar multiplication on an elliptic curve: priv × G. That’s it. One-way street with no U-turn, no cheat code, no “formula bridge.”

Trying to reverse it is like watching an explosion and thinking you can reconstruct the uranium atom by reading the smoke.

And this “gravity leaking from parallel dimensions” talk? That’s like you took Randall Sundrum brane world physics and started remixing it like it explains Bitcoin.

In real theory, extra dimensions could in principle let gravity spread through a higher dimensional bulk while the Standard Model forces stay locked on a 3 plus 1 dimensional brane. That is part of why these models exist in the first place, to try to explain why gravity is so weak compared to the other forces.

But even in that setup it still does not line up cleanly with real observations like gravitational lensing, the cosmic microwave background, and large scale structure without adding extra assumptions on top of extra assumptions.

So trying to connect that kind of physics to Bitcoin is just category error. Bitcoin does not sit in some hidden geometry you can probe. It sits on deterministic elliptic curve multiplication where the private key is random entropy and the public key is just a one way transformation of it.

Different universe of problems entirely. 

The curve with the equation y^2=x^3(mod p) lends itself perfectly to decryption. I have ready‑made formulas for this curve and I can derive the private key from the public key for this curve. For the Bitcoin curve y^2=x^3+7(mod p), I haven’t found a solution. But it exists, because it cannot not exist. You don’t know a lot of things

[Diaghilev]

Tell me, does Mara send the client code to the email address specified when requesting the code?

[SecretAdmirere]

Tell me, does Mara send the client code to the email address specified when requesting the code?

No, the CEO personally comes to your door and delivers it via USB drive with "MARA" engraved on it

[mjojo]

It’s a long explanation. I’ve provided examples of formulas and set the task goal. If we derive the formula, we’ll work only with the x coordinate — the y coordinate isn’t needed.

Following your example:

x is the base point GP;

x1 is the public key;

2^n is the private key with the number 2 raised to an unknown power.

In the first and second examples, you can substitute any value for x — formula 2 will always return x. However, formulas 3 and 4 won’t work in this case.

In general, the point is that if you know the degree n, you can easily adjust the private key to match the x-coordinate of the base point GP through sequential increase, using formula 4.

Code:

from decimal import Decimal, getcontext

# set high precision
getcontext().prec = 100

# x value
x = Decimal("55066263022277343669578718895168534326250603453777594175500187360389116729240")

# -----------------------------------
# Calculate x1
# Formula:
# ((x**2 / sqrt(x**3)) * 1.5)**2 - 2*x
# -----------------------------------

x1 = (((x**2 / (x**3).sqrt()) * Decimal("1.5")) ** 2) - (2 * x)

# -----------------------------------
# Calculate x

x = (2**2) * x1

print("x1 =")
print(format(x1, 'f'))

print("\nx =")
print(format(x, 'f'))

this python for your formula no. 1 and 2, my question what P number and Gen X for the curve y^2=x^3(mod p) is random or specific number like in Secp256k1 curve,because without Bcurve point is not valid again. then how you derive the pubkey to pvkey from the  the curve y^2=x^3(mod p), did you reduce bit by bit or just with one reverse. thank you if you share anymore

[And24r]

It’s a long explanation. I’ve provided examples of formulas and set the task goal. If we derive the formula, we’ll work only with the x coordinate — the y coordinate isn’t needed.

Following your example:

x is the base point GP;

x1 is the public key;

2^n is the private key with the number 2 raised to an unknown power.

In the first and second examples, you can substitute any value for x — formula 2 will always return x. However, formulas 3 and 4 won’t work in this case.

In general, the point is that if you know the degree n, you can easily adjust the private key to match the x-coordinate of the base point GP through sequential increase, using formula 4.

Code:

from decimal import Decimal, getcontext

# set high precision
getcontext().prec = 100

# x value
x = Decimal("55066263022277343669578718895168534326250603453777594175500187360389116729240")

# -----------------------------------
# Calculate x1
# Formula:
# ((x**2 / sqrt(x**3)) * 1.5)**2 - 2*x
# -----------------------------------

x1 = (((x**2 / (x**3).sqrt()) * Decimal("1.5")) ** 2) - (2 * x)

# -----------------------------------
# Calculate x

x = (2**2) * x1

print("x1 =")
print(format(x1, 'f'))

print("\nx =")
print(format(x, 'f'))

this python for your formula no. 1 and 2, my question what P number and Gen X for the curve y^2=x^3(mod p) is random or specific number like in Secp256k1 curve,because without Bcurve point is not valid again. then how you derive the pubkey to pvkey from the  the curve y^2=x^3(mod p), did you reduce bit by bit or just with one reverse. thank you if you share anymore

The number P(mod p) can be chosen arbitrarily — it doesn’t matter for the example. However, first you need to generate the points of this curve using this modulus. You can derive a private key from a public key on the curve y^2=x^3
  using any modulus P. I cannot provide you with the formulas for this, as I have spent a lot of time searching for them. But if you share your work with me, I will be happy to share my findings with you.

Here is an example with modulus p=59 and the curve y^2=x^3:

1- x=29 y=9
2- 22 38
3 - 36 20
4 - 35 49
5 - 46 35
6 - 9 32

[mjojo]

Here is an example with modulus p=59 and the curve y^2=x^3:

1- x=29 y=9
2- 22 38
3 - 36 20
4 - 35 49
5 - 46 35
6 - 9 32

this full coordinate with P=59

1G=(29,9)
2G=(22,38)
3G=(36,20)
4G=(35,49)
5G=(46,35)
6G=(9,32)
7G=(3,26)
8G=(53,43)
9G=(4,51)
10G=(41,56)
//
51G=(53,16)
52G=(3,33)
53G=(9,27)
54G=(46,24)
55G=(35,10)
56G=(36,39)
57G=(22,21)
58G=(29,50)
[Finished in 93ms]

With Bcurve=0 the point is valid

Left side  (y² mod P):
22
Right side ((x³ + A*x + B) mod P):
22
VALID: Point lies on secp256k1 curve

X HEX:
000000000000000000000000000000000000000000000000000000000000001d
Y HEX:
0000000000000000000000000000000000000000000000000000000000000009

So you derive this pubkey 56G=(36,39) to pvkey 56 just with one reverse or bit by bit