During my test with a modified KeyQuest (cpu), I found instant the private key of puzzle 65. I guess I got lucky but it's really surprising! So I think random can also bring luck sometimes.

Using Hamming negative adaptive bias with cross-over population HoF ascension for delta weight minimization, this came up.


132 bits matched, 16 leading. That's around 62.2 bits of classical difficulty.

Let's see if anyone can find the key.

You can be first lol, I mean, as far as I know you are. But being first at something like this, I dunno. I mean it's cool and all, just doesn't help us out with these challenges or any real life scenario.

Range was not needed.

0xe51e464444fc4495991b7852b855

So I sat here the last 10 minutes to create my own script to do what you are doing.

I was successful.

Knowing the range speeds up the script, in that you can eliminate certain numbers, but the range is not needed.

 Your project seems amazing. Do you have success so far? If you need more accurate calculator just let me know. I have a very fast karatsuba calculator that can solve up to 500 digit integers and up to 100 decimals. Can you tell more about about your project?

I have already provided the two parameters you need, and now you’re asking me for extra information.

Alright, you say that everything must be done exactly according to your steps, but the whole process is predicated on the fact that I need to know the private key.

Without the private key, I wouldn’t be able to get the string mentioned in step 6.

So, since I already know the private key, why do we have to engage in these tedious number transformation games?

Stop playing your number games—it's like saying that your birth year plus your age equals the current year.

Okay, let's do this.

decimal hash160 string:

range boundary:

Pls give me the private key immediately. 

What are you talking about? Doesn’t the result of step five still end up being the decimal string of hash160?


I feel like you're just messing with us.

Look, I am not even going to attempt to recreate what you are doing or saying, so I will not say it does or does not work. That is not what I was saying nor said.

I am asking you, in what way does this help with anything? How will it help solve a puzzle / challenge? How does it help if I lost my WIF / private key?

How does it help, in any way?

How will it replace any equipment??

Forget math, let's use Magic-Based Private Key Recovery!

1️⃣ Adobe Illustrator & The Sacred Geometry of Bitcoin ✨

Open an SVG of the secp256k1 curve in Illustrator.
Draw random arrows in different font sizes.
Connect "mystical" hex numbers inside squares, triangles, and circles.
Eventually, a private key will just "appear" (probably in Comic Sans).
2️⃣ Numerology & The Power of Hex 🔢🔺

Take the Bitcoin address, convert it into hex.
Rearrange the numbers like a Sudoku puzzle.
Add a "magic" constant (random prime number).
Somehow, a private key pops out—if not, add another hex and pretend it worked.
3️⃣ Horoscope-Based Key Recovery 🔭

If Mercury is in retrograde, your hash160 might decrypt itself.
 

Are you kidding us or are you serious? Trying to get a private key from a hash160 is like trying to unblend a smoothie back into whole fruits. The bananas, strawberries, and crypto magic are gone forever. No matter how hard you try, you’re not getting that banana back!" 🍌😂

Go back a few pages and see how the kangaroo script works. See the sequence of operations it performs.

modulo = gmpy2.mpz(0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F)
order = gmpy2.mpz(0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141)
Gx = gmpy2.mpz(0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798)
Gy = gmpy2.mpz(0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8)
PG = (Gx, Gy)

def add(P, Q):
    if P == (0, 0): return Q
    if Q == (0, 0): return P
    Px, Py = P
    Qx, Qy = Q
    if Px == Qx:
        if Py == Qy:
            inv_den = gmpy2.invert(Py << 1, modulo)
            m = (3 * Px * Px * inv_den) % modulo
        else:
            return (0, 0)
    else:
        inv_dx = gmpy2.invert(Qx - Px, modulo)
        m = ((Qy - Py) * inv_dx) % modulo
    x = (m * m - Px - Qx) % modulo
    y = (m * (Px - x) - Py) % modulo
    return (x, y)

def mul(k, P=PG):
    R0, R1 = (0, 0), P
    for i in reversed(range(k.bit_length())):
        if (k >> i) & 1:
            R0, R1 = add(R0, R1), add(R1, R1)
        else:
            R1, R0 = add(R0, R1), add(R0, R0)
    return R0

def X2Y(X, y_parity, p=modulo):
    X3_7 = (pow(X, 3, p) + 7) % p
    if gmpy2.jacobi(X3_7, p) != 1:
        return None
    Y = gmpy2.powmod(X3_7, (p + 1) >> 2, p)
    return Y if (Y & 1) == y_parity else (p - Y)

The function mul(k, P=PG) is performing scalar multiplication on the elliptic curve:

It takes a number k (the private key).
It multiplies it by the generator point PG (which is a fixed point on the curve).
This gives a new point (X, Y), which is the public key.
Since elliptic curves don't have normal multiplication, we use point addition (add()) repeatedly to get k * PG.

👉 This is a one-way operation!

Given k, we can compute (X, Y).
But given (X, Y), finding k is impossible (this is the ECDLP problem).

Bitcoin doesn’t store (X, Y) directly; instead, it compresses the public key using X2Y(X, y_parity).

A full public key is (X, Y).
A compressed public key is just X and y_parity (0 or 1 for even/odd Y).
The function X2Y(X, y_parity) helps recover Y from X.
After this, Bitcoin:

Hashes the compressed public key (SHA-256 → RIPEMD-160 → hash160).
Encodes it into a Bitcoin address with  →  (Base58Check).

👉 So, when someone says they can reverse hash160 back to a private key, they’re skipping all these steps!
That’s why it’s impossible. 😆

No, that is no joke.
That is a tool that may make Your $$$ equipment useless when corretly used. Test me.