[Samsung Galaxy S20 / S20+ / S20 Ultra Questions & ] adfree:
For tiny exercise(S) I have created """RSA 20 """ example

For this I used Tool...

Public Exponent HEX:
010001

Bit length: 20

P = 038B
Q = 02DD
Exp1 = 65
Exp2 = 022D
Coeff = 0296

Priv Exp = 018CC5
Publ Exp = 010001
Modulus = 0A24FF

I need:

A

Modulus aka public key

0A24FF

B

Public Exponent

010001

DD

"Signature" is encrypted by Private Exponent aka Private Key

03DBFB

EE

DEcrypted Signature by Modulus and Public Exponent

018738

ZZZ

Private Exponent aka Private Key

018CC5

For simplify... ALL are 3 Byte in length:

A is 3 Byte

B is 3 Byte

DD is 3 Byte

EE is 3 Byte

ZZZ is 3 Byte

In this example we allready know priv key aka Private Exponent...

for this Baby example IMHO maximal attempts for Brute Force:

FFFFFF

Now I take Calculator and "convert" HEX FFFFFF in:

16777215

IMHO MAXIMUM 16777215 attempts to bruteforce...

But we know we are lucky... HEX

018CC5

Means after 101573 stupid + 1 Operation...

If we start with 3 Bytes ZERO:

000000

000001

000002

.
.
.

In my lucky example first start with 0... so here is starting from 0 and operation + 1 cool... IMHO...

But maybe for other Hacking attempts start from end... and operation -1

FFFFFF

FFFFFE

FFFFFD

.
.
.

And we could split... to Brute Force on more then 1 Computer...

Baby Example...

Computer 1 start with:

000000

Computer 2 start with:

100000

Computer 3 start with:

200000

And so on....

IMHO we could do something like this with RSA 2048...

256 Byte is bigger then 3 Bytes... I know harhar...

BUT we have all...

We can identify Signature and decrypt... and we have the decrypted content...

And if sombody looked into le.k... key.xml

<?xml version="1.0"?>
<KeyPair>
<Modulus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odulus>
<PublicExponent>00010001</PublicExponent>
<PrivateExponent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rivateExponent>
</KeyPair>

Private Exponent is 256 Byte... Modulus is 256 Byte...

I can see Sig is 256 Byte... decrypted Sig is 256 Byte...

We are talking about HEX Value... FF is max... no FG FZ...

Maybe in this way not all RSA keys breakable... BUT we have more then 1 key to break...

So chance is increased this is not total wrong IMHO...

Best Regards

Edit 1.

If i have time i will check Tools/Scripts like this:

GitHub - ZeroBone/PollardRsaCracker: RSA cracking algorithm based on Pollard factorization (Pollard's p-1 method).

RSA cracking algorithm based on Pollard factorization (Pollard's p-1 method). - ZeroBone/PollardRsaCracker

github.com

With my ""RSA 20"" example