There are 25 characters, each of which has 36 possible values. So 36^25 possibilities, and log2(36^25) = 129.2. There are basically 129 bits of entropy in there, so good luck bruteforcing it.
This makes me think of a shareware app (I think an icon editor) for Windows 3.1 back in 1994 or so... I could find a valid registration key by entering random numbers by hand in around 2 minutes. And I wasn't lucky as I tried and succeeded several times ;-) But the rule (figured out after I had 10 or so valid keys) was simple maths with the digits, no crypto behind.
> There are 25 characters, each of which has 36 possible values. So 36^25 possibilities, and log2(36^25) = 129.2. There are basically 129 bits of entropy in there, so good luck bruteforcing it.
Kinda depends what is encrypted there. If it is just "magic number + SKU + licence ID" and there is no online check whether that combination is valid then you're "just" trying to hit one that's valid and that cuts few bits off equation as there is spectrum of keys that will be valid but not generated by Microsoft.
This makes me think of a shareware app (I think an icon editor) for Windows 3.1 back in 1994 or so... I could find a valid registration key by entering random numbers by hand in around 2 minutes. And I wasn't lucky as I tried and succeeded several times ;-) But the rule (figured out after I had 10 or so valid keys) was simple maths with the digits, no crypto behind.