Let's try to do a 'lil bit of math.

[EnforcerKiller]

So, say we got the AAAAA-AAAAA-AAAAA format of keys, right? And there's 26 letters in the english alphabet, while there's 10 numbers that can be use total (0-9). I tried doing 36 x 15 and then x 15 again, it gave 8100. But when I tried 36x36 15 times, it gave me 2.21073920E+23, which leaves me to wonder what's the correct way to do the math and what's the correct answer to this. Now I'm not an expert at math, but maybe there's one person who can do the math right and yield the answer?

[Zireth]

Look it up on the internet

[Shmip]

36^15 = 2.2107392E+23

[mellow]

sounds hard. i dont know how lol ;p

[Washell]

4 digits (0 to 9) has 10,000 options (10 x 10 x 10 x 10) or 10^4
So a 15 character long alphanumeric code is 36^15 which is 221073919720733357899776

Good luck trying them all. ;)

[EnforcerKiller]

Thanks for the replies!

[Spawn of Totoro]

This on is for 8 digits, but you have the process and forumula to figure out more.

https://www.quora.com/How-many-8-digit-passwords-can-be-generated-using-combination-of-10-single-digit-numbers-ex-1-2-3-9-and-26-Alphabets

[math]2.612182843\times10^{12}[/math]

As mentioned in the description it is a rather simple permutation and combination problem. Consider three sets A, B and C defined as:

A = set of all possible combinations 8 letter combinations
B = set of all 8 digit combinations containing only alphabets
C = set of all 8 digit combinations containing only numerals

now the required set [math]S = A\setminus(B \cup C)[/math]

also since [math]B,C \subset A[/math], and [math]B \cap C = \emptyset[/math]
[math]|S|=|A| - (|B \cup C|) = |A| - (|B| +|C|)[/math]

Now,
[math]|A| = (26+10)^{8} = 36^{8}[/math],
[math]|B| = 26^{8}[/math]
[math]|C| = 10^8[/math]

this gives us the answer [math]36^{8} - 26^{8} - 10^{8}[/math] which evaluates to [math]2.612182843\times10^{12}[/math]

[Azza ☠]

Taking the format: AAAAA-AAAAA-AAAAA

36^{8} - 26^{8} - 10^{8}

a2+kc2/a2=8π Gp+Ac2/3 a/a=-4πG/3(p+3p/c2)+Ac2/3

Answer: FHCZJ-Q26F9-AGIC6

[𝕙.𝕓𝕒𝕣𝕜𝕒𝕤]

https://www.quora.com/Can-Steam-run-out-of-keys-How-many-can-exist

[bobes sponja]

Messaggio originale di Azza ☠:

Taking the format: AAAAA-AAAAA-AAAAA

36^{8} - 26^{8} - 10^{8}

a2+kc2/a2=8π Gp+Ac2/3 a/a=-4πG/3(p+3p/c2)+Ac2/3

Answer: FHCZJ-Q26F9-AGIC6

dude... thanks for the code :D

[EnforcerKiller]

Messaggio originale di FαɳƈყBLYAT:

Messaggio originale di Azza ☠:

Taking the format: AAAAA-AAAAA-AAAAA

36^{8} - 26^{8} - 10^{8}

a2+kc2/a2=8π Gp+Ac2/3 a/a=-4πG/3(p+3p/c2)+Ac2/3

Answer: FHCZJ-Q26F9-AGIC6

dude... thanks for the code :D

What was it?

[bobes sponja]

Messaggio originale di EnforcerKiller:

Messaggio originale di FαɳƈყBLYAT:

dude... thanks for the code :D

What was it?

eh... I dunno,says game associated with the code is already in my library
so I gave it to a friend

[Johnny-Boy]

All i know is one pile of dung + one pile of dung is one BIG pile of dung.
Das how we do eet on da farm

[bobes sponja]

Messaggio originale di EnforcerKiller:

Messaggio originale di FαɳƈყBLYAT:

dude... thanks for the code :D

What was it?

apparently 5 of my friends already own it, even my alt acc that has no games.

[EnforcerKiller]

Messaggio originale di FαɳƈყBLYAT:

Messaggio originale di EnforcerKiller:

What was it?

apparently 5 of my friends already own it, even my alt acc that has no games.

Probably a free game then. Could be anything, even dota 2. It used to be a paid game. Just a hunch.

[Azza ☠]

Messaggio originale di FαɳƈყBLYAT:

Messaggio originale di Azza ☠:

Taking the format: AAAAA-AAAAA-AAAAA

36^{8} - 26^{8} - 10^{8}

a2+kc2/a2=8π Gp+Ac2/3 a/a=-4πG/3(p+3p/c2)+Ac2/3

Answer: FHCZJ-Q26F9-AGIC6

dude... thanks for the code :D

It's a magic anti-leech key (was an early access key for a now free game), it will tell anyone who attempts to use it that they already have the game... but congratz you won and said the magic word "Thanks" so get a real steam game :)

I'll add you (sent Legend of Grimrock, enjoy).