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[-] LeFrog@discuss.tchncs.de 29 points 5 months ago* (last edited 5 months ago)

Wow I never thought about that.

But it is always like this:

let there be any base "b"
That can represent a number by the sum of their positional digits:

number = sum(d_i * b ^ i)
   where i is the position index and d_i is the digit at this position. (note: index starts with 0, from the least digit farthest to the right)

So the (decimal) number 4 in base 4 is then

1×4¹ + 0×4^0 = 10

And (decimal) number 8 in base 8 is

1×8¹ + 0×8^0 = 10

And 10 in base 10:

1×10¹ + 0×10^0 = 10
[-] TheBat@lemmy.world 36 points 5 months ago
[-] witty_username@feddit.nl 26 points 5 months ago

All your base are belong to 10

[-] AtariDump@lemmy.world 3 points 5 months ago

You have no chance to survive make your time.

[-] teft@lemmy.world 3 points 5 months ago* (last edited 5 months ago)

Someone set us up the base!

[-] Tar_alcaran@sh.itjust.works 6 points 5 months ago

Which is why to write hexadecimal, we added letters to it, because there isn't a digit for "12".

this post was submitted on 24 May 2024
1186 points (97.6% liked)

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