the trick for multiplying things larger that 10 by 11 is one of those things thats supposed to be elegant and mindblowing but actually it sucks ass and makes no sense

actually, lets get serious about this. lets put our balls on the table. i think if your adding up the digits of a number with each other you definitely ought to go to the hague

numbers are numbers. once you bring the digits that numbers our made out of into it your breaking the fourth wall and fucking around with the simulation. its unsafe

231 is not fucking 2+1=3 you fucking counting slut. 231 is the number after 230. you do not get to impose your conspirational ideology onto me. this isnt the fucking x files over here

@garfiald "all numbers in base 10 whose digits sum to a number divisible by 3 are themselves divisible by 3" what the fuck??? WHY

@byttyrs exactly!!! how can you deny that maths is fake with shit like this. their making maths up on the fly and rubbing it in our faces

@garfiald extremely, pathologically, provokingly self-consistent forms of bullshit

@byttyrs @garfiald Fun fact: The sum-of-digits-divisible-by-3 rule also works in hexadecimal.

serious answer to maybe non-serious question

@byttyrs @garfiald
In base 10, when you add 1 to a number, one of its digits increases by 1, and some 9s may become 0s. So adding 1 to a number always increases its sum-of-digits by 1 (maybe minus a multiple of 9). This means the difference between a number and its sum-of-digits is always a multiple of 9. So if a number is divisible by 9 then so is its sum-of-digits and vice-versa. In hexadecimal the same thing works for 15 (replace 9 with 0xF).
(1/2)

serious answer to maybe non-serious question

@byttyrs @garfiald
The divisible-by-3 rule works for both base-10 and base-16 because both 9 and 15 are divisible by 3. In base 10, the difference between a number and its sum-of-digits is a multiple of 9; in base 16 it's a multiple of 15. Either way that's also a multiple of 3. Since the difference between a number and its sum-of-digits is a multiple of three, then a number is divisible by three exactly when its sum-of-digits is.
(2/2)

@byttyrs @garfiald i guess yall dont want to know about how if you add up the two-digit groups of a number you can tell if something is divisible by 11:

121 -> 01 + 21 = 22 (121 is 11^2)
275 -> 02 + 75 = 77 (275 is 11*5^2)

@byttyrs @garfiald i believe that the reason both of these work is that the remainder of 10 is 1 when divided by 3, so every power of 10 is also equivalent to 1 mod 3:

420 = 4*100 + 2*10 + 0
= 4*(1+99)+2*(1+9)+0
= (4*1 + 2*1) + (4*99 + 2*9)

the right-hand group is a multiple of 3 for sure, so the left hand group is going to be the remainder when you divide by 3.

@byttyrs @garfiald it works in two-digit groups for 11 because it's not 10 but 100 that has a remainder of 1 when divided by 11, so you only break up by the even powers of 10

@FirstProgenitor @garfiald that's actually the result of thatcher-era cuts. there used to be like 5 more numbers between 230 and 231

@garfiald I can't believe I've been subtooted by THE garf eel

@garfiald I'm going to adopt like a bunch of kids and name each of the a different word from this

@garfiald what the fuck is x

x isn't a number it's a letter

@garfiald i am reblogging this solely for balls on the table. please enjoy this fanart of the concept

@garfiald when inspiration strikes u just gotta draw u kno Server run by the main developers of the project It is not focused on any particular niche interest - everyone is welcome as long as you follow our code of conduct!