The Pigeonhole Principle: If You Have N Pigeons And N+1 Holes, Then There Must Be Some Pigeon With At

The pigeonhole principle: If you have n pigeons and n+1 holes, then there must be some pigeon with at least two holes in it. That poor pigeon. Don't ask me why they made this principle so gruesome.

It's hypothetically possible that I may have mixed up the order, but ehhhh, whatever.

1. it's always easier to work in modular arithmetic (finite addition table!), so find the first primorial that's at least the larger number = 210 = 2*3*5*7

every number up to 210 is uniquely determined by its values mod each of these primes.

156 = 2(mod7),1(mod5),0(mod3),0(mod2)

25 = 4,0,1,1

now to add them you only need pointwise modular addition

2,1,0,0 + 4,0,1,1 = 6,1,1,1

and then just find the smallest number with these modular values = 181

the only edge case is when this system overflows, so in that case you take the smallest number with these values that's also larger than both summands.

2. evaluating multiplication is harder than addition, so we approximate 2*45+15 as 2*(45+15)=2*60. then by distributivity the error is 2*15-15, which is (2-1)*15=1*15. doing this simpler multiplication by hand, we have 1*15=15, so 2*45+15 = 2*60±15. better approximations can be made but this is good enough for most purposes.

3. for division we combine our last two ideas of modular arithmetic and approximations. start with the smallest prime greater than 104 = 107, and note that there is a finite field under addition and multiplication mod 107. so in particular, it has an element 1/8, which we can find easily by noticing that 1/8=(1/2)³. 1/2 is easy to find, since 1 equals 108 mod 107, which is even, and so 1/2 = 54 mod 107. now we compute 54³ mod 107 and get 67. so we take 104*1/8 ≈ 104*67 mod 107 = 13 as a first approximation. now we check 13*8 = 104 (w/o modding!) and we're done! if the numerator weren't divisible by 8 we would've gotten a wrong answer (for example, (105*67)%107=80, 8*80≠105) and in that case we repeatedly subtract 67 (mod 107) from the answer until we have (8*answer+#times we've subtracted 67) = numerator. the answer is the integral division and the number of times we subtracted 67 is the remainder.

hope that helps:)

ok so for personal reasons I need some of the dumbest, goofiest, most elaborate cutting-your-toast-with-a-chainsaw type ways to do elementary school arithmetic problems. the type of stuff we did to that "whats 48+7" poll

my first thought for calculating 104/8 is to convert into base 8, then bitshift one to the right and then convert back but I feel like thats too simple

so, hit me with your best shot at solving the following problems:

whats 156 + 25

what's 2*45 + 15

what's 104/8

it’s a common misconception that maths is all theoretical; they actually keep the 0 in a vault in France and u can go look at it if u got connections.

Bitches love me for my proofs of formal equivalence between two seemingly unrelated things

if you give me an arbitrary computer program with a certain input i can tell whether it halts or not. rip to you but i’m different