A Maths Puzzle: Balls, Solution and The Chinese Remainder Theorem

Now, what I saw was that if you put your balls into piles of "n", you will have precisely "n - 1" left over. This is true regardless of n. I also noted that your n was each prime number from 2 to 11. I didnt use modular arithmetic. I simply multiplied the prime numbers 2 through 11 together and subtracted 1. Seems simpler to do it that way, though likely not a general solution technique.

"All the way down"... like turtles? I just thought that line was funny, given the context

this is actually one of those videos as you feel like a retard after watching;) jesus crist that was way to much

Um...??? The Chinese Thereom just confused me... I think I got smarter??!!

damn i got 2099 balls...it was soooo close...

Nice work though, I am reading your comments :)

thnx...k i`ll tell u how i tried to solve..becoz i had no idea of chinese theorem, i tried the trial n error method....u said for piles of 11, 10 balls remain...so i tuk all multiples of 11 n simply added 10 to them so naturally the remainder will b 10...now for piles of 2, 1 ball remains which means the final answer is an odd number..so i rejected all even multiples of 11 n multiples which end in 5 coz they r completely divisible by 5...then i got 2099 as anwer..small calculation mistake may b

Your method was fine. Don't worry about the Theorem, just thought I'd show people something new.

I thought you had 170945114945190252226559 balls

if there were 21 then there would be none left over from 7