A Maths Puzzle: Balls

Dang, my method was so complicated compared to your guys' method *cries*

Okay, that 11 piles part of my comment was wrong I think, Imma find the first 4 ones, 69 + 70 = 139 + 70 = 209 / 30 = 6 with a remainder of 29! There we go, now the 2, 3, 5, and 7 parts work...I don't really know what to do now.....Now I think I should add maybe a number divisible by 2, 3, 5, and 7, just until I get a number that is divisible by 11 with 10 left over, well 209 + 1 = 210, divisible by those 4. Well 210 x 11 = 2310, Now to subtract 11, but now it is even by 11, so subtract 1 = 2309

This is how I did it:
Piles of 2 with 1 left over makes it an odd number (Even numbers would be evenly divided by 2)
Piles of 5 with 4 left over means the last digit has to be 4 or 9, because it has to be odd it is a nine
Piles of 3 with 2 left over...Well if it ends in nine, what is divided by three to get 2 left over? 29!
Piles of 7 with 6 left over, that would make 69, keep adding seventy to make 29 left over
Piles of 11 with 10 left over, only way is 11 x 10x -1 to make a possible number.

The number just has to be one less than a number that has 2, 3, 5, 7 and 11 as prime factors. 2309 is just the lowest the number can be.

let see, 3x5x7x11=2310 BUT theres a remainder when you group them into those number. so, 2310-1=2309?

lets me see.... hm. let x be the smallest possible number. so x +1 can be divided by 2,3,5,7 and 11. so x+1 = 2x3x5x7x11=2310. so x= 2309?

Excellent.

mmm, my solution did not work out; 1/2 x 2/3 x 4/5 x 6/7 x 10/11 = 8623/2310= 1 1693/2310 wich means 1693 balls, but when I cheched it it already failed by 3...

Multiplying the denominators (the numbers on the bottom of the fraction) to get 2310 was right. We just needed a remainder of 1 on top.

Oke, thanks I get it now. It was a nice riddle.