A Maths Puzzle: Balls

Thanks to arthritismutilans http://www.youtube.com/user/arthritismutilans
and DeeDoubleYou81 http://www.youtube.com/user/DeeDoubleYou81 for the puzzle.

If I put my juggling collection into piles of 2 I have one juggling ball left over.
If I put my collection into piles of 3 I have 2 left over.
If I put my collection into piles of 5 I have 4 left over.
If I put my collection into piles of 7 I have 6 left over.
If I put my collection into piles of 11 I have 10 left over.

Have many juggling balls do I own? Is there only one solution?

Solution video here http://www.youtube.com/watch?v=zB2nbCCeSRI

Category:

Education

Tags:

• maths
• math
• mathematics
• problem
• puzzle
• juggling
• balls
• chinese
• remainder
• theorem
• number
• theory

[Namatok]

The box on it's side saying "This way up >>>>" was annoying me

[singingbanana]

@Namatok It's very likely I did that in purpose.

[leekeewei]

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?

[singingbanana]

Excellent.

[Strijdparel]

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...

[singingbanana]

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.

[skizle1337]

hahaha ''i got more balls than most, i guses im just blessed that way'' made my day lool

[palmomki]

I don't feel like making the operation, but the solution is to find any Common Multiple of 3, 5, 7 AND 11 and decrease it by 1. So, because of the Infinity of the Natural Numbers, there are endless solutions

[KamiKagutsuchi]

Chinese remainder theorem ftw

[Namatok]

@singingbanana You evil man!

[jesmurf]

My way of getting it was bad. It worked out eventually but it took too long. I figured that since it had to be devidable by 5 +4 it had to end with 9. So then I tried to multiply all kinds of numbers that ended with 9 with 11 and added 10. Then I eventually found out that only 1 in 3 numbers would work out and also only one in 7. then I figured out the same could be said for all numbers as long as it wasn't equal, after that I just used the method leekeewei used in the first place.

[PCGamerPortal]

chinese remainder theorem :D

[RandomNinjaOfEvil]

If you put them in piles of seven, you have a lethal combination.