A Maths Puzzle: Balls

[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]

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

[singingbanana]

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

[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

[skizle1337]

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

[RandomNinjaOfEvil]

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

[TheAnonymousParadox]

I got x congruent -1 congruent 2309 mod 2310. Hence you have 2309 balls.

[maankindj]

math AND julling? man you're great

[TehMathWiztard]

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

[TehMathWiztard]

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

[TehMathWiztard]

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.

[zyzzyzus]

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.

[Bella500000]

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

[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.

[Strijdparel]

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

[MaskedYoYoGuy]

solve this code

tyejkyftyeheuityekly

[TUPELOBOY19]

twenty one

[lusmat]

2309

[logolou]

twenty one

[Xxsgtnutty231xX]

if he has 2 he has 1, so if he added all the digits and ad a numeric division for

e=11 and adding the numbers in the square root of 28 which is the sum of all numbers, when giving a selectible intiger to the number line , multiplying 2, 3, 5, 7, and 11 giving us an even number will be the ANSWER: 420!!!

[Moorich]

it can't be 420, because if he puts all of them into piles of 2, he'll have 0 left over instead of 1. The very first condition fails.

[IrkIr]

This is easy. 11*7*5*3*2-1=2309 that's the first number. 11*7*5*3*2*x-1=another number. You need to replace x and you get all the other numbers. There is infinitive amount of numbers you can get because you can replace x by any other number.

[wampireoffog]

2309, 4619 ...

There's so much soliton.

[m123bl]

it's hilarious when the balls fall down on him! xD

[JoNJoShY]

2309

[dave597]

there are infinite solutions in the form n=2310*a-1 where a is a positive integer

[JOZZIE52]

2309

4619

6929

9239

11549

13859

16169

Thats all i could fit into this comment but there is a lot more i found a pattern so now could have a big list

[JOZZIE52]

just sent singingbannan the full list with the first 1000 entries must have taken me atleast 2 mins

[seansean35]

2309...3hrs to do this

[JOZZIE52]

wow did u do it all by hand? caus excel makes thigs a lost easier i got list of 1000 in like 2 mins

[seansean35]

yhus,yhus I did*faints*

[swimmersrsick1]

dont know

[LegitimateLegenD]

21?

[JellyfishPie00]

huh?

[yuskei3]

There's more than one, but the easy one is 2309.

[ghosty4ever]

DUUUUUCCCKKKK!!!!!!!!!!!!

[magician2242]

2309 taht's easy i need a chalange but there is more then one solution

[Ilopsen]

My GOD!

76 balls gets past PO11 and PO7, but the others won't work!

[KiranGlitch]

do you notice that he only puts them in piles of prime numbers?

[konopong]

You would have one less than a common multiple of 2,3,5,7,and 11.

[victoriscool1]

23!!!!

[MrMasterMagician]

A very good math puzzle cant really figure it out post solution

[RonnielikesRealrock]

when you putem in piles of even numbers the number thats gets left over is odd... and visa versa thats easy

[chrisstorm4901]

21 balls

[Everything96]

that couldn't be it

[TH3GHOST2]

a fair few

[mebethatthing]

cant get it:P

[Pluvialis]

I've got it: there are infinite solutions, given by this formula for all natural numbers "n":

2310n-1.

[FlySpades11]

lol nice hope you don't mind what I did

[scotland7yard]

I got 2309.

I just realised that instead of fixing on how many balls are leftover it was better to think of how many balls shy you are from the next multiple, and it's always 1! so by multiplying all the numbers and subtracting 1 you get the answer... :)

I'm glad you're back SB keep'em videos coming!

[Pluvialis]

Well I must admit I brute forced it, so I'd like to see the proper solution, but I do know that 2309 AND 4619 are solutions! So it's not unique. I see no reason to suppose there are less than an infinite number of solutions.

[Pluvialis]

In fact, looks like you can just double any one solution and add one to find another.

Although 6929 is the third solution, which is the first, times three, add two. So maybe you can get all the solutions by starting with the first and timesing by any of 2, 3, 5, 7 and 11 then adding 1, 2, 4, 6, or 10 respectively.

[mediteight]

I agree: 2309 + 2310 n are solution for all natural numbers n.

It has to be an odd number and 4 mod 5, so must end in a 9

Using 10 mod 11, I checked though the remainders mod 7 for:

109, 4 219, 2 329, 0 439, 5 549, 3 659, 1 769, 6

and then found they repeat...

so answer must be 769 + (7*110) m

Now to satisfy 2 mod 3, the m in this answer must be 2 mod 3

which means the answer must be

769 + (7 * 110) * 2 + n * (7 * 110 * 3)

[mediteight]

Hmm - definitely prefer scotland7yard's way of doing this to mine!

:)

[Error081688]

oh, someone got it already. Good job!

[Error081688]

RATS! I got 769, but that works for all of the conditions except 3 remander 2....

I'll get it!

[robgrime1]

32???????

the mice have worked it out to 42

but that's their answer to everything

[singingbanana]

Boom boom :)

[higafujiyoshi]

@singingbanana i went with 21 because if you have 1 pile off 11 remainder 10 you have 21 balls

[Error081688]

hahaha

[DKDABU]

The 1st thought that I had was it had to end on 9 because in 2 piles it leaves 1, makes it non even number, 5 leaves four makes it only 9 as end one, but it still have to be 10 above 11*x, and then it's like 21-32-43-54-65-76-87-98-109, but ended up with the result 2309 balls, that's ALOT mate.. xD

[MrCrownKing]

32!!!

[DKDABU]

32 doesn't stack with the 7 pile, then it's not good..

[froboykujo69]

my answer is 2309

[froboykujo69]

the trick is multiply the numbers and 2,3,5,7,11 then subtract one. There is more than one solution. Multiply 2,3,5,7,11, the next prime number(13) then subtract 1. I'm sure that pattern will stand.

[DKDABU]

Well the pattern didn't stand,but hell yeah it was the easiest way to do it, I multiplyed 11 with all the numbers from 1- 209 ( addd with 10 everytime ofc) and just took all the numbers ending on 9.. "2x3x5x7x11-1" is the fastest way to the solution.. :P

[grande1899]

I managed to work out that the last digit must be 9, but I have no clue what the whole number is =p

[omaristheman]

8459?

[DeathRazorSk8ing]

this is hard to figure out...give me a minute or so...

[comownage]

when I read the tittle I was like 0.0

[wishmastter]

I like your little Puzzles..

[CletusMcPhee]

is it not 21

[Moorich]

love your english accent