balls and box

Balls in a box problem

A box is assumed to be a right angled prism with a rectangular base. The base is assumed to have dimensions equal to an integral multiple (n) of ball diameters by another integral multiple (m) of ball diameters. All tiers of balls are assumed to be complete.

The ball tier inventory for a first and odd rows is n*m and for a nested second and even rows is (n-1)*(m-1)

Now look at the total inventory
to is the number of odd rows & te is the number of even rows
so
to*n*m+te*(n-1)(m-1)=99
& consider that 99 is an odd number & odd#=even#+odd#
so either (to*n*m) or (te*(n-1)(m-1)) is odd
consider odd and evens
even*even*even==even*even=even
odd*odd*odd=odd*odd=odd
odd*odd*even=odd*even=even
odd*even*even=odd*even=even
Two cases
1) base tiers are odd, and nested tiers are even
2) base tiers are even and nested tiers are odd

Case1)
so if (to*n*m) is odd n,m,to are all odd and n&m>2
and to=te+1, to=te, or to=te-1

try n=3, m=5 nm=15 & (n-1)(m-1)=8
so can to*15+te*8=99?
try to=te+1
Then 15+te(23)=99 and te=64/23 & te is not an integer
Try to=te
23te=99, te=99/23 & is not an integer
try to=te-1
te(23)-15=99, te=114 & te not an integer

try n=5 and m=7, nm=35, (n-1)(m-1)=24
35+49te=99 te not an integer
49te=99 te not an integer
49te-35=99 te not an integer

try n=7 and m=9, nm=63, (n-1)(m-1)=48
63+48>99 won't work

Conclusion Case 1 won't work

Case 2)
n-1=1 & m-1=3 so n=2 and m=4
8to+3t1=99
let to=t1+1
8+11t1=99, t1=91/11 not an integer
let t0=t1
11t1=99, t1=9 & t1 is an integer
8*9+3*8=72+27=99

So you can put all 99 balls in a single box.

Are three whole numbers chosen at random divisible by six?

Use a similar odd vs even style analysis

6 is an even number & any number that is divisible by an even number is even
three is an odd number and three whole numbers chosen at random are all odd, all even, one odd two even, or two even one odd
&
odd+odd+odd=odd
even+even+even=even
even+even+odd=odd
even+odd+odd=even
since the sum of three whole numbers chosen at random can be odd, which is not evenly divisible by an even number.

So three whole numbers chosen at random are not universally divisible by six.

Rap c∫;?/

raprap

wrong
try again
did you understand what i mean
it may be my mistake

ofcourse you can put all the balls in the box
i never menion the dimensions

but i mean if you get out from the box any 3 balls the sum of their numbers must be devidble by 6

i can put only 3 ball 1.2.3 and if i choose any 3 balls from the box they will be also 1.2.3 and devidable by 6
so a solusion may be 3 balls only

but i ask about the maximum numbers

No!

Reread the second part of my above post.

Rap

So three whole numbers chosen at random are not universally divisible by six.

that will be if you put all balls in the box

don't put them all

If you put any odd numbered balls in the box, three chosen at random will not universally be divisible by six.

Rap

I get 16 two different ways:
1: 6x, x = 1 to 16
2: 6x+2, x = 1 to 16

The first one is obvious since all numbers are multiples of six.

For the second one:
(6x+2)+(6y+2)+(6z+2) = 6(x+y+z)+6 = 6(x+y+z+1)

6x+4 would also work, but the 16th value is 100.

If you put any odd numbered balls in the box, three chosen at random will not universally be divisible by six.


false

I get 16 two different ways:
1: 6x, x = 1 to 16
2: 6x+2, x = 1 to 16

The first one is obvious since all numbers are multiples of six.

For the second one:
(6x+2)+(6y+2)+(6z+2) = 6(x+y+z)+6 = 6(x+y+z+1)

6x+4 would also work, but the 16th value is 100.

excellent markr
16 balls is a good number
but it may be dlightly higher


hint.

----------------------------------------

Is this strictly a math problem and not a trick problem?

For example can I assume that 99 and 81 can't be rotated to produce 66 and 18?

If this is strictly a math problem, then I think this is a valid proof that 16 is the maximum:

Assume there are at least four numbers in the set.
Select any three (X, Y, Z). These sum to a multiple of six.
According to the problem statement, we can substitute a fourth number (W) for X, Y, or Z and still have a sum that is a multiple of six.
Therefore, W must be congruent to X, Y, and Z modulo 6.
If W is congruent to X, Y, and Z, then X, Y, and Z must be congruent to each other.
Therefore the whole set must consist of an equivalence class modulo 6.
There are three such sets that satisfy the problem:
6x+0, 6x+2, and 6x+4.
In the range 1-99, these have 16, 16, and 15 members respectively.

it's a math problem

There are three such sets that satisfy the problem:
6x+0, 6x+2, and 6x+4.

here's your false statement

try again

Oops, correct logic, bad arithmetic.

That should read:
In the range 1-99, these have 16, 17, and 16 members, respectively.

So, the answer is 17 using the numbers 6x+2 where 0 <= x <= 16.

Don't tell Try I screwed up again. :wink:

By the way, I didn't understand your hint (a bunch of dashes?).

What combinations will guarantee that regardless of what the first 2 balls are, the 3rd ball will always form the right combination?

Let the numbers of the 3 balls picked be A, B & C. This means
T = A + B + C is divisible by 6.

A solution would be for A, B & C each to be divisible by 6. That is
T = 6I + 6J + 6K = 6 (I + J + K)

Where I, J & K are 3 different integers. These integers may range from 1 through 16. 16 is the largest number we can use because 6x16 = 96, the largest number from 1 through 99 that is divisible by 6. In this case, the balls are numbered as:

1*6, 2*6, 3*6, 4*6, 5*6, 6*6, 7*6, 8*6, 9*6, 10*6, 11*6, 12*6, 13*6, 14*6, 15*6, 16*6
which equals
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96
In this case there are 16 balls in total.

What if each ball is numbered with 1 more than what would be divisible by 6? That is;
T = (6I + 1) + (6J + 1) + (6K + 1) = 6 (I + J + K) + 3
This is not divisible by 6.

What if each ball is numbered with 2 more than what would be divisible by 6? That is;
T = (6I + 2) + (6J + 2) + (6K + 2) = 6 (I + J + K) + 6
This is divisible by 6.

As before I, J & K are 3 different integers. These integers may range from 0 through 16. We can start from 0 this time because the smallest number on any ball will then be 2. In this case, the balls are numbered as
0*6+2, 1*6+2, 3*6+2, 4*6+2,…, 15*6+2, 16*6+2
which equals
2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74, 80, 86, 92, 98
In this case there are 17 balls in total.

Now what if each ball is numbered with 2 less than what would be divisible by 6? That is;
T = (6I - 2) + (6J - 2) + (6K - 2) = 6 (I + J + K) - 6
This is also divisible by 6.

The I, J & K in this case may range from 1 through 16. In this case, the balls are numbered as:
1*6-2, 2*6-2, 3*6-2, 4*6-2,…,15*6-2, 16*6-2
which equals
4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94
In this case there are again 16 balls.

So the maximum number of balls we can have in the box is 17.


Damn, got beat again

Mea Culpa I misunderstood this problem.

But yes if
(3n)mod6=0
the (n)mod6=0,2,4
since
(3*0)mod6=0
(3*2)mod6=0
(3*4)mod6=0
so the question becomes how many n=5i+0,2, or 4 for i an integer s.t. 0<n<100
if n=6i then 0<i<17 or the number of ball is n
if n=6i+2 then 0<=i<17 or the number of balls is 17
if n=6i+4 thern 0<=i<16 or the number of balls is 16
so the maximum number of balls in the box is for balls numbered 2,8,14,20,26,32,38,44,50,56,62,68,74,80,86,92,98
Note none of the balls are odd numbers
Again mea culpa I gloriously misunderstood the problem.

Rap c∫;?/

you are all right

markr:
my bad hint was built on misunderstood of your solution

i meant you add 2 and 4 to every ball number
what about replacing the + with - to substract 4 from each ball number so the total substract at any random 3 ball would be 12 but you will find a way to the 17 th ball


i hope the riddle was not bad

The riddle was fine. I just blew it.

try again

So the maximum number of balls we can have in the box is 17.


too late


Mea Culpa I misunderstood this problem.
never mind it's my bad english