# Three-d scattering of numbers?

1. Dec 7, 2004

### EG

has anyone investigated the three-d scattering of numbers?

FYI:

for any base

take the symbols that represent the numbers

arrange them in a rectangular or square grid

from a starting point, follow them in a square, rectangle, or diamond until you construct a number of 2n digits

divide by a number consisting of n "1" ' s

and there is no remainder.

Probably falls out of some other findings, but an interesting construction.

EG

2. Dec 7, 2004

### EG

This construction works in any base

I conjecture in any dimension as well

Anybody got a proof?

Is the construction understood?

3. Dec 7, 2004

### Hurkyl

Staff Emeritus
I don't understand the construction.

4. Dec 8, 2004

### HallsofIvy

"take the symbols that represent the numbers"

Do you mean digits?

"arrange them in a rectangular or square grid"
Like 0 1 2 3 4
5 6 7 8 9 ?

"from a starting point, follow them in a square, rectangle, or diamond until you construct a number of 2n digits."

I Think you mean like 78901234 with n= 4.

"divide by a number consisting of n "1" 's and there is no remainder"

Dividing that by1111 gives 71018 with remainder 236 so apparently I don't understand.

5. Dec 8, 2004

### EG

1 2 3
4 5 6
7 8 9

construct 12369874
or 123654
or 5698

divide 12369874 by 1111 = no remainder
divide 123654 by 111 = no remainder
divide 5698 by 11 = no remainder

or, arrange them in a cube:

5 6
1 2
7 8
3 4

construct 1562 / 11 = no remainder
construct 2684 / 11 = no remainder
etc

is this a trivial result of n-dimensional matrix arithmetic?

6. Dec 8, 2004

### matt grime

What apparent nonsense:

1245 and 1254 are both seemingly constructible in the first example, and they can't both be divisible by 11, can they?

7. Dec 8, 2004

### matt grime

Oh, the examples you picked work for obvious reasons once you know that a number is divisible by 11 iff the sum of the 1st, 3rd, 5th etc digits minus the sum of hte 2nd, 4th, 6th etc digits is also divisible by 11.

8. Dec 8, 2004

### Hurkyl

Staff Emeritus
what about 123658 in your example with 9 numbers in a square?

9. Dec 8, 2004

### Gokul43201

Staff Emeritus
I still don't understand the construction.

10. Dec 8, 2004

### Ethereal

Actually 1254 is divisible by 11. And it appears to me that 1245 cannot be constructed from his diagram.

From the first diagram, he says draw an imaginary line through the numbers until you form a square, starting from any digit, either clockwise or anti-clockwise:

123
456
789

The chosen numbers are bolded. The line passes through 4 digits, hence it passes through 2n digits where n=2. The starting digit does not seem to matter. Divide the number formed by a digit consisting of n "1s", in this case 11. 11|1254

I tried some other combinations:

123
456
789

whether anti-clockwise or clockwise it does not seem to matter, just like the starting digit. All the combinations of the six chosen numbers above are divisible by 111 (n=3, since it passes through six digits). Taking it diamond-wise:

123
456
789

Going clockwise or anti-clockwise yields 4 digit numbers alll divisible by 11. The only problem I have is how the digits in the construction of the diagram is to be arranged. Can someone explain how this works? I have no idea.

EDIT: The following is his cube:
Going through the the same process seems to yield numbers divisible by n "1s".

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11. Dec 9, 2004

### matt grime

why is 1245 not allowed? the construction is completely permissible in the sense of he has said nothing to disallow it. "follow them in a square" is ungrammatical and ambiguous - and who's to say how the digits should be writtend out in the first place? Why is

543
612
789

Not acceptable?

The arrangements you are allowing have their entries as arithmetic progressions (mod 10) You should see why that makes it work in the special cases given.

12. Dec 9, 2004

### Ethereal

This is something he would have to reply to. I have no idea how the diagram is to be constructed to begin with.

13. Dec 9, 2004

### EG

The allowed constructions are only those which form a "closed loop"

123
456
789

123654 is allowed
123658 is not

1254 is allowed
1245 is not

etc.

14. Dec 10, 2004

### Ethereal

matt grime wasn't referring to the loops that can be constructed, but how the diagram is to be drawn in the first place.

15. Dec 10, 2004

### matt grime

Oh, I was referring to all of those things including what shapes can be constructed seeing as the post was ambiguous to say the least. As it is what you're doing is disguising some obvious modulo arithmetic.

16. Dec 10, 2004

### Hurkyl

Staff Emeritus

17. Dec 11, 2004

### okidream

1 2 3
4 5 6
7 8 9
10 11 12
And I say my number system is base 13. Then it will not work.

But if I re-represent 10 as 1+0 = 1, 11 as 1+1 =2, and 12 as 3, and so on,
it works.

EG, is this a numerology trick/question?

18. Dec 12, 2004

### Rogerio

It works for 1254 , it works for 5698 , it works for 2684, and so on.

It seems any arragement that works in base 10 , works in base 13, too.

19. Dec 12, 2004

### StatusX

Here's a pretty simple proof of the case for rectangular loops. This is specific to base 10, but it can be easily generalized:

The array of numbers looks like this:

1 2 3 ... d
d+1 d+2 d+3 ... 2d
...

Where there are d numbers in each row. Now the rectangular loop, which has length 2n, looks something like this:

[x] [x+1] [x+2] ... [x+k]
[x+d] x+d+1 x+d+2 ... [x+d+k]
[x+2d] x+2d+1 x+2d+2 ... [x+2d+k]
...
[x+l*d] [x+l*d+1] [x+l*d+2]... [x+l*d+k]

Where the bracketed terms make up the number, with x being the first digit and x+d the last. k and l are the width and height of the rectangle. The (n+1)th digit is x+l*d+k, and the sum of the first and (n+1)th digit is 2x+l*d+k. The sum of the second and (n+2)th is (x+1) + (x+l*d+k-1), which is 2x+l*d+k. It is easy to see that the sum of the ith digit and the (i+n)th digit is always 2x+l*d+k, which from here on I'll call 'm'. Also, from here on I'll call the ith digit di. So d1=x, and di+di+n = 2x+l*d+k = m.

Now the number, which I'll call N, can be put in the form:

$$N = 10^n x + y$$

where:

$$x = d_1 10^{n-1} + d_2 10^{n-2} + ... + d_n$$

$$y= d_{n+1} 10^{n-1} + d_{n+2} 10^{n-2} + ... + d_{2n}$$

We know that 111...11 with n 1's divides 10n-1. So if it divides N, it must also divide:

$$N -(10^n-1)(x)= 10^n x + y-(10^n-1)x = x + y$$

which is:

$$(d_1 10^{n-1} + d_2 10^{n-2} + ... + d_n ) + (d_{n+1} 10^{n-1} + d_{n+2} 10^{n-2} + ... + d_{2n})$$

$$= ((d_1+d_{n+1}) 10^{n-1} + (d_2+d_{n+2}) 10^{n-2} + ... +(d_n +d_{2n} )$$

$$= m 10^{n-1} + m 10^{n-2} + ... + m$$

$$= m (111...11)$$

which completes the proof.

edit: this also works if the string starts somewhere else in the loop or goes around the otherway. all this changes is which sums are in which decimal places, but since theyre all m anyway, you get the same result.

edit 2: I just thought of a few generalizations. This will work for any shape constructed like so: Make a contour with n squares. Copy it, rotate it by 180 degrees, and put the second piece wherever you want (use it to form a closed shape if you want, but you don't need to). The number is formed by starting at any point on one of the contours and continuing in one direction around it and in the same direction around the rotation until you get back to where you started. This covers the diamond shape.

I admit this is a little confusing. Here's an example: if your shapes are Ls, you could start at the corner of one L, go up the long side, than across the short side of the other one, down its longs side, and finally across the short side of the original one to get back to the start. The second L must be rotated by 180 degrees, but could be anywhere you want it. They could form a rectangle, not touch at all, or overlap, just as long as you go around the right way.

This number will have 2n digits, and is divisible not just by 111...11 with n ones, but also with k ones for any k that divides n.

Last edited: Dec 13, 2004
20. Dec 13, 2004

### matt grime

Like I said a page ago, it's simply a fact about arithemetic sequences mod 10, or whatever base you chose to work in.