> Yesterday, I discovered something else, but related, while sitting on a train playing "Snake" on my Nokia 3210 cell phone on highest speed, so you get 9 points for every food piece thingy the snake eats. I got a high score, 567 points. I wondered how many thingies my snake had eaten, so I divided it by 9 and found it to be 63. Somehow, I wound up discovering that if you juggle the digits around, it'll always be a multiple of 9.
>
> 567 / 9 = 63
> 576 / 9 = 64
> 657 / 9 = 73
> 675 / 9 = 74
> 756 / 9 = 84
> 765 / 9 = 85
>
> Then I thought that it's just a little bit odd. It's not odd that any of those numbers are multiples of three (since the digit sum (18) also is -- this we learn in early school years), but every one of them are also multiples of 3*3.
>
> Then I thought: What if I add a couple of digits?
>
> The digit sum of 234567 is 27, so this number, too, and all other arrangements of those digits, will be divisable by 3.
>
> And the same probably also goes for 123456789.
>
> But I think every arrangement of these numbers (without having tested all of them), will be a multiple of 9. So did I just find out something we didn't learn in school? That if the digit sum of a number is a multiple of 9, the number will also be?
>
> Trav"likes to juggle numbers but lacks the knowledge to deduce things like this on his own"holt.


I think I also learned that rule in school.

I, too, began to wonder these things while playing Snake on my Nokia (mine is 6150). I used to play with lower speeds, and my record, 1366, comes from that time.

I thought about this during a boring mathematics lecture (funny, I had a test on mathematics just a few minutes ago). I noticed that every number with three digits that are next to each other, is divisble by three. For example, 123, 234, 324 and so on. This is because one of them is always divisible by three, and the other two cancel each other out. And, if the middle one is divisible by three, the whole number becames divisible by nine. 234, 423, 756, for example.

It can be a pain to add up the sum of digits of a horribly long number. It is also difficult to see whether that number is divisible by three (or nine). You can use the same rule here: Just keep adding up the digits until you have a single number. I also found out (I can't prove this, but it seems to work always) that it doesn't matter at all in which order you do the adding. Example:

846513654983217 (a long number, isn't it?)

8+4+6+5+1+3+6+5+4+9+8+3+2+1+7 = 72, 7+2 = 9

But it might be easier to keep the numbers as small as possible, like this:

846513654983217

8+4=12 (the first two digits)
1+2=3

3+6=9 (add the third digit)

9+5=14 (the fourth)
1+4=5

5+1=6 (the fifth)

6+3=9 (and so on..)

9+6=15
1+5=6

6+5=11
1+1=2

2+4=6

6+9=15
1+5=6

6+8=14
1+4=5

5+3=8

8+2=10
1+0=1

1+1=2

2+7=9

Seems a bit long, but it's relatively easy when you have to do the check without a calculator or even a pen.


Juho