On Nov 8, 2010, at 7:49 PM, Greg wrote:
> So I'm unclear on what 3 (mod 3) means...
I may have answered my own question, let me know:
6 = 3 (mod 3)
That means that *both* sides are modulo 3, in which case 0 = 0.
Whereas,
(a + c) = 3 != 2 (mod 3)
Makes sense because:
0 != 2
If so, then now the only thing is I'm not sure how you saw that the wheels
satisfied that equation in the first place...
- Greg
> On Nov 8, 2010, at 6:29 PM, Ken Wesson wrote:
>
>> On Mon, Nov 8, 2010 at 9:08 PM, Greg <g...@kinostudios.com> wrote:
>>> Ah, I see that this wasn't immediately clear from my explanation: they
>>> change by 1, but 3 wraps to 1. I.e. the chain is:
>>>
>>> 3 => 1 => 2 => 3 => 1 => etc...
>>
>> I assume only the first two and last two can be changed together.
>
> Correct.
>
>> But if only the first two and last two can be changed together, then
>> anytime one of the end ones rotates by one, the middle one does as
>> well. Note that the initial condition 3 3 3 satisfies (a + c) = b (mod
>> 3) where the numbers are a b c. And if you raise the middle one and
>> one of the other two mod 3, then you raise b mod 3 by 1 and (a + c)
>> mod 3 by 1. So (a + c) = b (mod 3) is an invariant of the system. And
>> 2 2 1 violates it since (a + c) = 3 != 2 (mod 3).
>
> Woah, that seems pretty cool, but I'm not sure I understand it completely
> because I'm unfamiliar with the notation... perhaps you could clear up my
> understanding:
>
> 3 3 3 -> a b c
>
> Then,
>
> (a + c) = 3 + 3 = 6 = b (mod 3) = 3 (mod 3)
>
> I'm not sure I understand what you mean by: 3 (mod 3)
>
> I understand modulus in this way: 3 % 3 = 0
>
> So I'm unclear on what 3 (mod 3) means...
>
> - Greg
>
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