Question

What's the most general form of \(\large a \) if it obeys this relationship?
\(\Large a \mod c =b \\ \Large a \mod b = c \)

Answer 1

Wait what lol

Answer 2

well I can say that \[a=qc+b\]and\[a=rb+c\]For some integers \(q,r\)

Answer 3

Oh

Answer 4

So let's see how far that gets me. We then have \[qc+b=rb+c.\]Solving for \(c\) I then get\[c=b\frac{r-1}{q-1}\]Plug that back in, \[a=b\left(q\frac{r-1}{q-1}+1\right)\]and a similar statement if you solve for \(b\) instead.

Answer 5

Unfortunately, I'm not entirely sure where to approach this next.

Answer 6

Hey thanks for giving it a shot, I am not entirely sure where to take this either, my approach was a little different and I kind of found my way into this problem through the back door I think but thought it was interesting.

I don't think I've seen you around in like 3 years, kind of shocked haha.

Answer 7

It's definitely been a few months since I last logged on. I was feeling bored and there was a troll that needed bopping earlier tonight though :P

Answer 8

Haha, well I guess I should go ahead and reveal the answer since it's kinda tricky, but really not incredibly amazing. I like your approach of how you started it so I'll kinda follow you and diverge a bit.

\[\Large a = qc+b = rb+c \\ \Large a \equiv qc \equiv c \mod b \\ \Large a \equiv b \equiv rb \mod c\]
So here I see that we have qc equivalent to c and similar equation following it, so I can see that q has a pretty specific form: \[\Large q = nb+1 \\ \Large (nb+1)c \equiv c \mod b\] Do something similar with r, \[\Large r = mc+1 \] Let's go back and plug these into the original two equations we started off with: \[\Large a = (nb+1)c+b = (mc+1)b+c \\ \Large a = nbc+b+c = mbc+b+c \\ \Large \therefore n=m\] and the most general form is: \[\Large a=nbc+b+c\]

Answer 9

@ganeshie8 !!!!!!!!!!!!!!! mod ^^^^

Answer 10

Now I feel like I should have finished that off. I was thinking along those lines but screwed up when I was simplifying in my head and ended up getting nonsense :(

Answer 11

Well honestly it wasn't until you said a few things that I realized my own mistake. I actually go to this while thinking about this formula: \[\Large a+1=(b+1)(c+1)\] and forgot the arbitrary value that could be on the ab term.

Answer 12

And that came out of looking at the divisors of numbers, trying to find out relationships between numbers with the same number of divisors, and sort of ended up here...\[\Large \tau (p^a)=\tau(q^br^c)\]
So back to thinking about this original idea I guess haha.

Answer 13

I've got a friend who's interested in things like this. If I remember I'll run this by him and see what he knows.

Answer 14

Yeah that would be awesome. I guess I'll elaborate a little further on my half-baked idea.

Suppose you make a bunch of circles on the ground in a line and label them 0,1,2,3,4, etc... Now starting at 0, drop a stone in each circle so every circle has a stone in it. Now start again from 0 and only put in a stone every other one at 0,2,4,6,8, etc... Now start at 0 and put one stone in every third pile, 0,3,6,9, etc... Keep doing this cyclic and increasing process and what do we have?

Each labeled circle has just as many stones in it as it has divisors. So accordingly, the 4th circle has 3 stones, each representing 1,2, and 4 while the 7th circle has a pile with only 2 stones in it since it's prime.

So one of my ideas is what if we disconnect the concept of divisibility for a while and still call these things "factors" but change how we distribute the stones among the circles. So we could shift how we place the stones in the circles by displacing every third cycle by say just 1, so instead of starting at 0 for the ones placed apart by 3 we lay them in 1,4,7,10,etc... instead of 0,3,6,9. We could lay them in some kind of fibonacci sequence or something even instead of consistent periods.

Anyways, just playing around.

Answer 15

Definitely could be interesting. I'll let you know if I find anything that seems to actually be useful.

Answer 16

In a sense though, what you're doing is just creating arithmetic progressions. And I know there's been some interesting stuff done with those in the past century or so.

Answer 17

Hmm. I guess the way I want to think of it as throwing away multiplication and division and trading them for some new operation that's defined to take us between these new definitions of factors. So we still have the same numbers, it's just the form of the peiodicity between them is thrown away for something new.

I guess I'll look into arithmetic progressions, that sounds like it could help me in seeing if what I'm talking about is different or like everything else, already discovered by someone and I'm just enjoying rediscovering it hahaha.

Answer 18

So one result I came up with that seems unique and probably too arbitrary to really mean anything to anyone, is to change the definitions of factors so that doing the stone shifting thing I spoke about we change 0,3,6,9,12 to 1,4,7,10,13, then we change 0,6,12,18,24 to 1,7,13,19,25, and continue again with 0,9,18. down by 1 then we will have slightly different factors.

Specifically in the new system, where \(\ \large \tau \prime \) represents the new divisor counting function and \(\ \large 3 \nmid n\) we can express it in terms of the old system. \[\Large \tau \prime (n3^a)=\tau(n3^a)-a=a*(\tau(n)-1)+\tau(n) \\ \Large \tau \prime (n3^a+1)=\tau(n3^a+1)+a\\ \Large \tau \prime(n3^a+2)=\tau(n3^a+2)\] Notice what we have here is we are taking "factors" out of the\(\ \large n3^a\) terms and putting them into the next term up \(\ \large n3^a+1\) while the \(\ \large n3^a+2\) terms are completely untouched! This means we have a kind of "conservation equation" since the total factors remains unchanged in this instance (this is not something I am really concerned about in general, I don't really have an idea what possible things we can do in generality, it's just purely for fun right now). So the conservation equation is: \[\Large \sum_{k=0}^{3-1}\tau \prime (n3^a +k)=\sum_{k=0}^{3-1}\tau (n3^a +k)\] A single change like this generalizes clearly to: \[\Large \sum_{k=0}^{p-1}\tau \prime (np^a +k)=\sum_{k=0}^{p-1}\tau (np^a +k)\] So now we can stack on multiple shifts of this type since now we have found the equation for the shift of a single prime by only one space, all other shifts can stack like this.