| New Reply |
Solving Equations With Modulos |
Share Thread | Thread Tools |
| Dec31-12, 07:24 AM | #1 |
|
|
Solving Equations With Modulos
Are there general methods for solving equations of the form
a+bx = mod(c+dx, m), where, in the notation I have made up here, mod is the modulo function which resets the argument to zero when it reaches m. I hope it's clear what I mean here. |
| Dec31-12, 10:37 AM | #2 |
|
Mentor
|
Every solution of that equation will also satisfy 0 = mod(c-a+(d-b)x, m), or, in a more conventional notation, f=g x mod m where f=a-c and g=d-b. This is a simple modular equation, and general methods to find all solutions exist.
All solutions which satisfy 0<=a+bx<m are solutions to your initial equation. |
| Dec31-12, 10:59 AM | #3 |
|
|
I dont quite follow. When you switch from my made up notation to the real notation (sorry about that), it looks like a completely new equation. Unless you moved terms to the other side, which I didn't think was allowed. I could get a better idea of the solution by considering:
mod(x,n) = x- n*floor(x/n) so that for my equations: a+bx = c+dx - m*floor((c+dx)/m) But what are the general methods for finding the solutions here? I should be clear here that I'm considering x as a real number and not necessarily and integer here. |
| Dec31-12, 11:33 AM | #4 |
|
Mentor
|
Solving Equations With Modulos0 = mod(c-a+(d-b)x, m) switch notation 0 = c-a+(d-b)x mod m add a-c (for mathematical details: you can do this as addition is a group in Z/nZ, and it works for non-integer values as well) a-c = (d-b)x mod m Note that "mod m" refers to the whole equation in mathematics. It is used differently in programming languages. |
| Dec31-12, 12:11 PM | #5 |
|
|
add(x,y) = x+y-m*floor((x+y)/m), which you didn't follow. Is this not the case? |
| Dec31-12, 12:14 PM | #6 |
|
Mentor
|
|
| Dec31-12, 12:30 PM | #7 |
|
|
Ok I'm starting to see where you're coming from now. However, I still dont know what these general methods you talk about are. Could you point me towards an explanation of the methods involved?
Edit: So x = f/g + k*m/g is the general solution for some integer k? Plus the constraint that 0<=a+bx<m. |
| Dec31-12, 01:26 PM | #8 |
|
Mentor
|
Looks correct.
|
| New Reply |
| Thread Tools | |
Similar Threads for: Solving Equations With Modulos
|
||||
| Thread | Forum | Replies | ||
| Solving for variables using 3 different equations (simultaneous equations) | Precalculus Mathematics Homework | 1 | ||
| Modulos one | Calculus | 0 | ||
| modulos raised to phi | Introductory Physics Homework | 1 | ||
| palindromes and modulos | Introductory Physics Homework | 6 | ||
| modulos | Linear & Abstract Algebra | 16 | ||