https://github.com/MyJestor/An-Alternative-to-Chinese-Remainder-Theorem-
# An Iterative Alternative to Classical Chinese Remainder Theorem

## Background
The classical CRT requires pairwise coprime moduli. I've developed
an iterative algorithm that generalizes this.

## Main Result
The algorithm solves any system of linear congruences by:
1. Combining congruences pair-wise iteratively
2. Using GCD and LCM properties
3. Requiring only that solutions exist (no coprimality condition)

## Claim
This approach is:
- More general (handles non-coprime cases)
- More intuitive (step-by-step logic)
- Yields optimal period (LCM vs product)

## Questions for the Community
1. Is this algorithm novel in the literature?
2. What is its computational complexity vs classical CRT?
3. Does it have known applications?

## References
[Link to GitHub implementation]
https://github.com/MyJestor/An-Alternative-to-Chinese-Remainder-Theorem-

https://math.stackexchange.com/questions/733754/visually-stunning-math-concepts-which-are-easy-to-explain

By Lee Sallows: If the letters BJFGSDNRMLATPHOCIYVEU are assigned to the integers -10 to 10, then:

J+A+N+U+A+R+Y = -9+0-4+10+0-3+7 = 1
F+E+B+R+U+A+R+Y = -8+9-10-3+10+0-3+7 = 2
M+A+R+C+H = -2+0-3+5+3 = 3
A+P+R+I+L = 0+2-3+6-1 = 4
M+A+Y = -2+0+7 = 5
J+U+N+E = -9+10-4+9 = 6
J+U+L+Y = -9+10-1+7 = 7
A+U+G+U+S+T = 0+10-7+10-6+1 = 8
S+E+P+T+E+M+B+E+R = -6+9+2+1+9-2-10+9-3 = 9
O+C+T+O+B+ER = 4+5+1+4-10+9-3 = 10
N+O+V+E+M+B+E+R = -4+4+8+9-2-10+9-3 = 11
D+E+C+E+M+B+E+R = -5+9+5+9-2-10+9-3 = 12
Similarly, if -7 to 7 are assigned SROEMUNFIDYHTAW, then SUNDAY to SATURDAY take on ordinal values. See Alignment.

(David Morice, "Kickshaws," Word Ways 24:2 [May 1991], 105-116.

Gerrymandering is more sophisticated than it once was. Mathematicians are pioneering new ways to determine electoral fairness — and their work is ending up in court. https://www.quantamagazine.org/how-math-has-changed-the-shape-of-gerrymandering-20230601/

\(Given:Any\ two\ relatively\ prime\ positive\ integers\ a,and\ b,\ such\ that\)
\(\ a+b=c \ and\ gcd(a,b)=1\)
\(To\ Prove:\ (radabc)^q<(radabc)^{\frac{5}{3}}\)
\(Proof:\ \because a+b=c\)
\( \implies\ radab\neq radc\implies\ Either\ ,radc<radab\ ,OR\ \ radab<radc \)
\(so\ \therefore\ there\ exists\ a\ positive\ integer\ k\ge1\ such\ that\ ,(radc)^k<rad(ab)<(radc)^{(k+1)}\)
\(\{Here\ we\ will\ not\ consider\ the\ case,\ rad(ab)<rad(c)\}\)
\(……Without\ loss\ of\ generality\ for\ some\ integer\ k\ such\ that\ 1\le k\)
\(\because (radc)^k<rad(ab)<(radc)^{(k+1)}\implies (radc)^{(k+1)}<rad(abc)<(radc)^{(k+2)}\)
\((radc)^{(k+1)}\le rad(abc)^{(1+\epsilon)}=(radc)^{(k+2)}\implies n(1+\epsilon) =q(k+2)\)
\( \because (radc)^k<rad(ab)\ after\ multiplying\ by\ (rad(ab))^k\ we\ get\ \)
\(\therefore (radabc)^k<rad(ab)^{k+1}<rad(abc)^{k+1}\)
\(\therefore (radab)<rad(abc)<rad(abc)^\frac{k+1}{k}\)
\(\implies (rad(ab))^{(\frac{k+1}{k})}\le rad(ab)^2\)
\(rad(ab)<(radabc)<(rad(abc))^{(1+\epsilon')}\approx(rad(ab))^{(\frac{k+1}{k})}\le (rad(ab))^2\implies m(1+\epsilon')\le2q\)
\(\therefore \)
\((radabc)^{(1+\epsilon')}=(rad(ab))^{(\frac{k+1}{k})}\le rad(ab)^2\implies mk(1+\epsilon')=q (k+1)\)
\( q(1+k\epsilon'+\epsilon)=m(1-(k\epsilon'+\epsilon)^2)\implies(rad(ab))^{(1+k\epsilon'+\epsilon)}=rad(abc))^ {(1-(k\epsilon'+\epsilon)^2)}\)
\(m\left(3+(k-1)k\epsilon'+(k-2)-(k\epsilon'+\epsilon)^2\right) = q(4+k\epsilon'+\epsilon)\)
\(m\left(2-(3k\epsilon'+\epsilon+(\epsilon)^2-(k\epsilon')^2)\right) = q(2+\epsilon-k\epsilon')\)
\(m\left(5+(k-4)k\epsilon'-\left\{3-k+2(k\epsilon'+\epsilon)\right\}\epsilon\right) = q(6+2\epsilon)\)
\(m\left(3-(4k\epsilon'+2\epsilon+(\epsilon)^2-(k\epsilon')^2)\right) = q(3+\epsilon-k\epsilon')\)

\( It’s\ better\ to\ be\ approximate\ than\ to\ be\ exact\ ,
nature\ prefers\ to\ be\ approximate\ \)
\(mathematics\ spans\ more\ by\ being\ approximate\ ,I\ prefer\ \le or\ge over\ equality=.\)
\(\ What\ a\ paradox\ ,“Mathematics\ is\ an\ exact\ science” \)

\( “\Truth\ is\ not\ afraid\ of\ being\ challenged\ where\ as\ \falsehood\ does\ not\ like\ to\ be\\\ questioned \ ”.\)