Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2009-01-28 03:08:40

integer
Member
Registered: 2008-02-21
Posts: 79

Chinese Remainder Theorem with multiple unknowns: HOW DO I SOLVE IT?

Need some (a lot of) help with Chinese Remainder Theorem

How can the Chinese Remainder Theorem be used to solve this:

Sorry for the poor form above.
Obviously, LaTex not fully understood, but I have downloaded the LaTeX tutorial.
Background
(not necessary, but you might be interested in how the above was found.

A middle school student (7th grade)
wanted to know why the Perrin, Lucas and
Fibonacci sequences do not have negative
numbers.  Since they are defined as such
they do not. The negative values only occur
when moving in the negative direction.

However, other sequences with different
initial values can and do have negative
values on the positive side of zero.


The formula for the Perrin Sequence:

P(n+0) + P(n+1) = P(n+3)

with the intial values of:  P(0)= 3, P(1)= 0, P(2)= 2


The Perrin Sequence also satisfies the equation:

P(n+0) + P(n+1) + P(n+2) + P(n+3) + P(n+4) = P(n+8)  {1}
   A   +    B   +    C   +    D   +    E   =    I


If we assign random values to five consecutive terms
in the squence, and using only equation {1} assign
values to all other terms in a quasi Perrin sequence,
we should have enough data to compute the intermediate terms:

P(n+5),   P(n+6),   P(n+7)
   F         G         H 


Our sequence defined:
P(-3) P(-2) P(-1) P(0) P(1) P(2) P(3) P(4) P(5) P(6) P(7) P(8) P(9)
   X     Y     Z    A    B    C    D    E    F    G    H    I    J

X,Y,Z are unknown
A,B,C,D,E are assigned random numeric values

it follows that
F = X+Y+Z+A+B
G = Y+Z+A+B+C
H = Z+A+B+C+D
I = A+B+C+D+E
J = B+C+D+E+F


By substituting, combining, reducing we derived these equations:

X +  2Y +  3Z == v1 mod 11
2X +  3Y +  4Z == v2 mod 13
6X + 10Y + 13Z == v3 mod 17
9X + 16Y + 22Z == v4 mod 19

This is where we are -- enveloped in a fog.

The Chinese Remainder Theorem is easy enough to use on single
unknown values:
x == a mod b
x == c mod d

But with three unknowns we are lost.

An internet search only shows methods to solve for a single
unknown value.

Any one have an idea as to how we need to proceed so that
we may determine the smallest positive value for X?

Y and Z would then be defined in relation to X.

Any help, hints, pointers, references, or thoughts would
be appreciated.

Thanks.

Last edited by integer (2009-01-28 03:11:09)

Offline

Board footer

Powered by FluxBB