Modular Algorithm for Solving Equations with Unknowns Modulo 100

[booney1983]

how can i solve this problem?

[ x1= a (mod 100) , a= 20 (mod 37) ]


[ x2= b (mod 100) , b= 15 (mod 37) ]


[ x3= c (mod 100) , c= 18 (mod 37) ]

must be ; x2= a.k + y (mod100)

and

x3= b.k + y (mod100)

i need find b and c.. thank you best regards..

 

[HallsofIvy]

Looks to me like "Chinese remainder theorem". In any case, it is certainly not "Calculus and Analysis. I am moving this to the number theory forum.

 

[CRGreathouse]

must be ; x2= a.k + y (mod100)

and

x3= b.k + y (mod100)

Are these the same y and k, the same y, but some arbitrary k, or what?

 

[booney1983]

yes same y and k

 

[robert Ihnot]

booney1983: yes same y and k

That's not especially helpful since no effort to define them was given. As for y I see that you use it in the two final equations. So I am wondering if it could not simply be dropped and is meaningless?

What is b.k? I have guessed it means b times k. The symbol "*" is sometimes used for multiplication. Maybe some countries think differently.

 

[CRGreathouse]

Here's how I'm interpreting the question:

Find b and c such that the following equations are satisfied in terms of the other variables.
1. x_1\equiv a\pmod{100}
2. a\equiv20\pmod{37}
3. x_2\equiv b\pmod{100}
4. b\equiv15\pmod{37}
5. x_3\equiv c\pmod{100}
6. c\equiv18\pmod{37}
7. x_2\equiv a\cdot k+y\pmod{100}
8. x_3\equiv b\cdot k+y\pmod{100}It's trivial to determine the values of b and c mod 100: b\equiv ak+y, c\equiv bk+y. The CRT could then be used to determine b and c mod 3700.

 

[robert Ihnot]

One simple solution is X1=X2=X3=18 Mod 100. a=2018, b=718, c=18. k=1, y=0. which renders the last two equations redundant, while 5 and 6 are taken care of since X3=c.