Modular Algorithm for Solving Equations with Unknowns Modulo 100

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booney1983
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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..
 
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yes same y and k
 
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.
 
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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. [tex]x_1\equiv a\pmod{100}[/tex]
2. [tex]a\equiv20\pmod{37}[/tex]
3. [tex]x_2\equiv b\pmod{100}[/tex]
4. [tex]b\equiv15\pmod{37}[/tex]
5. [tex]x_3\equiv c\pmod{100}[/tex]
6. [tex]c\equiv18\pmod{37}[/tex]
7. [tex]x_2\equiv a\cdot k+y\pmod{100}[/tex]
8. [tex]x_3\equiv b\cdot k+y\pmod{100}[/tex]It's trivial to determine the values of b and c mod 100: [itex]b\equiv ak+y[/itex], [itex]c\equiv bk+y[/itex]. The CRT could then be used to determine b and c mod 3700.
 
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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.
 
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