Solving Modular Equations with Multiple Variables

[krispiekr3am]

5x = 1 (mod 13)
13x = 2 (mod 23)
37x = 5 (mod 13)

 

[radou]

5x = 1 (mod 13)
13x = 2 (mod 23)
37x = 5 (mod 13)

Start off with the definition of 'mod': http://mathworld.wolfram.com/Congruence.html" .

 

[0rthodontist]

5x = 1 (mod 13)
13x = 2 (mod 23)
37x = 5 (mod 13)

You could use the chinese remainder theorem if you can put it in the appropriate form. For example if you find some k such that 5k = 1 mod 13 (Euclid's algorithm), then the first equation is x = k (mod 13).

 

[krispiekr3am]

i started with this
5x-1=13y
x=(13y+1)/5
y=1
x then =12/5
is that correct?
-------------------------------
13x-2=23y
x=(23y+2)/13
y=0
x then = 2/13?
-------------------------------

37x-5=13y
x=(13y+5)/37
y=1
x then = 18/37

is that right?

 

[krispiekr3am]

i have created an excel and create a table like this
x 3x module7
1 3 3
2 6 6
3 9 2
4 12 5
5 15 1

What should i do with this graph to do with 13x=2(mod23)

 

[matt grime]

As has been said, use the Chinese remainder theorem (whcih is Euclid's algorithm but dressed up).

Of course if you typed it correctly you're looking for 5x=1 (13) and 37x=5 (13). Are those even compatible? The fact that you've written 37 and not 11 means that either you've mistyped the 13, or you're not happy with modulo arithmetic, and don't see that you can always replace something with something else congruent mod m if it helps.

 

[CRGreathouse]

Are those even compatible?

Nope, 5\not\equiv8\pmod{13}.

 

[matt grime]

Yes, I know. It was rhetorical/supposed to make the OP think about it, not someone for whom the question is easy.

 

[krispiekr3am]

the answer to 5x = 1 (mod 13) is x = 8 (mod 13)
and the answer to 13x = 2 (mod 23) is x = 9 (mod 23)
and the answer to 37x = 5 (mod 13) is x = 4 (mod 13)

 

[matt grime]

So those aren't sumultaneous equations then? It would greatly benefit you (and would stop you driving your teachers mad) if you wrote things unambiguously, and in full sentences.