# How do you solve for X?

by krispiekr3am
Tags: solve
 P: 23 5x = 1 (mod 13) 13x = 2 (mod 23) 37x = 5 (mod 13)
HW Helper
P: 3,224
 Quote by krispiekr3am 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.
P: 1,253
 Quote by krispiekr3am 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).

 P: 23 How do you solve for X? 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?
 P: 23 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)
 Sci Advisor HW Helper P: 9,397 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.
Nope, $5\not\equiv8\pmod{13}$.