Hi, Im have trouble figuring out how to solve simultaneous equations of the form:
4a + b (mod 26) = 20
19a + b (mod 26) = 25
I know how to do simultaneous equations normally but with the modulus I keep messing up somewhere.
I multiplied everything in the top equation by + 1 and everything in the bottom equation by -1 to get:
4a + b (mod 26) = 20
-19a - b (mod 26) = -25
then add them to get:
-15a = -5
It seems obvious to me that Im doing something wrong but I've had trouble finding out how to do this properly. Thank you.
See post number #3 here to see how it is done
http://www.mathhelpforum.com/math-he...tml#post646421
If does but if they are corpime then youcan express 1 as a linear combination 15 and 26
now if you mod out by 26 you get
so b must be 15 inverse.
Sorry, I'm having trouble understanding this. I looked at the other post but couldn't figure out where everything was coming from.
I looked at some other tutorials and was equally confused. I know how to do the extended euclidean alogrithm but in your post and
the other tutorials it looks very different from the way I was shown how to do it. Everything else made perfect sense but getting the
inverse has me stumped.
okay so you want to find 15 inverse mod 26
So you start with 26 and divide it by 15 to get
Now that we have the remainder of 1 we want to express it in terms of 26 and 15.
we back substitute to work our way back to the top.
So now we have 1 as a linear combination of 15 and 26
so we can see that 15 inverse is 7