*** Discrete Math ***
posted by Jane on .
I would like to know if my answers are correct. I also need help solving some exercises the right way.
(a) Explain why [13] is invertible in
Z50.
Ans: it's invertible because
there's some [t]E Z50 such that
[13][t]=1
(b) Use the usual algorithm to find
[13]^(1)and give your anwer
mod50.
Ans: [13]^(1) = ? mod50
I wasn't sure how to use the usual
algorithm to solve this so I
used the matrix form, hopefully I
did it right. But I know that
I won't get full credit for it.
So, I would like to know the
right way of doing it.
__ __ __ __
 1 0  50  ~  1 3  11  ~
 0 1  13   0 1  13 
   
__ __ ___ ___
 1 3  11  ~ 6 23  1 
 1 4  2   1 4  2 
   
1=6[50] 23[13] ==> [1]= 23[13]
==> [13]^(1) = [23]
(Can i just leave it like that or do I
have to continue?)
long division :
2 So, 23+ remainder =
____  23 + 4= 27
23  50 Therefore.
46 
  [t]=[27] mod 50
4  ===> [13][27]=1
(c) Use your answer in part (b) to solve the congruence 13x= 20mod 50.
and give your answer in mod50.
Ans: I wasn't sure how to solve it, so I did it the long way, one by one.
a =bmodn
13X=20mod 50 ==> 50  13x 20
When x=40 ==>
50  13(40)20 => it has
remainder 20 when
I do long division.
But I would like to know how to do it the right way.

The matrix method is the most efficcient way.
23 is the same as 27 mod 50.
13x= 20 (everything = mod 50)>
x = 13^(1)*20 = 27*20 = (25 + 2)*20 = 40