Respuesta :
Answer:
a) 5+13k where k is integer
b) 20+13k where k is integer
c)12+13k where k is integer
Step-by-step explanation:
(a)
[tex]8x \equiv 1 (mod 13) \text{ means } 8x-1=13k[/tex].
8x-1=13k
Subtract 13k on both sides:
8x-13k-1=0
Add 1 on both sides:
8x-13k=1
I'm going to use Euclidean Algorithm.
13=8(1)+5
8=5(1)+3
5=3(1)+2
3=2(1)+1
Now backwards through the equations:
3-2=1
3-(5-3)=1
3-5+3=1
(8-5)-5+(8-5)=1
2(8)-3(5)=1
2(8)-3(13-8)=1
5(8)-3(13)=1
So compare this to:
8x-13k=1
We see that x is 5 while k is 3.
Anyways 5 is a solution or 5+13k is a solution where k is an integer.
b)
[tex]8x \equiv 4 (mod 13)[/tex]
8x-4=13k
Subtract 13k on both sides:
8x-13k-4=0
Add 4 on both sides:
8x-13k=4
We got this from above:
5(8)-3(13)=1
If we multiply both sides by 4 we get:
8(20)-13(12)=4
So x=20 and 20+13k is also a solution where k is an integer.
c)
[tex]99x \equiv 5 (mod 13)[/tex
99x-5=13k
Subtract 13k on both sides:
99x-13k-5=0
Add 5 on both sides:
99x-13k=5
Using Euclidean Algorithm:
99=13(7)+8
13=8(1)+5
Go back through the equations:
13-8=5
13-(99-13(7))=5
8(13)-99=5
99(-1)+8(13)=5
Compare this to 99x-13k=5 and see that x=-1 or -1+13=12 or 12+13k is a solution where k is an integer.
Answer:
a) x = 5 mod 13.
b) x = 7 mod 13.
Step-by-step explanation:
a) 8x = 1 mod 13
x = 2, 16 = 3 mod 13
x = 3, 24 = 11 mod 13
x = 4, 32 = 6 mod 13
x = 5 , 40 = 1 mod 13
8x = 40
x = 5 mod 13.
b) 8x = 4 mod 13
x = 7, 56 = 4 mod 13.
7 = 4 mod 13
x = 7 mod 13.
