Respuesta :
Answer:
(a) (34+83t,-43-105t) where t is an integer
(b) (272+83t,-344-105t) where t is an integer.
(c) 62
Step-by-step explanation:
a)
We are going to perform Euclidean's Algorithm.
Let's begin with seeing how many times 83 goes int 105.
105=83(1)+22 (eq1)
83=22(3)+17 (eq2)
22=17(1)+5 (eq3)
17=5(3)+2 (eq4)
5=2(2)+1 (eq5)
Now let's go backwards through those equations.
5-2(2)=1 (eq5 rewritten so that the remainder was by itself)
5-2[17-5(3)]=1 (replaced the 2 in ( ) with eq4 solved for the remainder)
5-2(17)+5(6)=1 (distributive property was performed)
-2(17)+5(7)=1 (combined my 5's)
-2(17)+7(5)=1 (multiplication is commutative)
-2(17)+7(22-17)=1 (used eq3)
-2(17)+7(22)-7(17)=1 (distribute property was performed)
-9(17)+7(22)=1 (combined my 17's)
-9(83-22(3))+7(22)=1 (used eq2)
-9(83)+22(27)+7(22)=1 (distributive property was performed)
83(-9)+22(34)=1 (multiplication is commutative and combined my 22's)
83(-9)+34(105-83)=1 (used eq1)
105(34)+83(-43)=1 (after distributive property and reordering)
So we have a point on the line being (x,y)=(34,-43).
We can use the slope to figure out all the other integer pairs from that initial point there.
The slope of ax+by=c is -a/b.
So the slope of 105x+83y=1 is -105/83.
So every time we go down 105 units we go right 83 units
This says we have the following integer pairs on our line:
(34+83t,-43-105t) where t is an integer.
Let's verify:
Plug it in!
105[34+83t]+83[-43-105t]
105(34)+105(83)t+83(-43)-83(105)t
105(34)+83(-43)
1
We are good!
(b)
We got from part (a) that 105(34)+83(-43)=1.
Multiply both sides we get 8 on the right hand side:
105(34*8)+83(-43*8)=8
Simplify:
105(272)+83(-344)=8
So the integer pairs is (272+83t,-344-105t) where t is an integer.
Let's verify:
105[272+83t]+83[-344-105t]
105(272)+105(83)t+83(-344)-83(105)t
105(272)+83(-344)
8
(c)
Let u=83^(-1) mod 105.
Then 83u=1 mod 105.
This implies:
83u-1=105k for some integers k.
Add 1 on both sides:
83u=105k+1
Subtract 105k on both sides:
83u-105k=1
Reorder:
105(-k)+83u=1.
We found all (x,y) integer pairs such that 105x+83y=1.
We go (34+83t,-43-105t) where t is an integer.
So k=-34-83t while u=-43-105t.
Since we want to find an integer t such that u is between 0 and 104, we could solve 0<-43-105t<104.
Add 43 on all sides:
43<-105t<147
Divide all sides by -105:
-43/105>t>-147/105
-147/105<t<-43/105
This says t is approximately between -1.4 and -0.4 . This includes only the integer -1.
When t=-1, we have u=-43-105(-1)=-43+105=62.
