(a)Find all integer solutions to the equation 105x + 83y = 1.
(b) Find all integer solutions to the equation 105x + 83y = 8.
(c) Find 83-1 mod 105. (Note: Answer must be in between 0 and 104, inclusive.)

(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.

the answer is f(x)=2x-1

step-by-step explanation:

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.