hairyberry

Anything from 0-11 is possible (0 is only possible when p = 11, otherwise p wont be prime)

but the general form 11n+a where 0<=a<=10 should satisfy any prime number

You can easily factor into

\(x(9x^3+z^2-12)+21\)

So the remainder is 21

69*29 = 2001, you can easily prime factor

Therefore, a = 6, b = 2, or the other way around works

\(x^3+y^3+(x+y)(x^2-xy+y^2)=(x+y)(x^2+y^2-\frac{(x+y)^2-x^2-y^2}{2}) \)

\(\text{set} \,x^2+y^2=a\)

\(5=(2)(a-\frac{4-a}{2})\)

\(a=x^2+y^2=3 \)

To find the total sequences just find how many ways there are to order 1, 2, 3, 4, 5, 6, 6.

This is 7!/2! = 2520

Pick any two points, I chose (3, 0) and (-1, -1).

Calculate slope:

\(\frac{-1 - 0}{-1 - 3}=\frac{1}{4}\)

From slope-intercept form, you know the equation is

y = (1/4)x + b

Plug in any point to find b. I chose (3, 0)

0 = (1/4)*3 + b

b = -(3/4)

Therefore the equation is 

y = (1/4)x - (3/4)

To get into the desired form, subtract (1/4)x from both sides then multiply by lcm

-(1/4)x + y = -(3/4)

x - 4y = 3, is the equation in standard form.

Correct me if I'm wrong, but SSA congruence is not a thing, meaning the triangle is nonunique

Therefore there are multiple possible values for the area.

To find all possible combinations without ordering, we just set the number of 3s to be 1, 2, 3, or  4 and set the rest to 1s.

We can split into these following cases:

#1: 4 3s and 0 1s

#2: 3 3s and 3 1s

#3: 2 3s and 6 1s

#4: 1 3 and 9 1s

#5: 0 3s and 12 1s

Now we find the number of ways to order each case:

Case 1: 

1 way to order this:

Case 2:

\(\frac{6!}{3!3!}=20\) ways to order this

Case 3:

\(\frac{8!}{6!2!}=28\) ways to order this

Case 4:

\(\frac{10!}{1!9!}=10\) ways to order this

Case 5:

1 way to order this.

Adding, 1+20+28+10+1 = 60 total numbers.

The only possible degree of p+q is 11.

No matter how you cancel out, you can't cancel out the coefficient of the ^11 term because the highest degree of q is 7. Because degree is defined by the highest exponent the degree is always 11.