Rom

I just answered this yesterday

Apply the rational root theorem

\(\text{let }p(x) = a_0 + a_1 x + \dots + a_n x^n\)

\(\text{then if }x = \dfrac{p}{q} \text{ written in lowest terms is a root of p(x), }\\ \text{it must be that p divides }a_0 \text{and that q divides }a_n\)

\(\text{here we have }a_0 = -5,~a_3=1\)

\(\text{the only possible p's are }\pm 1,~\pm 5\\ \text{and the only possible q's are }\pm 1\)

\(\text{thus the only possible rational roots are }\pm 1,~\pm 5 \\ \text{and we just have to try them all to see if any are actually roots.}\)

\(p(-1)=14, ~p(1) = 22,~p(5)=710,~p(-5)=430 \\ \text{i.e. none of these are in fact roots and thus p(x) is irreducible over the rational numbers}\)

There are no solutions over the real numbers.

Over the complex numbers the two solutions are

x = +/- 6i

where i is the square root of -1

\(x-(11+\sqrt[3]{13}) = 0 \\ x - 11 = \sqrt[3]{13} \\ \text{now cube both sides} \\ x^3-33 x^2+363 x-1331 = 13 \\ x^3-33 x^2+363 x-1318 = 0\)

I had mathematica draw the lines for 5, 6, and 7, points.

Counted intersections by eye for each.  Worked out the pattern.

Suppose person 1 is telling the truth.

Then person #2 says that at least 2/3 of person 1 has told a lie which I will take to mean he says person 1 lied.

But we know that person 1 told the truth so this means that person #2 lied.

But person #1 truthfully said that none of them lied so we have a contradiction.

Thus person 1 must have lied.

If you work through the responses and the count of liars you see that 2/3 of the people are liars i.e. knaves.

\(\dfrac 2 3 \cdot 2019 = 673\)

By looking at images of symmetrically placed points and the lines connecting them I see that the number of intersections in the interior of the circle given n points is given by

\(ni = 2^{n-2}-3,~\forall n \geq 4 \\ \text{and for }n=9\text{ this results in 125 intersections.}\)

have fun verifying this.

There's no magic to this.

There are  7 pairs that multiply to a multiple of 196.

There are 21 unordered pairs that can be made from the set.

P[product is a multiple of 196]=7/21 = 1/3

as we know a triangle has 180 degrees we know that angles B and C must sum to 90.

so we have

\(\cos(C) = \cos(90^\circ - B) = \\ \cos(90^\circ)\cos(B) + \sin(90^\circ)\sin(B) = \\ \sin(B) = \dfrac 4 7\)

\(10^n \text{ has n zeroes}\\ 10^{10^{23}}\text{ has }10^{23} \text{ zeroes}\)