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Let n be a natural number with exactly 2 positive prime divisors.

If n^2 has 27 divisors, how many does n have?

\(\text{ Let $ n = p^n\times q^m $ and $p$ and $q$ are prime }\)

The number of devisors of \(n\) is \((n+1)(m+1)\)

\(\text{ Let $ n^2 = p^{2n} \times q^{2m} $ } \)

The number of devisors of \(n^2\) is \((2n+1)(2m+1)\)

\(\begin{array}{|rcll|} \hline (2n+1)(2m+1) &=& 27 \quad & | \quad 27=3^3 \\\\ \underbrace{(2n+1)}_{=3}\underbrace{(2m+1)}_{=3^2} &=& 3^3\\\\ 2n+1 &=& 3 \\ 2n &=& 2 \\ \mathbf{n} &\mathbf{=} & \mathbf{ 1 } \\\\ 2m+1 &=& 3^2 \\ 2m+1 &=& 9 \\ 2m &=& 8 \\ \mathbf{m} &\mathbf{=} & \mathbf{4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && (n+1)(m+1) \quad & | \quad n=1 \text{ and } m = 4 \\ &=& (1+1)(4+1) \\ &=& 2\cdot 5 \\ &\mathbf{=} & \mathbf{10} \\ \hline \end{array}\)

n has 10 divisors.

Example:

\(\text{Let $n = 162 = 2\times 3^4$ } \\ \text{Divisors $162: 1,2,3,6,9,18,27,54,81,162\ $ ($10$ divisors)}\)

\(\text{Let $n^2 = 162^2 = 26244 = 2^2\times 3^8 $} \\ \text{Divisors 162^2:$\\1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, 243, 324, 486,\\ 729, 972, 1458, 2187, 2916, 4374, 6561, 8748, 13122, 26244\ $ ($27$ divisors)}\)

A right cylindrical tank with circular bases is being filled with water at a rate of 20π cubic meters per hour.

As the tank is filled, the water level rises four meters per hour. What is the radius of the tank, in meters?

Express your answer in simplest radical form.

\(\begin{array}{|rcll|} \hline V &=& \pi r^2\cdot h \quad & | \quad : \text{hour} \\\\ \dfrac{V}{\text{hour}} &=& \pi r^2\cdot \dfrac{\text{height}} {\text{hour}} \\\\ 20\pi\dfrac{m^3}{\text{hour}} &=& \pi r^2\cdot 4\dfrac{m}{\text{hour}} \\\\ 20\pi\ m^2 &=& \pi r^2\cdot 4 \quad & | \quad :4\pi \\\\ 5\ m^2 &=& r^2 \\\\ r &=& \sqrt{5}\ m \\ \hline \end{array}\)

The radius of the tank is \(\sqrt{5}\) meters

Advanced Quadratic

Find the product of the \(y\)-coordinates of all the distinct solutions \((x,y)\) for the two equations
\(y=x^2-8\) and \(y^2=-5x+44\).

\(\begin{array}{|lrcll|} \hline (1) & y &=& x^2-8 \\\\ (2) & y^2 &=& -5x+44 \quad | \quad y^2 = (x^2-8)^2 \\ & (x^2-8)^2 &=& -5x+44 \\ & x^4-16x^2+84 &=& -5x+44 \\ & x^4-16x^2+5x+20 &=& 0 \\ & x^2(x^2-16)+5(x+4) &=& 0 \quad | \quad x^2-16 = (x-4)(x+4) \\ & x^2(x-4)(x+4)+5(x+4) &=& 0 \\ & (x+4)\left(x^2(x-4)+5\right) &=& 0 \\ \\ 1. & x_1+4 &=& 0 \\ & \mathbf{x_1} &\mathbf{=}& \mathbf{-4} \\ & y_1 &=& x_1^2-8 \\ & y_1 &=& (-4)^2-8 \\ & \mathbf{y_1} &\mathbf{=}& \mathbf{8} \\ & \text{solution $(-4,8)$ } \\\\ 2. & x^2(x-4)+5 &=& 0 \\ & x^3-4x^2 + 5 &=& 0 \\ \text{try } x=-1 & (-1)^3-4(-1)^2 + 5 &\overset{?}{=}& 0 \\ & -1-4 + 5 & = & 0\ \checkmark \\ & \mathbf{x_2} &\mathbf{=}& \mathbf{-1} \\ & y_2 &=& x_2^2-8 \\ & y_2 &=& (-1)^2-8 \\ & \mathbf{y_2} &\mathbf{=}& \mathbf{-7} \\ & \text{solution $(-1,-7)$ } \\\\ \hline \end{array}\)

Polynomial long division

\(\begin{array}{|rcll|} \hline x^2-5x+5 &=& 0 \\ x &=& \dfrac{5\pm\sqrt{25-4\cdot 5 }}{2}\\ x &=& \dfrac{5\pm\sqrt{5}}{2} \\\\ \mathbf{x_3} &\mathbf{=}& \mathbf{\dfrac{5+\sqrt{5}}{2}=3.618 } \\ y_3 &=& x_3^2-8 \\ y_3 &=& \left(\dfrac{5+\sqrt{5}}{2} \right)^2-8 \\ \mathbf{y_3} &\mathbf{=}& \mathbf{\dfrac{5\sqrt{5}-1}{2} = 5.09 } \\ \text{solution $(\dfrac{5+\sqrt{5}}{2},\dfrac{5\sqrt{5}-1}{2}) \\ =(3.618,5.09)$ } \\\\ \mathbf{x_4} &\mathbf{=}& \mathbf{\dfrac{5-\sqrt{5}}{2}=1.382 } \\ y_4 &=& x_4^2-8 \\ y_4 &=& \left(\dfrac{5-\sqrt{5}}{2} \right)^2-8 \\ \mathbf{y_4} &\mathbf{=}& \mathbf{-\left(\dfrac{5\sqrt{5}+1}{2}\right) = -6.09 } \\ \text{solution $(\dfrac{5-\sqrt{5}}{2},-\left(\dfrac{5\sqrt{5}+1}{2}\right)) \\ =(1.382,-6.09)$ } \\ \hline \end{array}\)

Find the product of the y-coordinates of all the distinct solutions (x,y):

\(\begin{array}{|rcll|} \hline && y_1y_2y_3y_4 \\ &=& 8\times(-7)\times \left(\dfrac{5\sqrt{5}-1}{2}\right) \times \left(-\left(\dfrac{5\sqrt{5}+1}{2}\right)\right) \\ &=& 8\times(-7)\times (-31) \\ &\mathbf{=}& \mathbf{1736} \\ \hline \end{array}\)

Number theory!

Find an integer x such that  \(\mathbf{0 \leq x < 527}\) and  \(\mathbf{x^{37} \equiv 3 \pmod{527}}\).

0 \leq x < 527\ \text{and} \ x^{37} \equiv 3 \pmod{527}.

answer see:

A cube has side 6 lengths. Its vertices are alternately colored black and purple, as shown below.
What is the volume of the tetrahedron whose corners are the purple vertices of the cube?
(A tetrahedron is a pyramid with a triangular base.)

A cube has side 6 lengths. \(\mathbf{\text{Let $s = 6$}}\)

The cube has volume \(\mathbf{V_{\text{cube}} = s^3}\)

The 4 right triangular pyramids that must be carved off the cube to produce the regular tetrahedron each have volume

\(\mathbf{V_{\text{right triangular pyramid}} = \frac13\ \times \frac{s^2}{2} \times s } \)

The volume of the regular tetrahedron is  \(\mathbf{ V_{\text{cube} }- 4\times V_{\text{right triangular pyramid}} }\)

\(\begin{array}{|rcll|} \hline && V_{\text{tetrahedron}} \\\\ &=& V_{\text{cube} }- 4\times V_{\text{right triangular pyramid}} \\\\ &=& s^3 - 4\times \dfrac13\ \times \dfrac{s^2}{2} \times s \\\\ &=& s^3 - \dfrac23 s^3 \\\\ &=& \dfrac13 s^3 \quad & | \quad s = 6 \\\\ &=& \dfrac{6^3}{3} \\\\ &=& \dfrac{216}{3} \\\\ &=& 72 \\ \hline \end{array}\)

The volume of the tetrahedron is 72

In how many different ways can
\(\frac{2}{15} \)
be represented as
\(\frac{1}{a} + \frac{1}{b}\),
if and are positive integers with
\(a \ge b\) ?

Because \(n = 15\) is odd:

we calculate \(n^2 = 225\)

And all divisors of \(n^2 = 225 \) are: \(1,3,5,9,15,25,45,75,225\)

\(\text{Let $n^2 = p\times q=225 $ } \)

\(\begin{array}{|r|r|r|r|r|r|r|r|} \hline p & q & s = \frac{p+q}{2} & t = \frac{p-q}{2} & r = \frac{t}{2} & k = \frac{n+\sqrt{n^2+t^2} }{2} & \mathbf{a} = k+r & \mathbf{b} = k-r \\ \hline 225 & 1 & 113 & 112 & 56 & 64 & 120 & 8 \\ 75 & 3 & 39 & 36 & 18 & 27 & 45 & 9 \\ 45 & 5 & 25 & 20 & 10 & 20 & 30 & 10 \\ 25 & 9 & 17 & 8 & 4 & 16 & 20 & 12 \\ 15 & 15 & 15 & 0 & 0 & 15 & 15 & 15 \\ \hline \end{array} \)

\(\begin{array}{|r|r|c|} \hline & \frac{1}{a} + \frac{1}{b} & a \ge b\ \\ \hline 1. & \frac{1}{120} + \frac{1}{8} & \checkmark \\ 2. & \frac{1}{45} + \frac{1}{9} & \checkmark \\ 3. & \frac{1}{30} + \frac{1}{10} & \checkmark \\ 4. & \frac{1}{20} + \frac{1}{12} & \checkmark \\ 5. & \frac{1}{15} + \frac{1}{15} & \checkmark \\ \hline \end{array} \)

Circle  is the incircle of triangle ABC and is also the circumcircle of triangle XYZ. The point X is on line BC, point Y is on overline AB, and the point Z is on line AC. If angle A=40 degrees, angle B=60 degrees, and angle C=80 degrees, what is the measure of angle AYX?

\(\text{Let the center $M$ of the incircle, $\\$called the incenter,$\\$can be found as the intersection of the three internal angle bisectors }\)

\(\begin{array}{|rcll|} \hline \angle XBM &=& \dfrac{60^{\circ}}{2} \\ &=& 30^{\circ} \\\\ \angle XMB &=& 90^{\circ}- \angle XBM \\ &=& 90^{\circ}-30^{\circ} \\ &=& 60^{\circ} \\\\ \angle XMY &=& 2\times \angle XMB \\ &=& 2\times 60^{\circ} \\ &=& 120^{\circ} \\\\ \angle MYX &=& \dfrac{180^{\circ}-\angle XMY}{2} \\ &=& \dfrac{180^{\circ}-120^{\circ}}{2} \\ &=& 30^{\circ} \\\\ \angle AYX &=& 90^{\circ} + \angle MYX \\ &=& 90^{\circ} + 30^{\circ} \\ & \mathbf{=}& \mathbf{120^{\circ}} \\ \hline \end{array}\)

The points $B(1, 1)$, $I(2, 4)$ and $G(5, 1)$ are plotted in the standard rectangular coordinate system to form triangle $BIG$.

Triangle $BIG$ is translated five units to the left and two units upward to triangle $B'I'G'$,

in such a way that $B'$ is the image of $B$, $I'$ is the image of $I$, and $G'$ is the image of $G$.

What is the midpoint of segment $B'G'$?

Express your answer as an ordered pair.

\(\begin{array}{|l|lrr|} \hline & \text{translation} \\ & x' =x-5\\ & y' = y +2 & \\ \hline \text{B} =(1,1) & \text{B'} =(1-5,1+2) = (-4,3) \\ \text{I} = (2,4) & \text{I'} = (2-5,4+2) = (-3,6) \\ \text{G} = (5,1) & \text{G'} = (5-5,1+2) = (0,3) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{midpoint of } B'G' &=& \dfrac{B'+G'}{2} \\\\ &=& \dfrac{\dbinom{-4}{3}+\dbinom{0}{3}}{2} \\\\ &=& \dfrac{\dbinom{-4+0}{3+3}}{2} \\\\ &=& \dfrac{\dbinom{-4}{6}}{2} \\\\ &\mathbf{=}& \mathbf{ \dbinom{-2}{3} } \\ \hline \end{array}\)

2)

How many different non-congruent isosceles triangles can be formed by connecting three of the dots in a 4x4 square array of dots like the one shown below?

Picture: https://latex.artofproblemsolving.com/c/b/0/cb06885af853019aa03cd8d3d48f31d63144e6c3.png

10 sin^2(x) + 10 sin (x) cos (x) - cos^2 (x) = 2.

Solve for all values of  x between 0 and 360 degrees

\(\begin{array}{|rcll|} \hline 10 \sin^2(x) + 10 \sin (x) \cos (x) - \cos^2 (x) &=& 2 \\ && \small{\boxed{\sin(x) = \tan(x)\cos(x)}} \\ 10 [\tan(x)\cos(x)]^2 + 10 \tan(x)\cos(x)\cos (x) - \cos^2 (x) &=& 2 \\ 10 \tan^2(x)\cos^2(x) + 10 \tan(x)\cos^2(x) - \cos^2 (x) &=& 2 \\ \cos^2(x) \Big( 10 \tan^2(x) + 10 \tan(x) - 1 \Big) &=& 2 \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 \cdot \dfrac{1}{\cos^2(x) } \\ && \small{\boxed{ \dfrac{1}{\cos^2(x) } = 1+\tan^2(x)} } \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 \cdot( 1+\tan^2(x)) \\ 10 \tan^2(x) + 10 \tan(x) - 1 &=& 2 +2\tan^2(x) \\ 8 \tan^2(x) + 10 \tan(x) - 3 &=& 0 \\ && \large{\boxed{\tan(x) = z}} \\ 8 z^2 + 10z - 3 &=& 0 \\\\ z&=& \frac{-10\pm \sqrt{100-4\cdot 8\cdot(-3)} } {2\cdot 8} \\ z&=& \frac{-10\pm \sqrt{196} } {16} \\ z&=& \frac{-10\pm 14 } {16} \\\\ z_1 &=& \dfrac{-10 + 14 } {16} \\ \mathbf{z_1} &\mathbf{=}& \mathbf{\dfrac{1} {4}} \\\\ z_2 &=& \dfrac{-10 - 14 } {16} \\ \mathbf{z_2} &\mathbf{=}& -\mathbf{\dfrac{3} {2}} \\ \hline \end{array}\)

solutions:

\(\begin{array}{|rcll|} \hline \tan(x) &=& z_1 \\ \tan(x) &=& \frac{1}{4} \\ \mathbf{x} &\mathbf{=}& \mathbf{\arctan(\frac{1}{4}) + n\cdot 180^{\circ} } \quad & | \quad n \in Z \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \tan(x) &=& z_2 \\ \tan(x) &=& -\frac{3}{2} \\ x &=& \arctan(-\frac{3}{2}) + n\cdot 180^{\circ} \\ \mathbf{x} &\mathbf{=}& \mathbf{-\arctan(\frac{3}{2}) + n\cdot 180^{\circ} } \quad & | \quad n \in Z \\ \hline \end{array}\)

 x between 0 and \( \mathbf{360^{\circ}}\)

\(\begin{array}{|rcll|} \hline x_1 &=& \arctan(\frac{1}{4}) \\ \mathbf{x_1} &\mathbf{=}& \mathbf{14.0362434679^{\circ}} \\\\ x_2 &=& -\arctan(\frac{3}{2}) + 180^{\circ} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{123.690067526^{\circ}} \\\\ x_3 &=& \arctan(\frac{1}{4})+ 180^{\circ} \\ \mathbf{x_3} &\mathbf{=}& \mathbf{194.036243468^{\circ}} \\\\ x_4 &=& -\arctan(\frac{3}{2}) + 2\cdot 180^{\circ} \\ \mathbf{x_4} &\mathbf{=}& \mathbf{303.690067526^{\circ}} \\ \hline \end{array}\)