heureka

What is the smallest positive integer n such that  \(3n \equiv 1356 \pmod{22}\)

\(\begin{array}{|rcll|} \hline 3n &\equiv& 1356 \pmod{22} \\ 3n &\equiv& 1356-61*22 \pmod{22} \\ 3n &\equiv& 14 \pmod{22} \\ n &\equiv& \dfrac{14}{3} \pmod{22} \\ n &\equiv& 14*3^{-1} \pmod{22} \\ && \boxed{ 3^{-1} \pmod{22} = 3^{\phi(22)-1} \pmod{22} \quad | \quad \phi(22)=22*\left( 1-\dfrac{1}{2}\right)*\left( 1-\dfrac{1}{11}\right)=10 \\ =3^{10-1} \pmod{22} \\ = 3^9 \pmod{22} \\ = 19683 \pmod{22} \\ = 19683-894*22 \pmod{22} \\ = 15 \pmod{22} \\\mathbf{3^{-1} \pmod{22}=15 \pmod{22}} } \\ n &\equiv& 14*15 \pmod{22} \\ n &\equiv& 210 \pmod{22} \\ n &\equiv& 210-9*22 \pmod{22} \\ \mathbf{n} &\equiv& \mathbf{12 \pmod{22}} \\ \hline \end{array} \)

The smallest positive integer n is \(\mathbf{12}\)

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The right triangle with sides 3, 4, 5 has an incircle with radius 1. 

There is another triangle with sides 3, 4, x that also has an incircle with radius 1. 

What is x?

The inradius  r of the incircle in a triangle with sides of length  a, b , c is given by

\(r = \dfrac{\sqrt{s(s-a)(s-b)(s-c)}}{s}\), where \(s=\dfrac{a+b+c}{2}\)

\(\begin{array}{|rcll|} \hline \mathbf{r} &=& \mathbf{\dfrac{\sqrt{s(s-a)(s-b)(s-c)}}{s},} & \mathbf{s=\dfrac{a+b+c}{2}} \\\\ && \boxed{r=1,\ a=3,\ b=4,\ c=x } \\\\ 1 &=& \dfrac{\sqrt{s(s-3)(s-4)(s-x)}}{s}, & s=\dfrac{3+4+x}{2} \\\\ s &=& \sqrt{s(s-3)(s-4)(s-x)}\quad \text{square both sides} , & s=\dfrac{7+x}{2} \\\\ s^2 &=& s(s-3)(s-4)(s-x) \\\\ s &=& (s-3)(s-4)(s-x) \\\\ \dfrac{7+x}{2} &=& \left(\dfrac{7+x}{2}-3 \right) \left(\dfrac{7+x}{2}-4 \right) \left(\dfrac{7+x}{2}-x \right) \\\\ \dfrac{7+x}{2} &=& \dfrac{(x+1)(x-1)(7-x)}{8} \\\\ 4 ( 7+x ) &=& (x+1)(x-1)(7-x) \\\\ 28+4x &=& (x^2-1)(7-x) \\\\ 28+4x &=& 7x^2-x^3-7+x \\ \mathbf{x^3-7x^2+3x+35} &=& \mathbf{0} \\ && \boxed{x=5, \text{ see the right triangle with sides }3,\ 4,\ \mathbf{5}} \\\\ (x-5)(\underbrace{x^2-2x-7}_{=0}) &=& 0 \\ \hline \mathbf{x^2-2x-7} &=& \mathbf{0} \\\\ x &=& \dfrac{2\pm \sqrt{4-4*(-7)}}{2} \\\\ x &=& \dfrac{2\pm \sqrt{32}}{2} \\\\ x &=& \dfrac{2\pm \sqrt{16*2}}{2} \\\\ x &=& \dfrac{2\pm 4\sqrt{2}}{2} \\\\ \mathbf{ x } &=& \mathbf{ 1\pm 2\sqrt{2} } \\ \hline \\ \mathbf{ x} &=& \mathbf{1+2\sqrt{2}} \\ x &=& 1-2\sqrt{2} \quad x < 0 !,\ \text{no solution!} \\ \hline \end{array}\)

x is \(\mathbf{1+2\sqrt{2}} = 3.82842712475 \)

If, \(\left(\sqrt[3]{7}\right)^{5x}=14\) what is the value of \(\left(\sqrt[3]{7}\right)^{10x-6}\)

\(\begin{array}{|rcll|} \hline \mathbf{ \left(\sqrt[3]{7}\right)^{5x}} &=& \mathbf{14} \quad | \quad \text{square both sides} \\ \left( \left(\sqrt[3]{7}\right)^{5x} \right)^2 &=& 14^2 \\ \left(\sqrt[3]{7}\right)^{2*5x} &=& 196 \\ \left(\sqrt[3]{7}\right)^{10x} &=& 196 \quad | \quad : \left(\sqrt[3]{7}\right)^6 \\\\ \dfrac{ \left(\sqrt[3]{7}\right)^{10x} } {\left(\sqrt[3]{7}\right)^6} &=& \dfrac{196}{\left(\sqrt[3]{7}\right)^6} \\\\ \left(\sqrt[3]{7}\right)^{10x}*\left(\sqrt[3]{7}\right)^{-6} &=& \dfrac{196}{\left(\sqrt[3]{7}\right)^6} \\\\ \left(\sqrt[3]{7}\right)^{10x-6} &=& \dfrac{196}{\left(\sqrt[3]{7}\right)^6} \\\\ \left(\sqrt[3]{7}\right)^{10x-6} &=& \dfrac{196}{ 7^{\frac{6}{3}} } \\\\ \left(\sqrt[3]{7}\right)^{10x-6} &=& \dfrac{196}{7^2 } \\\\ \left(\sqrt[3]{7}\right)^{10x-6} &=& \dfrac{196}{49} \\\\ \mathbf{\left(\sqrt[3]{7}\right)^{10x-6}} &=& \mathbf{4} \\ \hline \end{array}\)

Compute  
\(\large \sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ)\)

Formula:
\(\begin{array}{|lrcll|} \hline (1) & \cos(x-y) &=& \cos(x)\cos(y)+\sin(x)\sin(y) \\ (2) & \cos(x+y) &=& \cos(x)\cos(y)-\sin(x)\sin(y)\\ \hline (1)-(2): & \mathbf{2\sin(x)\sin(y)} &=& \mathbf{\cos(x-y)-\cos(x+y)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) \quad | \quad \sin(30^\circ)=\dfrac{1}{2} \\ &=& \dfrac{1}{2}* \sin10(^\circ) \sin(70^\circ) \sin(50^\circ) \\ \hline \\ && \quad 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y) \\ && \quad 2\sin(10^\circ)\sin(70^\circ)=\cos(60^\circ)-\cos(80^\circ) \quad | \quad \cos(60^\circ)=\dfrac{1}{2} \\ &&\quad 2\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{2}-\cos(80^\circ) \quad | \quad \cos(80^\circ)=\cos(90^\circ-10^\circ)=\sin(10^\circ) \\ &&\quad 2\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{2}-\sin(10^\circ) \\ &&\quad \mathbf{\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{4}-\dfrac{1}{2}*\sin(10^\circ)} \\ \\ \hline &=& \dfrac{1}{2}\left( \dfrac{1}{4}-\dfrac{1}{2}*\sin(10^\circ) \right) \sin(50^\circ) \\ &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{4}*\sin(10^\circ)\sin(50^\circ) \\ \hline \\ && \quad 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y) \\ && \quad 2\sin(10^\circ)\sin(50^\circ)=\cos(40^\circ)-\cos(60^\circ) \quad | \quad \cos(60^\circ)=\dfrac{1}{2} \\ && \quad 2\sin(10^\circ)\sin(50^\circ)=\cos(40^\circ)-\dfrac{1}{2} \quad | \quad \cos(40^\circ)=\cos(90^\circ-50^\circ)=\sin(50^\circ) \\ &&\quad 2\sin(10^\circ)\sin(50^\circ)=\sin(50^\circ)-\dfrac{1}{2} \\ &&\quad \mathbf{\sin(10^\circ)\sin(50^\circ)=\dfrac{1}{2}*\sin(50^\circ)-\dfrac{1}{4} } \\ \\ \hline &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{4}*\left(\dfrac{1}{2}\sin(50^\circ)-\dfrac{1}{4}\right) \\ &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{8}\sin(50^\circ)+\dfrac{1}{16} \\ &=& \mathbf{\dfrac{1}{16}} \\ \hline \end{array} \)

ACEG is a rectangle. If segment BE is 30, segment CG is 40, segment DF is 15 and \(\angle FDE=90^\circ\).
Find CE.

\(\text{Let $CE=x$} \\ \text{Let $GE=y$} \\ \text{Let $AB=z$} \\ \text{Let $BC=y-z$}\)

\(\begin{array}{|lrcll|} \hline 1): & x^2 + y^2 &=& 40^2 \\ & x &=& \sqrt{40^2-y^2} \\ \hline 2): & x^2+(y-z)^2 &=& 30^2 \quad | \quad x^2 = 40^2-y^2 \\ & 40^2-y^2+(y-z)^2 &=& 30^2 \\ & y^2 - (y-z)^2 &=& 40^2-30^2 \\ & \mathbf{y^2 - (y-z)^2} &=& \mathbf{700} \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3): & \text{area $\triangle $ CDE} &=& \text{area $\triangle $ CFE} + \text{area $\triangle $ FEG} \\ & \dfrac{xy}{2} &=& \dfrac{x*15}{2} + \dfrac{y*DE}{2} \\ & xy &=& x*15 + y*DE \\ & && \boxed{\dfrac{DE}{15} =\dfrac{x}{y-z}\\ DE = \dfrac{15x}{y-z} } \\ & xy &=& x*15 + y*\left( \dfrac{15x}{y-z} \right) \quad | \quad : x \\ & y &=& 15 + \left( \dfrac{15y}{y-z} \right) \\ & (y-15)(y-z) &=& 15y \\ & y-z &=& \dfrac{15y}{y-15} \\ & \mathbf{(y-z)^2} &=& \mathbf{\dfrac{225y^2}{(y-15)^2}} \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{y^2 - (y-z)^2} &=& \mathbf{700} \quad | \quad \mathbf{(y-z)^2=\dfrac{225y^2}{(y-15)^2}} \\ y^2 - \dfrac{225y^2}{(y-15)^2} &=& 700 \quad | \quad *(y-15)^2 \\\\ y^2(y-15)^2 - 225y^2 &=& 700(y-15)^2 \\ y^2(y^2-30y+225) - 225y^2 &=& 700(y^2-30y+225) \\ y^4-30y^3+225y^2-225y^2 &=& 700y^2-700*30y+700*225 \\ y^4-30y^3 &=& 700y^2-700*30y+700*225 \\ \mathbf{ y^4-30y^3 -700y^2+700*30y-700*225} &=& \mathbf{0} \\ \Rightarrow \mathbf{y}&=&\mathbf{36.6659934977741} \qquad \text{WolframAlpha} \\ \hline x &=& \sqrt{40^2-y^2} \\ x &=& \sqrt{40^2-36.6659934977741^2} \\ \mathbf{x}&=&\mathbf{15.9876490087} \\ \hline \end{array}\)

CE is \(\approx \mathbf{16}\)

How many ways are there to split 14 people into seven pairs?

The first person of the 14 people can form a pair with \({\color{red}13}\) other people. We now have one pair less.
The first person of the remaining 12 people can form a pair with \({\color{red}11}\) other people. We now have two pairs less.
The first person of the remaining 10 people can form a pair with \({\color{red}9}\) other people. We now have three pairs less.
The first person of the remaining 8 people can form a pair with \({\color{red}7}\) other people. We now have four pairs less.
The first person of the remaining 6 people can form a pair with \({\color{red}5}\) other people. We now have five pairs less.
The first person of the remaining 4 people can form a pair with \({\color{red}3}\) other people. We now have six pairs less.
The first person of the remaining 2 people can form a pair with \({\color{red}1}\) other people. We now have seven pairs less.

There are \({\color{red}1}*{\color{red}3}*{\color{red}5}*{\color{red}7}*{\color{red}9}*{\color{red}11}*{\color{red}13} = \mathbf{135135}\) ways to split 14 people into seven pairs

triangle

\(\begin{array}{|rcll|} \hline \mathbf{\vec{BK}} &=& \mathbf{\dfrac{ \vec{BC} }{2} + \dfrac{ \vec{DA} }{2}} \\\\ && \boxed{ \vec{DA} = \vec{BA} - \dfrac{ \vec{BC} }{2} \\ \dfrac{\vec{DA}}{2} = \dfrac{\vec{BA}}{2} - \dfrac{ \vec{BC} }{4} } \\\\ \vec{BK} &=& \dfrac{ \vec{BC} }{2} + \dfrac{\vec{BA}}{2} - \dfrac{ \vec{BC} }{4} \\\\ \mathbf{\vec{BK}} &=& \mathbf{ \dfrac{ \vec{BC} }{4} + \dfrac{\vec{BA}}{2} } \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\vec{BL}} &=& \mathbf{ \vec{BC} + \dfrac{ \vec{CE} }{2}} \\\\ && \boxed{ \vec{CE} = \dfrac{\vec{BA}}{2} - \vec{BC} \\ \dfrac{\vec{CE}}{2} = \dfrac{\vec{BA}}{4} - \dfrac{ \vec{BC} }{2} } \\\\ \vec{BL} &=& \vec{BC} + \dfrac{\vec{BA}}{4} - \dfrac{ \vec{BC} }{2} \\\\ \mathbf{\vec{BL}} &=& \mathbf{ \dfrac{ \vec{BC} }{2} + \dfrac{\vec{BA}}{4} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2A_{yellow} = 18 &=& \vec{BL} \times \vec{BK} \\ 18 &=& \left( \dfrac{ \vec{BC} }{2} + \dfrac{\vec{BA}}{4} \right) \times \left( \dfrac{ \vec{BC} }{4} + \dfrac{\vec{BA}}{2} \right) \\\\ 18 &=& \underbrace{ \left( \dfrac{ \vec{BC} }{2} \times \dfrac{ \vec{BC} }{4}\right) }_{=0} +\left( \dfrac{ \vec{BC} }{2} \times \dfrac{\vec{BA}}{2} \right) +\left( \dfrac{\vec{BA}}{4} \times \dfrac{ \vec{BC} }{4} \right) + \underbrace{\left( \dfrac{\vec{BA}}{4} \times \dfrac{\vec{BA}}{2} \right) }_{=0} \\\\ 18 &=& \left( \dfrac{ \vec{BC} }{2} \times \dfrac{\vec{BA}}{2} \right) +\left( \dfrac{\vec{BA}}{4} \times \dfrac{ \vec{BC} }{4} \right) \\\\ 18 &=& \left( \dfrac{ \vec{BC} }{2} \times \dfrac{\vec{BA}}{2} \right)-\left( \dfrac{\vec{BC}}{4} \times \dfrac{ \vec{BA} }{4} \right) \\\\ 18 &=& \dfrac{1}{4}*\left( \vec{BC} \times \vec{BA} \right)-\dfrac{1}{16}*\left( \vec{BC} \times \vec{BA} \right) \\\\ 18 &=& \dfrac{4}{16}*\left( \vec{BC} \times \vec{BA} \right)-\dfrac{1}{16}*\left( \vec{BC} \times \vec{BA} \right) \\\\ 18 &=& \dfrac{3}{16}*\left( \vec{BC} \times \vec{BA} \right) \\\\ \dfrac{16*18}{3} &=& \vec{BC} \times \vec{BA} \\\\ 16*6 &=& \vec{BC} \times \vec{BA} \\\\ 96&=& \vec{BC} \times \vec{BA} \\\\ \mathbf{ \vec{BC} \times \vec{BA} } &=& \mathbf{96} \quad | \quad 2\times \text{area } \triangle \text{ABC} = \vec{BC} \times \vec{BA} \\\\ 2\times \text{area } \triangle \text{ABC} &=& 96 \quad | \quad :2 \\\\ \mathbf{ \text{area } \triangle \text{ABC} } &=& \mathbf{48} \\ \hline \end{array}\)

The area of the triangle ABC is \(\mathbf{48\ \text{cm}^2}\)

Can someone help find \(\begin{bmatrix}\sqrt3/2 & -1/2 \\1/2 & \sqrt3/2\end{bmatrix}^{2018}\)

I assume the matrix is \(\begin{bmatrix} \dfrac{\sqrt3}{2} & -\dfrac{1}{2} \\\\ \dfrac{1}{2} & \dfrac{\sqrt3}{2} \end{bmatrix}^{2018}\), then this matrix is a rotation matrix counterclockwise through an angle of \(30^\circ\)

\(\begin{array}{|rcll|} \hline \begin{bmatrix} \dfrac{\sqrt3}{2} & -\dfrac{1}{2} \\\\ \dfrac{1}{2} & \dfrac{\sqrt3}{2} \end{bmatrix}^{2018} &=& \begin{bmatrix} \cos(30^\circ) & -\sin(30^\circ) \\\\ \sin(30^\circ) & \cos(30^\circ) \end{bmatrix}^{2018} \\ \hline \end{array} \) Rotation through an angle of \(30^\circ \times 2018\)

\(\begin{array}{|rcll|} \hline 30^\circ \times 2018 &=& 60540^\circ \\ &=& 60540^\circ-168\times 360^\circ \\ &=& \color{red}\mathbf{60^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \begin{bmatrix} \cos(30^\circ) & -\sin(30^\circ) \\\\ \sin(30^\circ) & \cos(30^\circ) \end{bmatrix}^{2018} &=& \begin{bmatrix} \cos( {\color{red}\mathbf{60^\circ}} ) & -\sin( {\color{red}\mathbf{60^\circ}} ) \\\\ \sin( {\color{red}\mathbf{60^\circ}} ) & \cos( {\color{red}\mathbf{60^\circ}} ) \end{bmatrix} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \begin{bmatrix} \cos( {\color{red}\mathbf{60^\circ}} ) & -\sin( {\color{red}\mathbf{60^\circ}} ) \\\\ \sin( {\color{red}\mathbf{60^\circ}} ) & \cos( {\color{red}\mathbf{60^\circ}} ) \end{bmatrix} &=& \begin{bmatrix} \dfrac{1}{2} & -\dfrac{\sqrt3}{2} \\\\ \dfrac{\sqrt3}{2} & \dfrac{1}{2} \end{bmatrix} \\\\ \begin{bmatrix} \dfrac{\sqrt3}{2} & -\dfrac{1}{2} \\\\ \dfrac{1}{2} & \dfrac{\sqrt3}{2} \end{bmatrix}^{2018} &=& \begin{bmatrix} \dfrac{1}{2} & -\dfrac{\sqrt3}{2} \\\\ \dfrac{\sqrt3}{2} & \dfrac{1}{2} \end{bmatrix} \\ \hline \end{array} \)

If the sum of the first n terms of an arithmetic progression is \(\dfrac{3n^2+13n}{2}\),
find the 25th term of this progression.

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &=& \mathbf{\dfrac{3n^2+13n}{2}} \\\\ s_{25} &=& s_{24} + a_{25} \\\\ a_{25} &=& s_{25}- s_{24} \\\\ a_{25} &=& \dfrac{3*25^2+13*25}{2}- \dfrac{3*24^2+13*24}{2} \\\\ a_{25} &=& 1100 - 1020 \\ \mathbf{a_{25}} &=& \mathbf{80} \\ \hline \end{array}\)

The 25th term of this progression is \(\mathbf{80}\)