heureka

If z is a complex number satisfying z+1/z=1
calculate z^10+z^(-10)

Answer see:

Hello can i get some help on how to transform
this equation from a polar coordinate system
to a rectangular coordinate system
step by step please

$r^2=4r\sin(\theta)+5$

Formula:
$\boxed{\sin(\theta)=\dfrac{y}{r}~ \text{or}~ y=r\sin(\theta)\\ r = \sqrt{x^2+y^2}}$

$\begin{array}{|rcll|} \hline r^2 &=& 4r\sin(\theta)+5 \\ x^2+y^2 &=& 4y+5 \\ \mathbf{x^2+y^2-4y-5} &=& \mathbf{0} \\ \hline \end{array}$

Geometry help please?

$\begin{array}{|rcll|} \hline A &=& \dfrac{5x}{2} + 5*\left(\dfrac{(5-x)+5}{2}\right) \\ A &=& \dfrac{5x}{2} + 5*\left(\dfrac{10-x}{2}\right) \\ A &=& \dfrac{5}{2}\left( x+10-x \right) \\ A &=& \dfrac{5}{2}*10 \\ \mathbf{A} &=& \mathbf{25} \\ \hline \end{array}$

what is the smallest positive integer having exactly 5 different positive integer divisors?

The smallest positive integer  having exactly 5 different positive integer divisors is 16

Divisors:
1 | 2 | 4 | 8 | 16 (5 divisors)

Number Theory

There is a number formed by n copies of 2020 and 1 in the unit place.
If this number is divisible by 17,
what is the smallest possible value of n?

Formula: $2020n+1 \equiv 0 \pmod{17}$

$\begin{array}{|rcll|} \hline 2020n+1 &\equiv& 0 \pmod{17} \quad &|\quad 2020 \equiv 14 \pmod{17} \\ 14n+1 &\equiv& 0 \pmod{17} \\ 14n &\equiv& -1 \pmod{17} \\ n &\equiv& (-1)\dfrac{1}{14} \pmod{17} \\ \hline \end{array}$

Modular multiplicative inverse using Euler's theorem:

$\begin{array}{|rcll|} \hline \dfrac{1}{14} \pmod{17} &\equiv& 14^{\phi(17)-1} \pmod{17} \quad &|\quad \phi(17)=16 \\ &\equiv& 14^{16-1} \pmod{17} \\ &\equiv& 14^{15} \pmod{17} \quad &|\quad 14 \equiv -3 \pmod{17} \\ &\equiv& (-3)^{15} \pmod{17} \\ &\equiv& -(3)^{15} \pmod{17} \\ &\equiv& -(3)^{5*3} \pmod{17} \\ &\equiv& -\left(3^5\right)^{3} \pmod{17}\quad &|\quad 3^5 \equiv 5 \pmod{17} \\ &\equiv& -(5)^3 \pmod{17} \\ &\equiv& -125 \pmod{17} \quad &|\quad 125 \pmod{17} \equiv 6 \pmod{17} \\ \mathbf{ \dfrac{1}{14} \pmod{17} } &\equiv& \mathbf{ -6 \pmod{17}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline n &\equiv& (-1)\dfrac{1}{14} \pmod{17} \quad &|\quad \mathbf{ \dfrac{1}{14} \pmod{17} \equiv -6 \pmod{17}}\\ n &\equiv& (-1)(-6) \pmod{17} \\ \mathbf{n} &\equiv& \mathbf{6 \pmod{17}} \\ n &=& 6+17k \qquad k\in \mathbb{Z} \\ \hline \end{array}$

The smallest possible value of $n$ is $\mathbf{6}$

The six-digit integer 978XYZ consists of six distinct digits and is divisible by

7, 8 and 9.
What is the three-digit integer XYZ?

$\begin{array}{|rcll|} \hline 7 | 978XYZ \\ 8 | 978XYZ \\ 9 | 978XYZ \\ \hline \end{array} \begin{array}{|rcll|} \hline XYZ =~ ? \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 7*8*9 &=& 504 \\ 978000 : 504 &=& 1940~ R~ 240 \\ 240+\mathbf{264} &=& 504 \\ \mathbf{XYZ} &=& \mathbf{264} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 978000 +264 &=& 978264 \\ 978264 : 7 &=& 139752 \\ 978264 : 8 &=& 122283 \\ 978264 : 9 &=& 108696 \\ \hline \end{array}$

What is the product of the two smallest prime factors of $2^{100} - 1$?

First prime factor 2 ?
$2^{100} - 1$ is an odd number so 2 is no prime factor in $2^{100} - 1$

next prime factor 3 ?
If 3 is a prime factor in $2^{100} - 1$, then $2^{100} - 1 \equiv 0 \pmod{3}$
$\begin{array}{|rcll|} \hline 2^{100} - 1 &\equiv& 0 \pmod{3} \quad ? \quad | \quad 2 \equiv -1 \pmod{3} \\ (-1)^{100} - 1 &\equiv& 0 \pmod{3} \quad ? \\ 1-1 &\equiv& 0 \pmod{3} \quad ? \\ 0&\equiv& 0 \pmod{3}~ \checkmark \quad | \quad {\color{red}3} ~\text{is a prime factor in}~ 2^{100} - 1 \\ \hline \end{array}$

next prime factor 5 ?
If 5 is a prime factor oin $2^{100} - 1$, then $2^{100} - 1 \equiv 0 \pmod{5}$
$\begin{array}{|rcll|} \hline 2^{100} - 1 &\equiv& 0 \pmod{5} \quad ? \\ 2^{2*50} - 1 &\equiv& 0 \pmod{5} \quad ? \\ \left(2^2\right)^{50} - 1 &\equiv& 0 \pmod{5} \quad ? \quad | \quad 2^2=4 \equiv -1 \pmod{5} \\ (-1)^{50} - 1 &\equiv& 0 \pmod{5} \quad ? \\ 1-1 &\equiv& 0 \pmod{5} \quad ? \\ 0&\equiv& 0 \pmod{5} ~ \checkmark \quad | \quad {\color{red}5} ~\text{is a prime factor in}~ 2^{100} - 1 \\ \hline \end{array}$

$3*5 = 15$

The product of the two smallest prime factors of $2^{100} - 1$ is 15

If

$\log_2(x)=7~\text{find}~ \log_8(x)=y$

$\begin{array}{|lrcll|} \hline (1) & x &=& 2^7 \qquad \Big(\log_2(x)=7\Big) \\\\ (2) & x &=& 8^y \qquad \Big(\log_8(x)=y\Big) \\ & x &=& 2^{3y}\\ \hline & x = 2^7 &=& 2^{3y}\\ & 2^7 &=& 2^{3y} \quad & | \quad \text{compare exponent}\\ & 7 &=& 3 y \\ & \mathbf{y} &=& \dfrac{7}{3} \\ & \mathbf{\log_8(x)} &=& \dfrac{7}{3} \\ \hline \end{array}$

Find all the solutions to
$\sqrt[3]{ \sqrt[3]{5\sqrt{2}+x}+ \sqrt[3]{5\sqrt{2}-x} }=\sqrt{2}$

$\small{ \begin{array}{|rcll|} \hline \sqrt[3]{ \sqrt[3]{5\sqrt{2}+x}+ \sqrt[3]{5\sqrt{2}-x} } &=& \sqrt{2} \quad & | \quad \text{cube both sides}\\ \sqrt[3]{5\sqrt{2}+x}+ \sqrt[3]{5\sqrt{2}-x} &=& \left(\sqrt{2}\right)^3 \quad & | \quad \text{cube both sides} \\ \left( \sqrt[3]{5\sqrt{2}+x}+ \sqrt[3]{5\sqrt{2}-x}\right)^3 &=& \left(\sqrt{2}\right)^9 \\ \boxed{(a+b)^3 = a^3+b^3+3ab(a+b) } \\ 5\sqrt{2}+x + 5\sqrt{2}-x +3\sqrt[3]{(5\sqrt{2}+x)(5\sqrt{2}-x)}( \underbrace{ \sqrt[3]{5\sqrt{2}+x}+ \sqrt[3]{5\sqrt{2}-x}}_{=\left(\sqrt{2}\right)^3} ) &=& \left(\sqrt{2}\right)^9 \\ 10\sqrt{2} +3\sqrt[3]{25*2-x^2}\left(\sqrt{2}\right)^3 &=& \left(\sqrt{2}\right)^9 \\ 3\sqrt[3]{50-x^2} &=& \dfrac{\left(\sqrt{2}\right)^9 -10\sqrt{2} } {\left(\sqrt{2}\right)^3} \\ 3\sqrt[3]{50-x^2} &=& \dfrac{\sqrt{2}\left( \left(\sqrt{2}\right)^8-10 \right) } {\sqrt{2} \left(\sqrt{2}\right)^2 } \\ 3\sqrt[3]{50-x^2} &=& \dfrac{\left(\sqrt{2}\right)^8-10} {\left(\sqrt{2}\right)^2 } \\ 3\sqrt[3]{50-x^2} &=& \dfrac{16-10} {4} \\ 3\sqrt[3]{50-x^2} &=& 3 \\ \sqrt[3]{50-x^2} &=& 1 \quad & | \quad \text{cube both sides} \\ 50-x^2 &=& 1 \\ x^2 &=& 49 \\ x &=& \pm 7 \\ \hline \end{array} }$

Solutions:
$x=-7 \\ x=7$

Geometry Problem:

$\text{Let the side of the square $= a$. $AD=DC=a$ } \\ \text{Let $DF = y$ } \\ \text{Let the area of triangle $[AED]=\dfrac{a^2}{2}$ } \\ \text{Let the area of triangle $[BCF]=\dfrac{(a-y)a}{2}=S_2+S_4+S_7$ } \\ \text{Let the area of triangle $[ADF]=\dfrac{ay}{2}=S_5+S_8$ }$

$\begin{array}{|rcll|} \hline \dfrac{(a-y)a}{2} &=& S_2+S_4+S_7 \\ \mathbf{S_4} &=& \dfrac{(a-y)a}{2} - S_2 - S_7 \\ \hline \dfrac{ay}{2} &=& S_5+S_8 \\ \mathbf{S_5} &=& \dfrac{ay}{2} - S_8 \\ \hline \mathbf{S_3} &=& \mathbf{[AED] - S_4 -S_5} \\ S_3 &=& \dfrac{a^2}{2} -\left( \dfrac{(a-y)a}{2} - S_2 - S_7 \right) - \left(\dfrac{ay}{2} - S_8 \right) \\ S_3 &=& \dfrac{a^2}{2}-\dfrac{a^2}{2} +\dfrac{ay}{2} + S_2 + S_7 - \dfrac{ay}{2} + S_8 \\ S_3 &=& S_2 + S_7 + S_8 \\ S_3 &=& 2 + 8 + 2 \\ \mathbf{S_3} &=& \mathbf{12} \\ \hline \end{array}$