heureka

Consider the following diagram:

What are the lengths of AE and CE?

$\begin{array}{|rcll|} \hline CD^2 + 3^2 &=& 9^2 \\ CD^2 &=& 81 - 9 \\ CD^2 &=& 72 \\ CD &=& \sqrt{72} \\ \hline \end{array} \begin{array}{|rcll|} \hline BD^2 + 3^2 &=& 9^2 \\ BD^2 &=& 81 - 9 \\ BD^2 &=& 72 \\ BD &=& \sqrt{72} \\ \hline \end{array} \begin{array}{|rcll|} \hline CB &=& CD+BD \\ CB &=& \sqrt{72}+\sqrt{72} \\ CB &=& 2\sqrt{72} \\ CB^2 &=& 4*72 \\ \mathbf{CB^2} &=& \mathbf{288} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline AE^2+CE^2 &=& 9^2 \\ \mathbf{CE^2} &=& \mathbf{81-AE^2} \\ \hline \end{array} \begin{array}{|rcll|} \hline (9+AE)^2+CE^2 &=& CB^2 \\ (9+AE)^2+81-AE^2&=& 288 \\ 81+18AE+AE^2+81-AE^2&=& 288 \\ 81+18AE+81&=& 288 \\ 18AE&=& 288-2*81 \\ 18AE&=& 126 \\ \mathbf{AE}&=& \mathbf{7} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{CE^2} &=& \mathbf{81-AE^2} \\ CE^2 &=& 81-7^2 \\ CE^2 &=& 81-49 \\ CE^2 &=& 32 \\ CE^2 &=& 16*2 \\ \mathbf{CE} &=& \mathbf{4*\sqrt{2}} \\ \hline \end{array}$

Determine the smallest non-negative integer a that satisfies the congruences:

$a \equiv 2 \pmod{ 3}\\ a \equiv 4 \pmod{ 5}\\ a \equiv 1 \pmod{ 7}\\ a \equiv 8 \pmod{ 9}$

1. join
$a \equiv 2 \pmod{ 3} \\ a \equiv 8 \pmod{ 9}$

$\begin{array}{|rcll|} \hline a&=&2+3n \\ a&=&8+9m \\ \hline 2+3n&=& 8+9m \\ 3n&=& 6+9m \quad | \quad : 3\\ n&=& 2+3m \\ \hline a&=&2+3n \quad | \quad n=2+3m \\ a&=&2+3(2+3m) \\ a&=&2+6+9m \\ a&=&8+9m \\ \mathbf{a} \equiv \mathbf{ 8 \pmod{ 9} } \\ \hline \end{array} \\ \begin{equation}\left . \begin{aligned} & a \equiv 2 \pmod{ 3} \\ & a \equiv 8 \pmod{ 9} \end{aligned}\right \rbrace \mathbf{a} \equiv \mathbf{ 8 \pmod{ 9} } \end{equation}$

2. join
$a \equiv 4 \pmod{ 5} \\ a \equiv 1 \pmod{ 7}$

$\begin{array}{|rcl|rcl|rcl|} \hline a&=&4+5r \\ a&=&1+7s \\ \hline 4+5r&=&1+7s \\ 5r &=& 7s-3 \\ r &=& \frac{7s-3}{5} \\ r &=& \frac{5s+2s-3}{5} \\ r &=& s+ \underbrace{ \frac{2s-3}{5} }_{=t} \\ \mathbf{r} &=& \mathbf{s+t} & t &=& \frac{2s-3}{5}\\ & & & 5t &=& 2s-3 \\ & & & 2s &=& 5t+3 \\ & & & s&=& \frac{5t+3}{2} \\ & & & s&=& \frac{4t+t+2+1}{2} \\ & & & s&=& 2t+1 + \underbrace{ \frac{t+1}{2}}_{=u} \\ & & & \mathbf{s}&=& \mathbf{2t+1 + u} \\ & & & & & & u&=&\frac{t+1}{2} \\ & & & & & & 2u&=& t+1 \\ & & & & & & t&=& 2u-1 \\ & & & s &=& 2(2u-1)+1 +u \\ & & & s &=& 5u-1 \\ r &= &5u-1+2u-1 \\ r &= &7u-2 \\ \hline a &=& 4+5r \\ a &=& 4+5(7u-2) \\ a &=& -6+35u \\ \mathbf{a} &\equiv& \mathbf{ -6 \pmod{ 35} } \\ \hline \end{array}\\ \begin{equation}\left . \begin{aligned} & a \equiv 4 \pmod{ 5} \\ & a \equiv 1 \pmod{ 7} \end{aligned}\right \rbrace \mathbf{a} \equiv \mathbf{ -6 \pmod{ 35} } \end{equation}$

3. join
$a \equiv 8 \pmod{ 9} \\ a \equiv -6 \pmod{35}$

$\begin{array}{|rcl|rcl|rcl|} \hline a&=&8+9v \\ a&=&-6+35w \\ \hline 8+9v&=&-6+35w \\ 9v &=& 35w-14 \\ v&=& \frac{35w-14}{9} \\ v &=& \frac{27w+8w-9-5}{9} \\ v &=& 3w-1+ \underbrace{ \frac{8w-5}{9} }_{=x} \\ \mathbf{v} &=& \mathbf{3w-1+x} & x &=& \frac{8w-5}{9}\\ & & & 9x &=& 8w-5 \\ & & & 8w &=& 9x+5 \\ & & & w&=& \frac{9x+5}{8} \\ & & & w&=& \frac{8x+x+5}{8} \\ & & & w&=& x + \underbrace{ \frac{x+5}{8}}_{=y} \\ & & & \mathbf{w}&=& \mathbf{x+y} \\ & & & & & & y&=&\frac{x+5}{8} \\ & & & & & & 8y&=& x+5 \\ & & & & & & x&=& 8y-5 \\ & & & w &=& 8y-5 +y \\ & & & w &=& 9y-5 \\ v &= &3(9y-5)-1+8y-5 \\ v &= &35y-21 \\ \hline a &=& 8+9v \\ a &=& 8+9(35y-21) \\ a &=& -181+315y \\ \mathbf{a} &\equiv& \mathbf{ -181 \pmod{ 315} } \\ \mathbf{a} &\equiv& \mathbf{ -181+315 \pmod{ 315} } \\ \mathbf{a} &\equiv& \mathbf{ 134 \pmod{ 315} } \\ \hline \end{array}\\ \begin{equation}\left . \begin{aligned} & a \equiv 8 \pmod{ 9} \\ & a \equiv -6 \pmod{35} \end{aligned}\right \rbrace \mathbf{a} \equiv \mathbf{ 134 \pmod{ 315} } \end{equation}$

The smallest non-negative integer a is 134

What is the smallest integer n, greater than 1, such that
$\dfrac{1}{n} \pmod{130}$ and $\dfrac{1}{n} \pmod {9}$

are both defined?

There is an invertible modulo 130, if $\gcd(130,n)=1~$(130 and n are relatively prime)
There is an invertible modulo 9, if $\gcd(9,n)=1~$ (9 and n are relatively prime)

$n > 1$

$\begin{array}{|c|c|c|c|} \hline n & \gcd(130,n) & \gcd(9,n) \\ \hline 2 & \gcd(130,2)=2 & \gcd(9,2)=\color{red}1 \\ 3 & \gcd(130,3)=\color{red}1 & \gcd(9,3)=3 \\ 4 & \gcd(130,4)=2 & \gcd(9,4)=\color{red}1 \\ 5 & \gcd(130,5)=5 & \gcd(9,5)=\color{red}1 \\ 6 & \gcd(130,6)=2 & \gcd(9,6)=3 \\ \mathbf{7} & \gcd(130,7)=\color{red}1 & \gcd(9,7)=\color{red}1 & \checkmark \\ \hline \end{array}$

The smallest integer is 7

$\dfrac{1}{7} \pmod{130} \equiv 93 \text{ and } \dfrac{1}{7} \pmod {9} \equiv 4$

What is the remainder when $17^7$ is divided by $35$?

$\small{ \begin{array}{|rcll|} \hline \mathbf{17^{7} \pmod{35}} &\equiv&\left( 17\right)^{2*3+1} \pmod{35} \\ &\equiv& \left( 17^2 \right)^3*17 \pmod{35} \quad &| \quad 17^2=289 \equiv 9 \pmod{35} \\ &\equiv& \left( 9 \right)^3*17 \pmod{35} \\ &\equiv& 729*17 \pmod{35} \quad &| \quad 729 \equiv 29 \pmod{35} \\ &\equiv& 29*17 \pmod{35} \\ &\equiv& 493 \pmod{35} \quad &| \quad 493 \equiv 3 \pmod{35} \\ &\equiv& \mathbf{ 3 \pmod{35} } \\ \hline \end{array} }$

The reminder is 3

What is the remainder when $2013^{13}$ is divided by $13$?

$\begin{array}{|rcll|} \hline && \mathbf{2013^{13} \pmod{13}} \quad &| \quad 2013 \equiv 11 \pmod{13} \\ &\equiv&11^{13}\pmod{13} \quad &| \quad 11^4 \equiv 3 \pmod{13} \\ &\equiv&\left( 11\right)^{4*3+1} \pmod{13} \\ &\equiv& \left( 11^4 \right)^3*11 \pmod{13} \\ &\equiv& \left( 3 \right)^3*11 \pmod{13} \\ &\equiv& 27*11 \pmod{13}\quad &| \quad 27 \equiv 1 \pmod{13} \\ &\equiv& 1*11 \pmod{13} \\ &\equiv& \mathbf{ 11 \pmod{13} } \\ \hline \end{array}$

The reminder is 11

What is the tens digit of $17^{1993}$?

$\small{ \begin{array}{|rcll|} \hline \mathbf{17^{1993} \pmod{100}} &\equiv& \left( 17^2 \right)^{996}+1 \pmod{100} \\ &\equiv& \left( 17^2 \right)^{996}*17 \pmod{100} \\ && \boxed{17^2 = 289 \equiv 89 \equiv 89-100 \equiv -11 \pmod{100}} \\ &\equiv& \left( -11 \right)^{996}*17 \pmod{100} \\ &\equiv& \left( (-11)^2 \right)^{498}*17 \pmod{100} \\ &\equiv& \left( 121 \right)^{498}*17 \pmod{100} \quad | \quad 121 \equiv 21 \pmod{100} \\ &\equiv& \left( 21 \right)^{498}*17 \pmod{100} \\ &\equiv& \left( 21 \right)^{5*99+3}*17 \pmod{100} \\ &\equiv& \left( 21^5 \right)^{99}*21^3*17 \pmod{100} \quad | \quad 21^5 \equiv 01 \pmod {100} \\ &\equiv& \left( 1\right)^{99}*21^3*17 \pmod{100} \\ &\equiv& 21^3*17 \pmod{100} \\ &\equiv& 157437 \pmod{100} \\ &\equiv& \mathbf{{\color{red}3}7 \pmod{100}} \\ \hline \end{array} }$

The tens digit is ${\color{red}3}$

add these binomials

$^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97}$

$\small{ \begin{array}{|rcll|} \hline ^{99}C_0+~^{99}C_1+~^{99}C_2+~^{99}C_3 + \ldots + ~^{99}C_{97}+~^{99}C_{98}+~^{99}C_{99} &=& 2^{99} \\\\ ~^{99}C_1+~^{99}C_3+~^{99}C_5 + \ldots + ~^{99}C_{97}+~^{99}C_{99} &=& \dfrac{2^{99}}{2} \\\\ ^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97} &=& \dfrac{2^{99}}{2} -~^{99}C_1-~^{99}C_{99} \\\\ ^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97} &=& 2^{98} -99-1 \\\\ \mathbf{ ^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97} } &=& \mathbf{2^{98} -100 } \\ \hline \end{array} }$

If each dimension of a rectangle decreases by 1,
its area will decrease from 2017 to 1917.
What will be the area of the rectangle
if each of its dimensions increases by 1?

$\text{Let $ab=2017$ }$

$\begin{array}{|lrcll|} \hline \text{decreases by $1$ :} & 1917 &=& (a-1)(b-1) \\ & 1917 &=& ab-(a+b) +1 \qquad (1) \\\\ \text{increases by $1$ :} & x &=& (a+1)(b+1) \\ & x &=& ab+(a+b) +1 \qquad (2) \\\\ \hline (1)+(2):& 1917+x &=& ab-(a+b) +1 + ab+(a+b) +1 \\ & 1917+x &=& 2ab+ 2 \\ & x &=& 2ab+ 2 -1917 \\ & x &=& 2ab -1915 \quad | \quad \mathbf{ab=2017} \\ & x &=& 2*2017 -1915 \\ & x &=& 2119 \\ \hline \end{array}$

The area of the rectangle
if each of its dimensions increases by 1 is 2119

The interior angles of a convex polygon are in an arithmetic progression.
If the smallest angle is 100 degrees and common difference is  10 degrees ,
then find the number of sides.

$\text{Let the sides of the polygon $n$ }$

Arithmetic progression:$\begin{array}{|rcll|} \hline \text{sum} &=& 100^\circ+110^\circ+120^\circ+130^\circ+\ldots+\Big( 100^\circ+(n-1)10^\circ\Big) \\ \text{sum} &=& \left(\dfrac{100^\circ + \Big( 100^\circ+(n-1)10^\circ\Big)}{2}\right) * n \\\\ \text{sum} &=& \left( \dfrac{100^\circ+100^\circ +10^\circ n -10^\circ}{2} \right) * n \\\\ \text{sum} &=& \left( \dfrac{190^\circ +10^\circ n}{2} \right) * n \\\\ \mathbf{\text{sum}} &=& \mathbf{(95^\circ +5^\circ n)*n} \\ \hline \end{array}$

Interior angles of a convex polygon: $\text{sum} = (n-2)*180^\circ$

$\begin{array}{|rcll|} \hline (95^\circ +5^\circ n)*n &=& (n-2)*180^\circ \\ 95 n +5 n^2 &=& 180n -2*180\\ \ldots \\ 5n^2-85n + 360&=& 0 \quad | \quad : 5 \\ \mathbf{n^2-17n + 72} &=& \mathbf{0} \\ \hline n &=& \dfrac{17 \pm \sqrt{17^2- 4 *72 } }{2} \\\\ n &=& \dfrac{17 \pm \sqrt{289-288 } }{2} \\\\ n &=& \dfrac{17 \pm 1 }{2} \\ \mathbf{n}=\mathbf{9} &\text{or}&\mathbf{n}=\mathbf{8} \\ \hline \end{array}$

Integral

$\int \dfrac{\sin^2(\theta) \cos(\theta) }{\sqrt{1+\sin(\theta) }}~d\theta$

$\begin{array}{|l|c|c|} \hline & \text{substitute} \\ \hline \int \dfrac{\sin^2(\theta) \cos(\theta) }{\sqrt{1+\sin(\theta)}}~d\theta & u = \sin(\theta)& du = \cos(\theta)~d\theta \\ & & d\theta =\dfrac{du}{\cos(\theta)} \\ =\int \dfrac{u^2 \cos(\theta) }{\sqrt{1+u}}*\dfrac{du}{\cos(\theta)} \\\\ =\int \dfrac{u^2 }{\sqrt{1+u}}~du & v=1+u & dv = du \\ & u=v-1& du = dv \\\\ =\int \dfrac{(v-1)^2 }{\sqrt{v}}~dv & w=\sqrt{v} & dw = \dfrac{1}{2\sqrt{v}}~dv \\ &v=w^2 & dv = 2\sqrt{v}~dw \\\\ =\int \dfrac{(w^2-1)^2 }{\sqrt{v}}~2\sqrt{v}~dw \\ =2\int(w^2-1)^2 ~dw \\\\ =2\int ( w^4-2w^2+1) ~dw \\\\ =2\left( \dfrac{w^5}{5}-\dfrac{2w^3}{3}+w\right) ~+c \\\\ =\dfrac{2w^5}{5}-\dfrac{4w^3}{3}+2w ~+c & w & \\ & =\sqrt{v} \\ & =\sqrt{1+u} \\ & =\sqrt{1+\sin(\theta)} \\\\ =\dfrac{2}{5}\Big(\sqrt{1+\sin(\theta)}\Big)^5 \\ -\dfrac{4}{3}\Big(\sqrt{1+\sin(\theta)}\Big)^3 \\ +2\sqrt{1+\sin(\theta)} ~+c \\\\ \hline \end{array}$