heureka

Alpha 15 grad, cos 15 grad = ??

$\begin{array}{|rcll|} \hline \cos(15^{\circ}) \\ &=& \tfrac14 \cdot (\sqrt{6}+\sqrt{2}) \\ &=& 0,96592582629 \\ \hline \end{array}$

For example:

log 4 ≈ 0.477

log 4 ~ 0.477

Which statement is more correct?

$\begin{array}{|rcll|} \hline log 4 \approx 0.477 \quad \text{is correct, it means approximately } \\ \hline \end{array}$

${\displaystyle \sim }$:
${\displaystyle a\sim b} \qquad$Equivalence relation between elements  ${\displaystyle a}$ and ${\displaystyle b}$
${\displaystyle a\sim b} \qquad$${\displaystyle a}$ is proportional to ${\displaystyle b}$

If tanØ=3/4 what is sinØ=?

$\begin{array}{|rcll|} \hline \sin(\phi) = \frac35 \quad \text{or} \quad \sin(\phi) = -\frac35 \\ \hline \end{array}$

Find the number of theater seats in an auditorium with 50 rows where
the first row has 20 seats and
each successive row has 2 more seats than the previous?

Formula:

$\begin{array}{|rcll|} \hline t_n &=& t_1 + (n-1)\cdot d \\ s_n &=& \left(\frac{t_1+t_n}{2} \right) \cdot n \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline n &=& 50 \\ t_1 &=& 20 \\ d &=& 2 \\\\ t_{50} &=& 20 + (50-1)\cdot 2 \\ t_{50} &=& 20 + 49\cdot 2 \\ t_{50} &=& 118 \\\\ s_{50} &=& \left(\frac{20+118}{2} \right) \cdot 50 \\ s_{50} &=& \left(\frac{138}{2} \right) \cdot 50 \\ s_{50} &=&69 \cdot 50 \\ s_{50} &=&3450 \\ \hline \end{array}$

The number of theater seats in an auditorium is 3450

A dice is rolled twice. What is the probability of observing at least one five?

$\color{green}{\text{In green dice first rolling }}\\ \color{blue}{\text{In blue dice second rolling }}\\ \color{red}{\text{In red number } 5 }$

$\begin{array}{|r|r|r|r|r|r|r|} \hline & \color{blue}1 & \color{blue}2 & \color{blue}3 & \color{blue}4 & \color{blue}5 & \color{blue}6 \\ \hline \color{green}1 & & & & & \color{red}X & \\ \hline \color{green}2 & & & & & \color{red}X & \\ \hline \color{green}3 & & & & & \color{red}X & \\ \hline \color{green}4 & & & & & \color{red}X & \\ \hline \color{green}5 &\color{red}X & \color{red}X &\color{red}X & \color{red}X & \color{red}X & \color{red}X \\ \hline \color{green}6 & & & & & \color{red}X & \\ \hline \end{array}$

$\text{The probability of rolling at least one five is } \dfrac{11}{36} \text{ or } 30.\bar{5}\ \%$

2. \int x^2sinxdx ?

Formula:

$\begin{array}{|rcll|} \hline \int(u\cdot v')\ dx = u\cdot v - \int(u'\cdot v)\ dx \\ \hline \end{array}$

$\begin{array}{|rclrclrcl|} \hline && \int (~x^2\cdot \sin(x) ~)\ dx\\ && & u &=& x^2 & u' &=& 2x \\ && & v' &=& \sin(x) & v &=& -\cos(x) \\ &=& -x^2\cdot \cos(x) + 2\cdot\int (x\cdot \cos(x) )\ dx \\ \\ \hline && \int (~x\cdot \cos(x) ~)\ dx\\ && & u &=& x & u' &=& 1 \\ & & & v' &=& \cos(x) & v &=& \sin(x) \\ &=& x\cdot \sin(x) - \int (1\cdot \sin(x) )\ dx \\ &=& x\cdot \sin(x) - \int (\sin(x) )\ dx & \int \sin(x) &=& -\cos(x) \\ &=& x\cdot \sin(x) + \cos(x) \\ \\ \hline && \mathbf{ \int (~x^2\cdot \sin(x) ~)\ dx }\\ &=& -x^2\cdot \cos(x) + 2\cdot[x\cdot \sin(x) + \cos(x) ] + c \\ &=& -x^2\cdot \cos(x) + 2\cdot x\cdot \sin(x)+ 2\cos(x)+ c \\ &=& \mathbf{ 2\cdot x\cdot \sin(x) -(x^2-2)\cdot \cos(x) + c } \\ \hline \end{array}$

Thank you very much, CPhill !

What is the smallest number that 

when divided by 31 leaves a remainder of 9, and

when divided by 41 leaves a remainder of 39?

$\begin{array}{|rcll|} \hline n \equiv {\color{red}9} \pmod {{\color{green}31}} \\ n \equiv {\color{red}39} \pmod {{\color{green}41}} \\ m=31\cdot 41=1271 \\ \hline \end{array}$

Because 31 and 41 are relatively prim ( gcd(31,41) = 1! ) we can go on:

$\begin{array}{rcll} n &=& {\color{red}9} \cdot {\color{green}41} \cdot [ \frac{1}{\color{green}41}\pmod {{\color{green}31}} ] + {\color{red}39} \cdot {\color{green}31} \cdot [ \frac{1}{\color{green}31}\pmod {{\color{green}41}} ] + {\color{green}31}\cdot {\color{green}41} \cdot k \quad & | \quad k\in Z\\ \end{array}$

$\begin{array}{rcll} && [ \frac{1}{\color{green}41}\pmod {{\color{green}31}} ] \quad &| \quad \text { modular inverse } 41\cdot \frac{1}{41} \equiv 1 \pmod {31} \\ &=& {\color{green}41}^{\varphi({\color{green}31})-1} \pmod {{\color{green}31}} \quad & | \quad \text{ Euler's totient function } \varphi(n) \quad \varphi(p) = p-1 \\ &=& 41^{30-1} \pmod {31} \\ &=& 41^{29} \pmod {31} \quad & | \quad 41 \pmod {31} = 10 \\ &=& 10^{29} \pmod {31} \\ &=& 28 \\\\ && [ \frac{1}{\color{green}31}\pmod {{\color{green}41}} ] \quad &| \quad \text { modular inverse } 31\cdot \frac{1}{31} \equiv 1 \pmod {41} \\ &=& {\color{green}31}^{\varphi({\color{green}41})-1} \pmod {{\color{green}41}} \quad & | \quad \text{ Euler's totient function } \varphi(n)\quad \varphi(p) = p-1 \\ &=& 31^{40-1} \pmod {41} \\ &=& 31^{39} \pmod {41} \\ &=& 4 \\ \end{array}$

$\begin{array}{|rcll|} \hline n &=& {\color{red}9} \cdot {\color{green}41} \cdot 28 + {\color{red}39} \cdot {\color{green}31} \cdot 4 + {\color{green}31}\cdot {\color{green}41} \cdot k \\ n &=& 10332 + 4836 + 1271 \cdot k \\ n &=& 15168 + 1271 \cdot k \quad & | \quad k\in Z \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline n_{min} &=& 15168 \pmod {1271 } \\ n_{min} &=& 1187 \\\\ n &=& 1187 + 1271 \cdot k \quad & | \quad k\in Z \\ \hline \end{array}$

The smallest number is 1187

What is the last digit of: 3^2046?

$\begin{array}{|rcll|} \hline && 3^{\varphi(10) } \equiv 1 \pmod{10} \quad | \quad \varphi(10) = 10\cdot(1-\frac12)\cdot(1-\frac15) = 4 \\ && 3^4 \equiv 1 \pmod{10} \\\\ && 3^{2046} \pmod{10} \\ &\equiv & 3^{4\cdot 511 +2} \pmod{10} \\ &\equiv & 3^{4\cdot 511}\cdot 3^2 \pmod{10} \\ &\equiv & (3^4)^{511}\cdot 3^2 \pmod{10} \\ &\equiv & (1)^{511}\cdot 3^2 \pmod{10} \\ &\equiv & 3^2 \pmod{10} \\ &\equiv & 9 \pmod{10} \\ &\equiv & 9 \\ \hline \end{array}$

The last digit is 9

What is the smallest number under 16,000 that

when divided by 127 leaves a remainder of 14,

and when divided by 131 leaves a ramainder of 66.

$\begin{array}{|rcll|} \hline n \equiv {\color{red}14} \pmod {{\color{green}127}} \\ n \equiv {\color{red}66} \pmod {{\color{green}131}} \\ m=127\cdot 131=16637 \\ \hline \end{array}$

Because 127 and 131 are relatively prim ( gcd(127,131) = 1! ) we can go on:

$\begin{array}{rcll} n &=& {\color{red}14} \cdot {\color{green}131} \cdot [ \frac{1}{\color{green}131}\pmod {{\color{green}127}} ] + {\color{red}66} \cdot {\color{green}127} \cdot [ \frac{1}{\color{green}127}\pmod {{\color{green}131}} ] + {\color{green}127}\cdot {\color{green}131} \cdot k \quad & | \quad k\in Z\\ \end{array}$

$\begin{array}{rcll} && [ \frac{1}{\color{green}131}\pmod {{\color{green}127}} ] \quad &| \quad \text { modular inverse } 131\cdot \frac{1}{131} \equiv 1 \pmod {127} \\ &=& {\color{green}131}^{\varphi({\color{green}127})-1} \pmod {{\color{green}127}} \quad & | \quad \text{ Euler's totient function } \varphi(n) \quad \varphi(p) = p-1 \\ &=& 131^{126-1} \mod {127} \\ &=& 131^{125} \mod {127} \\ &=& 32 \\\\ && [ \frac{1}{\color{green}127}\pmod {{\color{green}131}} ] \quad &| \quad \text { modular inverse } 127\cdot \frac{1}{127} \equiv 1 \pmod {131} \\ &=& {\color{green}127}^{\varphi({\color{green}131})-1} \pmod {{\color{green}131}} \quad & | \quad \text{ Euler's totient function } \varphi(n)\quad \varphi(p) = p-1 \\ &=& 127^{130-1} \mod {131} \\ &=& 127^{129} \mod {131} \\ &=& 98 \\ \end{array}$

$\begin{array}{|rcll|} \hline n &=& {\color{red}14} \cdot {\color{green}131} \cdot 32 + {\color{red}66} \cdot {\color{green}127} \cdot 98 + {\color{green}127}\cdot {\color{green}131} \cdot k \\ n &=& 58688 + 821436 + 16637 \cdot k \\ n &=& 880124 + 16637 \cdot k \quad & | \quad k\in Z \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline n_{min} &=& 880124 \pmod {16637 } \\ n_{min} &=& 15000 \\\\ n &=& 15000 + 16637 \cdot k \quad & | \quad k\in Z \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline k=0: & \mathbf{n} &\mathbf{=}& \mathbf{15000} \\ \hline \end{array}$

The smallest number is 15000