heureka

Define the sequence of positive integers $a_n$ recursively by
$a_1=3$ and $a_n=3a_{n- 1}$ for all $n\ge2$.
Determine the last two digits of $a_{2007}$.

$\begin{array}{|rcll|} \hline a_1 &=& 3 \\ a_2 &=& 3*a_1 \\ &=& 3*3 \\ &=& 3^2 \\ a_3 &=& 3 * a_2 \\ &=& 3*3^2 \\ &=& 3^3 \\ \dots \\ \mathbf{ a_{2007} } &=& \mathbf{ 3^{2007} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && 3^{2007} \pmod{100} \\ && \boxed { \text{Euler:}\\ 3^{\phi(100)} \equiv 1 \pmod{100} \quad \phi{100}=100*\left(1-\dfrac12 \right)*\left(1-\dfrac15 \right)\\ \phi{100}=40\\ 3^{40} \equiv 1 \pmod{100} }\\ &\equiv& \left(3^{40}\right)^{50}*3^7 \pmod{100} \\ &\equiv& 1^{50}*3^7 \pmod{100} \\ &\equiv& 3^7 \pmod{100} \\ &\equiv& 2187 \pmod{100} \\ \mathbf{ 3^{2007} \pmod{100} } &\equiv& \mathbf{ 87 \pmod{100} } \\ \hline \end{array}$

Recall that a $\text{Fibonacci sequence}$ is one in which all entries
from the third onward are the sum of the two entries before it.
If the $5th$ and $8th$ entries of a Fibonacci Sequence are $-19$ and $79$, respectively,
then what is the first entry of this sequence?

$\begin{array}{|rcrcrl|} \hline a_1 &=& &a_1& \\ a_2 &=& && &a_2\\ a_3 &=& &a_1& + &a_2 \\ a_4 &=& &a_1& + &2a_2 \\ a_5 &=& &2a_1& + &3a_2 & \qquad\mathbf{ a_5 = 2a_1+3a_2 = -19} \\ a_6 &=& &3a_1& + &5a_2 \\ a_7 &=& &5a_1& + &8a_2 \\ a_8 &=& &8a_1& + &13a_2 & \qquad\mathbf{ a_8 = 8a_1+13a_2 = 79} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 2a_1+3a_2 &=& -19 \\ 3a_2 &=& -19 -2a_1 \\ \mathbf{ a_2 } &=& \mathbf{ \dfrac{-19-2a_1}{3} } \\ \hline 8a_1+13a_2 &=& 79 \\ 8a_1 + 13 \left( \dfrac{-19-2a_1}{3} \right) &=& 79 \quad | \quad * 3 \\\\ 24a_1 +13(-19-2a_1) &=& 79*3 \\ 24a_1 -13*19-26a_1 &=& 237 \\ -2a_1 -247 &=& 237 \\ -2a_1 &=& 484 \quad | \quad :(-2) \\ \mathbf{a_1} &=& \mathbf{-242} \\ \hline a_2 &=& \dfrac{-19-2a_1}{3} \\\\ a_2 &=& \dfrac{-19-2(-242)}{3} \\\\ a_2 &=& \dfrac{-19+484}{3} \\\\ a_2 &=& \dfrac{465}{3} \\\\ \mathbf{a_2} &=& \mathbf{155} \\ \hline \end{array}$

$\begin{array}{|rcr|} \hline a_1 &=& -242 \\ a_2 &=& 155 \\ a_3 &=& -87 \\ a_4 &=& 68 \\ a_5 &=& -19 \\ a_6 &=& 49 \\ a_7 &=& 30 \\ a_8 &=& 79 \\ \hline \end{array}$

What is the coefficient of $\color{red}x$ in $(x^4 + x^3 + x^2 + x + 1)^5$?

What is the coefficient of $\color{red}x$ in $(x^4 + x^3 + x^2 + x + 1)^5$?
What is the coefficient of $\color{red}x^3$ in $(x^4 + x^3 + x^2 + x + 1)^5$?

$(x^4 + x^3 + x^2 + x + 1)^5 =(1 + x^2 + x^3 + x^4)^5$

$\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^2} \\ =(1 + x^2 + x^3 + x^4)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&x &+&x^2 &+&x^3+\dots \\ && & & & &+&x^2 &+&x^3+\dots \\ && & & & && &+&x^3+\dots \\ \hline &=& &1&+&2x&+&3x^2&+&4x^3+\dots \\ \hline \end{array}$

$\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^3} \\ =(1 + 2x+3x^2+4x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&2x &+&2x^2 &+&2x^3+\dots \\ && & & & &+&3x^2 &+&3x^3+\dots \\ && & & & && &+&4x^3+\dots \\ \hline &=& &1&+&3x&+&6x^2&+&10x^3+\dots \\ \hline \end{array}$

$\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^4} \\ =(1 + 3x+6x^2+10x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&3x &+&3x^2 &+&3x^3+\dots \\ && & & & &+&6x^2 &+&6x^3+\dots \\ && & & & && &+&10x^3+\dots \\ \hline &=& &1&+&4x&+&10x^2&+&20x^3+\dots \\ \hline \end{array}$

$\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^5} \\ =(1 + 4x+10x^2+20x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&4x &+&4x^2 &+&4x^3+\dots \\ && & & & &+&10x^2 &+&10x^3+\dots \\ && & & & && &+&20x^3+\dots \\ \hline &=& &1&+&{\color{red}5}x&+&15x^2&+&{\color{red}35}x^3+\dots \\ \hline \end{array}$

What is the rule of this sequence?
$10,~15,~25,~40,~60,~\dots$

how do u get the rule and what type of sequence is this

$\begin{array}{|rcll|} \hline n & a_n & d_0 & d_1 \\ \hline 1 & 10 & \\ & & +5 \\ 2 & 15 & & +5 \\ & & +10 \\ 3 & 25 & & +5\\ & & +15 \\ 4 & 40 & & +5\\ & & +20\\ 5 & 60 & \\ \hline \end{array}$

This is an Arithmetic Sequences Second order
$\mathbf{an^2 + bn + c = a_n}$ with a, b, and c integers

$\begin{array}{|lrcll|} \hline n=1: & a*1^2 + b*1 + c &=& 10 \\ n=2: & a*2^2 + b*2 + c &=& 15 \\ n=3: & a*3^2 + b*3 + c &=& 40 \\ \hline (1) & a+b+c &=& 10 \\ (2) & 4a+2b+c &=& 15\\ (3) & 9a +3b + c &=& 25 \\ \Rightarrow & a&=& \dfrac{5}{2} \\ \Rightarrow & b&=& -\dfrac{5}{2} \\ \Rightarrow & c&=& 10 \\ \hline & a_n &=& \dfrac{5}{2}*n^2 -\dfrac{5}{2}*n + 10 \\ \hline \end{array}$

Simplify

$\sqrt{15-6\sqrt6-10\sqrt2+8\sqrt3}$

$\begin{array}{|rcll|} \hline 15 &=& 2+3+4+6 \\ 15 &=& (\sqrt2)^2+(\sqrt3)^2+(\sqrt4)^2+(\sqrt6)^2 \\ &&\boxed{ \text{Let }a=\sqrt2 \\ \text{Let }b=\sqrt3 \\ \text{Let }c=\sqrt4 \\ \text{Let }d=\sqrt6 }\\15 &=& a^2+b^2+c^2+d^2 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline (a-b-c+d)^2 &=& a^2+b^2+c^2+d^2-2ab-2ac+2ad+2bc-2bd-2cd \\ &&\boxed{ 2ab =2\sqrt6 \\ 2ac = 4\sqrt2 \\ 2ad = 4\sqrt3 \\ 2bc = 4\sqrt3 \\ 2bd = 6\sqrt2 \\ 2cd = 4\sqrt6 } \\ ( \sqrt2 - \sqrt3-\sqrt4+\sqrt6)^2 &=& a^2+b^2+c^2+d^2-2\sqrt6 - 4\sqrt2 + 4\sqrt3 + 4\sqrt3 -6\sqrt2 - 4\sqrt6 \\ ( \sqrt2 - \sqrt3-\sqrt4+\sqrt6)^2 &=& 15-6\sqrt6 - 10\sqrt2 + 8\sqrt3 \\ ( -2+\sqrt2 - \sqrt3+\sqrt6)^2 &=& 15-6\sqrt6 - 10\sqrt2 + 8\sqrt3 \\ \mathbf{-2+\sqrt2 - \sqrt3+\sqrt6} &=& \mathbf{ \sqrt{15-6\sqrt6 - 10\sqrt2 + 8\sqrt3} } \\ \hline \end{array}$

The real numbers x and y satisfy $(x - 3)^2 + (y - 4)^2 = 18$.

Find the minimum value of $x^2 + y^2$.

$\text{Let the origin $O=(0,0)$ } \\ \text{Let the center of the circle $C=(3,4)$ } \\ \text{Let the radius of the circle $r=\sqrt{18}=3\sqrt{2}$ } \\ \text{Let $x^2+y^2=r_{\text{min}}^2$ }$

1. Distance between origin and center of the circle:

$\begin{array}{|rcll|} \hline \overline{OC} &=& \sqrt{(0 - 3)^2 + (0 - 4)^2} \\ \overline{OC} &=& \sqrt{9+16} \\ \overline{OC} &=& \sqrt{25} \\ \mathbf{\overline{OC}} &=& \mathbf{5} \\ \hline \end{array}$

2. $x^2+y^2 = r^2_{\text{min}}$

$\begin{array}{|rcll|} \hline r_{\text{min}} &=& \overline{OC} - r \\ r_{\text{min}} &=& 5 -3\sqrt{2} \\ \hline x^2+y^2 &=& r_{\text{min}}^2 \\ x^2+y^2 &=& \left(5 -3\sqrt{2}\right)^2 \\ \mathbf{ x^2+y^2 } &=& \mathbf{ 43 -30\sqrt{2} } \\ \text{The minimum value of }~\mathbf{ x^2+y^2 } &=& \mathbf{ 0.5735931288 } \\ \hline \end{array}$

The ratio of the number of magnets Aspilla and Barry had was 2 : 3.
When Aspilla bought 4 more magnets and Barry bought 13 more magnets,
the ratio became 1 : 2.
Find the number of magnets Barry had in the end.

$\begin{array}{|rcll|} \hline \dfrac{A}{B} &=& \dfrac{2}{3} \\\\ \mathbf{A} &=& \mathbf{ \dfrac{2}{3}B } \\ \hline \dfrac{A+4}{B+13} &=& \dfrac{1}{2} \quad &|\quad \mathbf{A= \dfrac{2}{3}B } \\\\ \dfrac{\dfrac{2}{3}B+4}{B+13} &=& \dfrac{1}{2} \\\\ 2*\left(\dfrac{2}{3}B+4 \right) &=& B+13 \\\\ \dfrac{4}{3}B+8 &=& B+13 \\\\ \dfrac{4}{3}B-B &=& 5 \\\\ \dfrac{1}{3}B &=& 5 \quad &|\quad *3 \\\\ \mathbf{B} &=& \mathbf{15} \\ \hline A &=& \dfrac{2}{3}B \quad & | \quad \mathbf{B=15} \\\\ A &=& \dfrac{2}{3} * 15 \\\\ \mathbf{A} &=& \mathbf{10} \\ \hline \end{array}$

The number of magnets Barry had in the end is $B + 13 = 15+13 = \mathbf{28}$

Ryan had some blue and green buttons.

3/7 of the buttons are blue and the rest were green.

He gave away 2/5 of his buttons.

80 of the buttons given away were blue.

2/3 of the buttons left were green.

How many buttons did Ryan have altogether at first?

Compute $999^{-1} \pmod{1000}$.

Express your answer as an integer from 0 to 999.

Modular multiplicative inverse using Euler's theorem

$\small{ \begin{array}{|rcll|} \hline && 999^{-1} \pmod{1000} \\ &\equiv& \dfrac{1}{999} \pmod{1000} \quad & | \quad \gcd(1000,999)=1! \\ &\equiv& 999^{\phi(1000)-1} \pmod{1000} \quad & | \quad 999 \equiv -1 \pmod{1000} \\ &\equiv& (-1)^{\phi(1000)-1} \pmod{1000} \quad & | \quad \phi(1000) = 400~\text{ (Euler torent function)} \\ &\equiv& (-1)^{400-1} \pmod{1000} \\ &\equiv& (-1)^{399} \pmod{1000} \\ &\equiv& -1 \pmod{1000} \\ &\equiv& -1+1000 \pmod{1000} \\ \mathbf{ 999^{-1} \pmod{1000} }&\equiv& \mathbf{ 999 \pmod{1000}} \\ \hline \end{array} }$