heureka

Where does $2^{99}$ come from?

Pascal's triangle:

See the sum of the each line:
$\small{ \begin{array}{|l|rclcl|} \hline n \\ \hline 0 &\dbinom{0}{0} &=& 1 &=& 2^0 \\ 1 &\dbinom{1}{0}+\dbinom{1}{1} &=& 1+1 &=& 2^1 \\ 2 &\dbinom{2}{0}+\dbinom{2}{1}+\dbinom{2}{2} &=& 1+2+1 &=& 2^2 \\ 3 &\dbinom{3}{0}+\dbinom{3}{1}+\dbinom{3}{2}+\dbinom{3}{3} &=& 1+3+3+1 &=& 2^3 \\ 4 &\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} &=& 1+4+6+4+1 &=& 2^4 \\ 5 &\dbinom{5}{0}+\dbinom{5}{1}+\dbinom{5}{2}+\dbinom{5}{3}+\dbinom{5}{4}+\dbinom{5}{5} &=& 1+5+10+10+5+1 &=& 2^5 \\ \ldots \\ \color{red}99 & \dbinom{99}{0} + \dbinom{99}{1} + \dots + \dbinom{99}{98}+\dbinom{99}{99} &=&1+99+\dots + 99 + 1 &=& \color{red}2^{99} \\ \hline \end{array} }$

$\begin{array}{|rcll|} \hline && \color{red}(1+1)^n \\\\ &=&\sum \limits_{k=0}^{n}\dbinom{n}{k} \\ \\ &=& \dbinom{n}{0}+\dbinom{n}{1} + \dots + \dbinom{n}{n-1}+\dbinom{n}{n} \\\\ &=& \color{red}2^n \\ \hline \end{array}$

Find the value of the sum
$\dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{96}+ \dbinom{99}{98}$.

$\text{Formula: } \qquad \dbinom{n}{k}= \dbinom{n}{n-k}$

$\begin{array}{|rcll|} \hline && \mathbf{ \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{96}+ \dbinom{99}{98}} \\\\ &=& \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{48} \\ && +\dbinom{99}{98}+ \dbinom{99}{96}+ \dbinom{99}{94} + \dots +\dbinom{99}{50}\\\\ &=& \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4}+ \dbinom{99}{6} + \dots + \dbinom{99}{48} \\ && + \dbinom{99}{1}+ \dbinom{99}{3} + \dbinom{99}{5} + \dbinom{99}{7}+ \dots +\dbinom{99}{49}\\\\ &=& \dbinom{99}{0} + \dbinom{99}{1} + \dbinom{99}{2}+ \dbinom{99}{3} + \dbinom{99}{4}+ \dots \\ && + \dbinom{99}{46} + \dbinom{99}{47} + \dbinom{99}{48}+\dbinom{99}{49}\\\\ &=& \dfrac{2^{99}}{2} \\\\ &=& \mathbf{ 2^{98} } \\ \hline \end{array}$

algebra
Define
$a@b = ab - b^2$ and
$a\#b = a + b - ab$.
What is $(6@2)/(6\#2)$?

$\begin{array}{|rcll|} \hline (6@2)/(6\#2) &=& \dfrac{6@2}{6\#2)} \\\\ (6@2)/(6\#2) &=& \dfrac{6*2-2^2}{6+2-6*2} \\\\ (6@2)/(6\#2) &=& \dfrac{12-4}{8-12} \\\\ (6@2)/(6\#2) &=& \dfrac{8}{-4} \\\\ \mathbf{(6@2)/(6\#2)} &=& \mathbf{-2} \\ \hline \end{array}$

Find the complex number z that satisfies
$(1 + i) z - 2 \overline{z} = -11 + 25i$.

$\text{Set $z=a+bi$} \\ \text{Set $\overline{z}=a-bi$}$

$\begin{array}{|rcll|} \hline \mathbf{ (1 + i) z - 2 \overline{z} } &=& \mathbf{ -11 + 25i } \\ (1 + i)(a+bi) - 2(a-bi) &=& -11 + 25i \\ (1 + i)a+(1+i)bi - 2a+2bi &=& -11 + 25i \\ a + ia+bi+bi^2 - 2a+2bi &=& -11 + 25i \quad | \quad i^2 = -1 \\ a + ia+bi-b - 2a+2bi &=& -11 + 25i \\ -(a+b) +i(a+3b)&=& -11 + 25i \qquad \text{compare}\\ \hline -(a+b) &=& -11 \\ a+b &=& 11 \\ \mathbf{a} &=& \mathbf{11-b} \\ \hline (a+3b) &=& 25 \\ 11-b+3b &=& 25 \\ 2b &=& 14 \\ \mathbf{b} &=& \mathbf{7} \\ \hline \mathbf{a} &=& \mathbf{11-b} \\ a &=& 11-7 \\ \mathbf{a} &=& \mathbf{4} \\ \hline z &=& a+bi \\ \mathbf{z} &=& \mathbf{4+7i} \\ \hline \end{array}$

What is

$\sqrt{12 + \sqrt{12 + \sqrt{12 + \dotsb}}}$?

$\begin{array}{|rcll|} \hline x &=& \sqrt{12 + \sqrt{12 + \sqrt{12 + \dotsb}}} \qquad \text{square both sides}\\ x^2 &=& 12 + \sqrt{12 + \sqrt{12 + \sqrt{12 + \dotsb}}} \\ x^2 &=& 12 + x \\ \mathbf{x^2-x-12} &=& \mathbf{0} \\ \hline x &=& \dfrac{ 1\pm \sqrt{1^2-4(-12)} } {2} \\\\ x &=& \dfrac{ 1\pm \sqrt{49} } {2} \\\\ x &=& \dfrac{ 1\pm 7 } {2} \\\\ \hline x_1 &=& \dfrac{ 1- 7 } {2} \\\\ x_1 &=& -3 \qquad \text{no solution $x$ must $> 0$ !}\\\\ x_2 &=& \dfrac{ 1+ 7 } {2} \\\\ \mathbf{x_2} &=& \mathbf{4} \\ \hline \end{array}$

$\sqrt{12 + \sqrt{12 + \sqrt{12 + \dotsb}}} = 4$

What is the ones digit of
$1^{2009} + 2^{2009} + 3^{2009} + \ldots + 2009^{2009} + 2010^{2009} + 2011^{2009} + 2012^{2009}$?

$\small{ \begin{array}{|rcll|} \hline && \mathbf{ 1^{2009} + 2^{2009} + 3^{2009} + 4^{2009} + 5^{2009} } \\ && \mathbf{ + 6^{2009} + 7^{2009} + 8^{2009} + 9^{2009} + 10^{2009} } \pmod {10} \\ && \boxed{ 6 \equiv -4\pmod {10} \\ 7 \equiv -3\pmod {10} \\ 8 \equiv -2\pmod {10} \\ 9 \equiv -1\pmod {10} \\ 10 \equiv 0\pmod {10} } \\ &=& 1^{2009} + 2^{2009} + 3^{2009} + 4^{2009} + 5^{2009} \\ && + (-4)^{2009} + (-3)^{2009} + (-2)^{2009} + (-1)^{2009} + 0^{2009} \pmod {10} \\ &=& 1^{2009} + 2^{2009} + 3^{2009} + 4^{2009} + 5^{2009} \\ && -4^{2009} -3^{2009} -2^{2009} -1^{2009} \pmod {10} \\ &=& \mathbf{ 5^{2009}\pmod {10} } \\ \hline && \mathbf{ 11^{2009} + 12^{2009} + 13^{2009} + 14^{2009} + 15^{2009} } \\ && \mathbf{ + 16^{2009} + 17^{2009} + 18^{2009} + 19^{2009} + 20^{2009} }\pmod {10} \\ &=& 1^{2009} + 2^{2009} + 3^{2009} + 4^{2009} + 5^{2009}\\ && -4^{2009} -3^{2009} -2^{2009} -1^{2009} \pmod {10} \\ &=& \mathbf{ 5^{2009}\pmod {10} } \\ && \ldots \\ \hline && 1^{2009} + 2^{2009} + 3^{2009} + \ldots \\ && + 2009^{2009} + 2010^{2009} \\ && + 2011^{2009} + 2012^{2009} \pmod {10} \\\\ &=& 1^{2009} + 2^{2009} + 3^{2009} + 4^{2009} + 5^{2009} \\ && + 6^{2009} + 7^{2009} + 8^{2009} + 9^{2009} + 10^{2009} \\ && + 11^{2009} + 12^{2009} + 13^{2009} + 14^{2009} + 15^{2009} \\ && + 16^{2009} + 17^{2009} + 18^{2009} + 19^{2009} + 20^{2009} \\ && \ldots \\ && + 2011^{2009} + 2012^{2009} + 2013^{2009} + 2014^{2009} + 2015^{2009} \\ && + 2016^{2009} + 2017^{2009} + 2018^{2009} + 2019^{2009} + 2020^{2009} \\ && + 2011^{2009} + 2012^{2009} \pmod {10} \\\\ &=& 201 * 5^{2009} + 2011^{2009} + 2012^{2009} \pmod {10} \\ && \boxed{ 5^{2009} \equiv 5\pmod {10} ~\text{circle } ~5,5,\ldots \\ 2011^{2009} \equiv 1^{2009} \equiv 1\pmod {10} \\ 2012^{2009} \equiv 2^{2009}\pmod {10} \equiv 2\pmod {10} ~\text{circle } ~2,4,8,6,2\ldots } \\ &=& 201 * 5 + 1 + 2 \pmod {10} \\ && \boxed{201 \equiv 1 \pmod{10} } \\ &=& 1*5 + 1 + 2 \\ &=& \mathbf{8}\pmod{10} \\ \hline \end{array} }$

Let A and B be real numbers such that
$\dfrac{A}{x-5}+B*(x+1)=\dfrac{-3x^2+12x+22}{x-5}$.
What is A+B?

$\small{ \begin{array}{|lrcll|} \hline & \dfrac{A}{x-5}+B*(x+1) &=&\dfrac{-3x^2+12x+22}{x-5} \quad | \quad *(x-5) \\\\ & \mathbf{ A+B*(x+1)(x-5) } &=& \mathbf{ -3x^2+12x+22 } \\ \hline 1.~ x=-1:& A + 0 &=& -3*(-1)^2 + 12*(-1) + 22 \\\\ & A &=& -3-12+22 \\ & \mathbf{A} &=& \mathbf{ 7 } \\ \hline 2.~ x =-2:& 7+B*(-1)(-7) &=& -3*(-2)^2 + 12*(-2) + 22 \\ & 7+7B &=& -12-24+22 \\ & 7B &=& -14-7 \\ & 7B &=& -21 \quad | \quad : 7 \\ & \mathbf{B} &=& \mathbf{ -3 } \\ \hline & A+B &=& 7-3 \\ & \mathbf{ A+B } &=& \mathbf{ 4 } \\ \hline \end{array} }$

Positive real numbers $x$, $y$  satisfy the equations $x^2 + y^2 = 1$ and
$x^4 + y^4 = \dfrac{17}{19}$.
Find $xy$.

$\begin{array}{|rcll|} \hline \left( x^2 + y^2\right)^2 &=& x^4 + 2x^2y^2 + y^4 \\ \left( x^2 + y^2\right)^2 &=& x^4 + y^4 + 2x^2y^2 \quad | \quad x^2 + y^2 = 1,~ \dfrac{17}{19} \\\\ 1^2 &=& \dfrac{17}{19} + 2x^2y^2 \\\\ 2x^2y^2 &=& 1 - \dfrac{17}{19} \\\\ 2x^2y^2 &=& \dfrac{2}{19} \quad | \quad :2 \\\\ x^2y^2 &=& \dfrac{1}{19} \quad | \quad \text{sqrt both sides} \\\\ \mathbf{ xy } &=& \mathbf{ \dfrac{1}{\sqrt{19}} } \\ \hline \end{array}$

If  $a + b = 7$ and $a^3 + b^3 = 44$,
what is the value of the sum $\dfrac{1}{a} + \dfrac{1}{b}$?
Express your answer as a common fraction.

$\begin{array}{|rcll|} \hline (a+b)^3 &=& a^3+b^3+3ab(a+b) \quad | \quad a + b = 7,~a^3 + b^3 = 44 \\ 7^3 &=& 44 +21ab \\ 21ab &=& 343 - 44 \\ \mathbf{ ab } &=& \mathbf{ \dfrac{299}{21} } \\ \hline \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{a+b}{ab} \quad | \quad a + b = 7,~ ab=\dfrac{299}{21} \\\\ &=& \dfrac{7*21}{299} \\\\ \mathbf{ \dfrac{1}{a} + \dfrac{1}{b} } &=& \mathbf{ \dfrac{147}{299} } \\ \hline \end{array}$

I pick two whole numbers x and y between 1 and 10 inclusive (not necessarily distinct). My friend picks two numbers x - 4 and 2y + 1. If the product of my friend's numbers is one greater than the product of my numbers, then what is the product of my numbers?