heureka

PUZZLE 1 

10,50,5,25,and 500. 

Use these numbers to make the number 575.

$575 = \frac{5\cdot 25\cdot 50-500}{10} \\ 575 = \frac{10\cdot 50}{5}-25 +500 \\ 575 = 500-50 +(10-5)\cdot 25 \\ 575 = 500 + [50- (10+25) ]\cdot 5 \\ $

Find the volume of this prism.

$\begin{array}{|rcll|} \hline V &=& 15\ cm \cdot 12\ cm \cdot 6\ cm - (15\ cm - 5\ cm) \cdot (12\ cm - 8\ cm) \cdot 6\ cm \\ V &=& 15\ cm \cdot 12\ cm \cdot 6\ cm - 10\ cm \cdot 4\ cm \cdot 6\ cm \\ V &=& 1080\ cm^3 - 240\ cm^3 \\ V &=& 840\ cm^3 \\ \hline \end{array}$

Search Results 102 + 101 + 84.3 + 83.8 + 82 + 81.51 + 65.30 + 61.91 + 53.4 + 49.10 + 32.93 + 30.75 + 30 + 24 + 24.42 + 13.56 + 12 + 10.62 + 10 + 11 + 9.26 + 9.13 + 4 + 3.4 + 3.5 + 3 + 3 + 3 + 2 + 2 + 1.5 + 1 + 1 + 1 + 1 = 1011

(tan15∘)x(tan25∘)x(tan35∘)x(tan85∘)

Formula:

$\begin{array}{|lcll|} \hline \sin (x) \; \sin (y) = \frac{1}{2}\Big(\cos (x-y) - \cos (x+y)\Big) \\ \cos (x) \; \cos (y) = \frac{1}{2}\Big(\cos (x-y) + \cos (x+y)\Big) \\ \sin (2x) = 2\cdot \sin(x)\cdot \cos(x) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \tan(15^{\circ})\cdot \tan(25^{\circ})\cdot \tan(35^{\circ})\cdot \tan(85^{\circ}) \\ &=& \frac{\sin(15^{\circ})\cdot \sin(25^{\circ})\cdot \sin(35^{\circ})\cdot \sin(85^{\circ})} {\cos(15^{\circ})\cdot \cos(25^{\circ})\cdot \cos(35^{\circ})\cdot \cos(85^{\circ})} \\ &=& \frac{ [\sin(85^{\circ})\cdot\sin(15^{\circ})]\cdot [\sin(35^{\circ})\cdot \sin(25^{\circ})] } { [\cos(85^{\circ})\cdot\cos(15^{\circ})]\cdot [\cos(35^{\circ})\cdot \cos(25^{\circ})] } \\\\ && \sin(85^{\circ})\cdot\sin(15^{\circ}) = \frac{1}{2}\Big(\cos (85^{\circ}-15^{\circ}) - \cos (85^{\circ}+15^{\circ})\Big) \\ && \mathbf{ \sin(85^{\circ})\cdot\sin(15^{\circ}) = \frac{1}{2}\Big(\cos (70^{\circ}) - \cos (100^{\circ})\Big) } \\\\ && \sin(35^{\circ})\cdot\sin(25^{\circ}) = \frac{1}{2}\Big(\cos (35^{\circ}-25^{\circ}) - \cos (35^{\circ}+25^{\circ})\Big) \\ && \mathbf{ \sin(35^{\circ})\cdot\sin(25^{\circ}) = \frac{1}{2}\Big(\cos (10^{\circ}) - \cos (60^{\circ})\Big) } \\\\ && \cos(85^{\circ})\cdot\cos(15^{\circ}) = \frac{1}{2}\Big(\cos (85^{\circ}-15^{\circ}) + \cos (85^{\circ}+15^{\circ})\Big) \\ && \mathbf{ \cos(85^{\circ})\cdot\cos(15^{\circ}) = \frac{1}{2}\Big(\cos (70^{\circ}) + \cos (100^{\circ})\Big) } \\\\ && \cos(35^{\circ})\cdot\cos(25^{\circ}) = \frac{1}{2}\Big(\cos (35^{\circ}-25^{\circ}) + \cos (35^{\circ}+25^{\circ})\Big) \\ && \mathbf{ \cos(35^{\circ})\cdot\cos(25^{\circ}) = \frac{1}{2}\Big(\cos (10^{\circ}) + \cos (60^{\circ})\Big) } \\\\ &=& \frac{ \frac{1}{2}\Big(\cos (70^{\circ}) - \cos (100^{\circ})\Big) \cdot \frac{1}{2}\Big(\cos (10^{\circ}) - \cos (60^{\circ})\Big) } { \frac{1}{2}\Big(\cos (70^{\circ}) + \cos (100^{\circ})\Big) \cdot \frac{1}{2}\Big(\cos (10^{\circ}) + \cos (60^{\circ})\Big) } \\\\ &=& \frac{ \Big(\cos (70^{\circ}) - \cos (100^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) - \cos (60^{\circ})\Big) } { \Big(\cos (70^{\circ}) + \cos (100^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) + \cos (60^{\circ})\Big) } \quad | \quad \cos(100^{\circ}) = \cos(90^{\circ}+10^{\circ})=-\sin(10^{\circ}) \\\\ &=& \frac{ \Big(\cos (70^{\circ}) + \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) - \cos (60^{\circ})\Big) } { \Big(\cos (70^{\circ}) - \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) + \cos (60^{\circ})\Big) } \quad | \quad \cos(70^{\circ}) = \cos(90^{\circ}-20^{\circ})=\sin(20^{\circ}) \\\\ &=& \frac{ \Big(\sin(20^{\circ}) + \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) - \cos (60^{\circ})\Big) } { \Big(\sin(20^{\circ}) - \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) + \cos (60^{\circ})\Big) } \quad | \quad \cos(60^{\circ}) = \frac{1}{2} \\\\ &=& \frac{ \Big(\sin(20^{\circ}) + \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) - \frac{1}{2}\Big) } { \Big(\sin(20^{\circ}) - \sin(10^{\circ})\Big) \cdot \Big(\cos (10^{\circ}) + \frac{1}{2}\Big) } \\\\ &=& \frac{ \sin(20^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2}\cdot \sin(20^{\circ})+\sin(10^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } { \sin(20^{\circ})\cdot \cos (10^{\circ}) + \frac{1}{2}\cdot \sin(20^{\circ})-\sin(10^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } \quad | \quad \sin(10^{\circ})\cdot \cos (10^{\circ}) = \frac{1}{2}\cdot \sin(20^{\circ}) \\\\ &=& \frac{ \sin(20^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2}\cdot \sin(20^{\circ})+\frac{1}{2}\cdot \sin(20^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } { \sin(20^{\circ})\cdot \cos (10^{\circ}) + \frac{1}{2}\cdot \sin(20^{\circ})-\frac{1}{2}\cdot \sin(20^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } \quad | \quad \frac{1}{2}\cdot \sin(20^{\circ})-\frac{1}{2}\cdot \sin(20^{\circ}) = 0 \\\\ &=& \frac{ \sin(20^{\circ})\cdot \cos (10^{\circ}) + 0 - \frac{1}{2} \cdot \sin(10^{\circ}) } { \sin(20^{\circ})\cdot \cos (10^{\circ}) + 0 - \frac{1}{2} \cdot \sin(10^{\circ}) } \\\\ &=& \frac{ \sin(20^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } { \sin(20^{\circ})\cdot \cos (10^{\circ}) - \frac{1}{2} \cdot \sin(10^{\circ}) } \\\\ &=& 1 \\ \hline \end{array}$

knowing that Tan(45)=1, find Sin(45) and Cos(45) with the fonamental relations

$\tan(45^{\circ}) = 1 \\ \cot(45^{\circ}) = \frac{1}{\tan(45^{\circ})} = \frac{1}{1}=1$

Formula:

$\begin{array}{|rcll|} \hline \sin^2(x)+\cos^2(x) &=& 1 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \sin^2(45^{\circ})+\cos^2(45^{\circ}) &=& 1 \quad & | \quad : \cos^2(45^{\circ}) \\ \frac{ \sin^2(45^{\circ})+\cos^2(45^{\circ}) } {\cos^2(45^{\circ})} &=& \frac{ 1 } {\cos^2(45^{\circ})} \\ \frac{ \sin^2(45^{\circ}) }{\cos^2(45^{\circ})} + \frac{ \cos^2(45^{\circ}) } {\cos^2(45^{\circ})} &=& \frac{ 1 } {\cos^2(45^{\circ})} \\ \tan^2(45^{\circ}) + 1 &=& \frac{ 1 } {\cos^2(45^{\circ})} \quad & | \quad \tan(45^{\circ}) = 1 \\ 1 + 1 &=& \frac{ 1 } {\cos^2(45^{\circ})} \\ 2 &=& \frac{ 1 } {\cos^2(45^{\circ})} \quad & | \quad \text{square root both sides} \\ \sqrt{2} &=& \frac{ 1 } {\cos(45^{\circ})} \\ \mathbf{ \frac{1}{\sqrt{2}} } & \mathbf{=} & \mathbf{ \cos(45^{\circ}) } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \sin^2(45^{\circ})+\cos^2(45^{\circ}) &=& 1 \quad & | \quad : \sin^2(45^{\circ}) \\ \frac{ \sin^2(45^{\circ})+\cos^2(45^{\circ}) } {\sin^2(x)} &=& \frac{ 1 } {\sin^2(45^{\circ})} \\ \frac{ \sin^2(45^{\circ}) }{\sin^2(45^{\circ})} + \frac{ \cos^2(45^{\circ}) } {\sin^2(45^{\circ})} &=& \frac{ 1 } {\sin^2(45^{\circ})} \\ 1+ \cot^2(45^{\circ}) &=& \frac{ 1 } {\sin^2(45^{\circ})} \quad & | \quad \cot(45^{\circ}) = 1 \\ 1+ 1 &=& \frac{ 1 } {\sin^2(45^{\circ})} \\ 2 &=& \frac{ 1 } {\sin^2(45^{\circ})} \quad & | \quad \text{square root both sides} \\ \sqrt{2} &=& \frac{ 1 } {\sin(45^{\circ})} \\ \mathbf{ \frac{1}{\sqrt{2}} } & \mathbf{=} & \mathbf{ \sin(45^{\circ}) } \\ \hline \end{array}$

Wie lange dauert es, bis sich bei 7,5 % Zinseszns p.a. en angelegtes Kapital vervierfacht?

K₀ ist das Anfangskapital

K_n ist das Kapital nach n Jahren $= K_0 \cdot 1,075^n$

$\begin{array}{|rcll|} \hline K_n &=& K_0 \cdot 1,075^n \quad & | \quad \text{angelegtes Kapital vervierfacht }:\ K_n = 4\cdot K_0 \\ 4\cdot K_0 &=& K_0 \cdot 1,075^n \\ 4 &=& 1.075^n \\ \log(4) &=& \log(1,075^n) \\ \log(4) &=& n\cdot \log(1,075) \\ n &=& \frac{\log(4)} {\log(1,075)} \\ n &=& \frac{0,60205999133} {0,03140846425} \\ n &=& 19,1687179133 \\ \hline \end{array}$

Das angelegte Kapital hat sich nach circa 19 Jahren vervierfacht.

|2x+13|=19

$\begin{array}{|rcll|} \hline |2x+13| &=& 19 \quad & | \quad \text{square both sides} \\ (2x+13)^2 &=& 19^2 \\ 4x^2+52x + 13^2 &=& 19^2 \\ 4x^2+52x + 13^2 -19^2 &=& 0 \\ 4x^2+52x -192 &=& 0 \quad & | \quad : 4 \\ x^2+13x -48 &=& 0 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x &=& \frac{-b \pm \sqrt{b^2-4ac}} {2a} \\\\ x^2+13x -48 &=& 0 \quad & | \quad a=1 \quad b=13 \quad c=-48 \\ x &=& \frac{-13 \pm \sqrt{13^2-4\cdot(-48)}} {2} \\ x &=& \frac{-13 \pm \sqrt{169+192}} {2} \\ x &=& \frac{-13 \pm \sqrt{361}} {2} \\ x &=& \frac{-13 \pm 19 } {2} \\\\ x_1 &=& \frac{-13 + 19 } {2} \\ x_1 &=& \frac{6} {2} \\ \mathbf{x_1} &\mathbf{=}& \mathbf{3} \\\\ x_2 &=& \frac{-13 - 19 } {2} \\ x_2 &=& \frac{-32 } {2} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{-16} \\ \hline \end{array}$

When multiplied out, the number 2017!=1x2x3....x2015x2016x2017 end with a striking of zeros.

How many zeros are there at the end of this number.

Let n = 2017

Legendre's Theorem - The Prime Factorization of Factorials:

$\begin{array}{|r|r|c|l|} \hline \text{p=factor} & \text{to base}_p & s=\text{ the sum of all the digits} & \text{exponent }= \frac{n-s} {\text{p }-1} \\ & & \text{in the expansion of n in base p } & \\ \hline 2 & 11111100001_2 & 7 & \frac{2017-7}{2-1} = 2010 \\ 3 & 2202201_3 & 9 & \frac{2017-9}{3-1} = 1004 \\ 5 & 31032_5 & 9 & \frac{2017-9}{5-1} = 502 \\ \cdots & \cdots & \cdots & \cdots \\ \hline \end{array}$

Prime factorization on 2017!:

$2^{2010}\times 3^{1004}\times 5^{502}\times 7^{334}\times 11^{200}\times \ldots \times 1999\times 2003\times 2011\times 2017$

Zeros at the end of 2017!:

$\begin{array}{|rcll|} \hline 2^{502} \times 5^{502} &=& (2\cdot5)^{502}=10^{502} \\ \hline \end{array}$

There are 502 zeros at the end of 2017!

Let ABCDEFGH be a cube of side length 5, as shown.

Let P and Q be points on line AB and line AE, respectively, such that AP = 2 and AQ = 1.

The plane through C, P, and Q intersects line DH at R. Find DR. 

$Let\ \vec{C} = \langle 0,0,0 \rangle\\ Let\ \vec{P} = \langle 5,3,0 \rangle\\ Let\ \vec{Q} = \langle 5,5,1 \rangle\\ Let\ DR = z \\ Let\ \vec{R} = \langle 0,5,z \rangle$

The plane through C, P, and Q:

$\begin{array}{|rcll|} \hline \vec{x} &=& \vec{C} + s \cdot(\vec{P}-\vec{C}) + t \cdot(\vec{Q}-\vec{C}) \quad & | \quad \vec{x} = \vec{R} \quad \vec{C} = \vec{0} \\ \vec{R} &=& \vec{0} + s \cdot(\vec{P}-\vec{0}) + t \cdot(\vec{Q}-\vec{0}) \\ \vec{R} &=& s \cdot \vec{P} + t \cdot \vec{Q} \quad & | \quad \vec{P} =\begin{pmatrix}5 \\ 3 \\ 0\end{pmatrix} \quad \vec{Q}=\begin{pmatrix}5 \\ 5 \\ 1\end{pmatrix} \quad \vec{R}=\begin{pmatrix}0 \\ 5 \\ z\end{pmatrix} \\ \begin{pmatrix}0 \\ 5 \\ z\end{pmatrix} &=& s \cdot \begin{pmatrix}5 \\ 3 \\ 0\end{pmatrix} + t \cdot \begin{pmatrix}5 \\ 5 \\ 1\end{pmatrix} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (1) & 0&=& 5s+5t \\ & 0&=& s+ t \\ & s&=& -t \\\\ (2) & 5&=& 3s+5t \\ & 5&=& 3(-t) + 5t \\ & 5&=& 2t \\ & t&=& \frac{5}{2} \\\\ (3) & z&=& 0\cdot s + t \\ & z&=& t \\ & z&=& \frac{5}{2} \\ \hline \end{array}$

DR = $\frac{5}{2}$

Given the two vectors below, find the angle between them. Round answers to the nearest hundredth.

v=⟨−13,−1⟩ w=⟨19,−3.1⟩

$\vec{v} = \binom{-13}{-1} \\ \vec{w} = \binom{19}{-3.1}$

$\begin{array}{|rcll|} \hline \tan(\varphi) &=& \frac{ |\vec{v} \times \vec{w}| } {\vec{v} \cdot \vec{w}} \\ &=& \frac{ \left|\binom{-13}{-1} \times \binom{19}{-3.1} \right| } {\binom{-13}{-1} \cdot \binom{19}{-3.1} } \\ &=& \frac{ (-13)\cdot(-3.1)-(-1)\cdot 19 } {(-13)\cdot 19 + (-1)\cdot(-3.1) } \\ &=& \frac{ 40.3+19 } {-247 + 3.1 } \\ &=& \frac{ 59.3 } {-243.9 } \quad & | \quad \frac{+}{-}\ \text{Quadrant }\ II. \\ &=& -\frac{ 59.3 } {243.9 } \\ \tan(\varphi) &=& -0.24313243132 \\ \varphi &=& \arctan( -0.24313243132 ) + 180^{\circ} \\ &=& -13.6653125958+ 180^{\circ} \\ \varphi &=& 166.334687404^{\circ} \\ \hline \end{array}$

The angle between $\vec{v}$ and $\vec{w}$ is $166.33^{\circ}$