heureka

Let \(n\) and \(k\) be positive integers such that \(\dbinom{n}{k}:\dbinom{n}{k + 1} = 4:11\). Find the smallest possible value of \(n\).

What is \(f(2010)\) if  \(3f(x)-5xf\left(\dfrac{1}{x}\right)=x-7\) 

\(\begin{array}{|lrcll|} \hline & \mathbf{3f(x) - 5xf\left(\dfrac{1}{x}\right)} &=& \mathbf{x-7} \qquad (1) \\ x\rightarrow \dfrac{1}{x}: & \mathbf{3f\left(\dfrac{1}{x}\right) - \dfrac{5}{x}f(x)} &=& \mathbf{\dfrac{1}{x}-7} \qquad (2) \\ & \boxed{\text{Let } a=f(x), \ b=f\left(\dfrac{1}{x}\right) } \\ & \mathbf{3a - 5xb} &=& \mathbf{x-7} \qquad (1) \\ & \mathbf{3b - \dfrac{5}{x}a} &=& \mathbf{\dfrac{1}{x}-7} \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (2): & \mathbf{3b - \dfrac{5}{x}a} &=& \mathbf{\dfrac{1}{x}-7} \\\\ & 3b &=& \dfrac{1}{x}-7 + \dfrac{5}{x}a \\\\ & \mathbf{b} &=& \mathbf{\dfrac{ \dfrac{1}{x}-7 + \dfrac{5}{x}a } {3}} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (1): & \mathbf{3a - 5xb} &=& \mathbf{x-7} \\\\ & 3a - 5x\left(\dfrac{ \dfrac{1}{x}-7 + \dfrac{5}{x}a } {3}\right) &=& x-7 \quad | \quad *3 \\\\ & 9a - 5x\left( \dfrac{1}{x}-7 + \dfrac{5}{x}a \right) &=& 3x - 21 \\\\ & 9a - 5 + 35x + 25a &=& 3x - 21 \\\\ & -16a+32x &=& -16 \quad | \quad :(-16) \\\\ & a-2x &=& 1 \\\\ & \mathbf{a} &=& \mathbf{2x+1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{2x+1} \quad | \quad x = 2010 \\\\ f(2010) &=& 2*2010+1 \\ \mathbf{f(2010)} &=& \mathbf{4021} \\ \hline \end{array}\)

Your friend has an egg collection comprising at least 200 eggs.
He wants to store them in dozen-egg containers.
After filling as many containers as possible, the last container had 1 egg left over.
He then decided to store his eggs in customized baker-dozen-egg containers, where each container can hold 13 eggs.
It turns out that, after filling as many of these containers as possible, he still has 1 egg left over.
What is the minimum number of eggs that your friend could have?

\(\begin{array}{|rcll|} \hline \text{eggs}_{\text{min.}} &=& 1+lcm(12,13)*n \qquad n\in \mathbb{N} \\ \text{eggs}_{\text{min.}} &=& 1+12*13*n \\ \text{eggs}_{\text{min.}} &=& 1+156*n \quad | \quad \text{eggs}_{\text{min.}} \mathbf{\geq 200 !},\ \text{so } n=2 \\ \text{eggs}_{\text{min.}} &=& 1+156* 2 \\ \mathbf{\text{eggs}_{\text{min.}}} &=& \mathbf{313} \\ \hline \end{array}\)

Sie können Mal, Geteilt, Plus und Minus in die Fragezeichen setzen.

Rechnung:

8 ? 4 ? 6 = 6 ? 7 ? 4

Zum Beispiel: \(8*4+6 = 6*7-4 \)

Daniel denkt sich eine Zahl, wenn er sie mit 7 multipliziert und dann 854 854 subtrahiert,

erhält er die kleinste fünfstellige Zahl.
Wie heißt die Zahl, die sich Daniel gedacht hat?

\(\begin{array}{|rcll|} \hline 7x - 854854 &>& 9999 \\ 7x &>& 9999 + 854854 \\ 7x &>& 864853 \\ x &>& \dfrac{864853}{7} \\ x &>& 123550.428571 \\ \mathbf{x} &=& \mathbf{123551} \\ \hline \end{array}\)

Die Zahl, die sich Daniel gedacht hat heißt 123551
7*123551-854854 = 10003

The system of equations \(\dfrac{xy}{x + y} = 1, \quad \dfrac{xz}{x + z} = 2, \quad \dfrac{yz}{y + z} = 3\)
has one ordered triple solution \((x,y,z)\).

What is the value of in this solution?

\(\begin{array}{|rcll|} \hline \dfrac{xy}{x + y} &=& 1 \\\\ \dfrac{x + y}{xy} &=& 1 \\\\ \dfrac{x}{xy}+\dfrac{y}{xy} &=& 1 \\\\ \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} &=& \mathbf{1} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{xz}{x + z} &=& 2 \\\\ \dfrac{x + z}{xz} &=& \dfrac{1}{2} \\\\ \dfrac{x}{xz}+\dfrac{z}{xz} &=& \dfrac{1}{2} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} &=& \mathbf{\dfrac{1}{2}} \qquad (2) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{yz}{y + z} &=& 3 \\\\ \dfrac{y + z}{yz} &=& \dfrac{1}{3} \\\\ \dfrac{y}{yz}+\dfrac{z}{yz} &=& \dfrac{1}{3} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} &=& \mathbf{\dfrac{1}{3}} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline -(1)+(2)+(3) : & -\dfrac{1}{y}-\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{y} &=& -1+\dfrac{1}{2}+\dfrac{1}{3} \\\\ & \dfrac{2}{z} &=& \dfrac{-1}{6} \\\\ & \dfrac{z}{2} &=& -6 \\\\ & \mathbf{z} &=& \mathbf{-12} \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)-(2)+(3) : & \dfrac{1}{y}+\dfrac{1}{x}- \dfrac{1}{z}-\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{y} &=& 1-\dfrac{1}{2}+\dfrac{1}{3} \\\\ & \dfrac{2}{y} &=& \dfrac{5}{6} \\\\ & \dfrac{y}{2} &=& \dfrac{6}{5} \\\\ & \mathbf{y} &=& \mathbf{\dfrac{12}{5}} \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)+(2)-(3) : & \dfrac{1}{y}+\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{x}-\dfrac{1}{z}-\dfrac{1}{y} &=& 1+\dfrac{1}{2}-\dfrac{1}{3} \\\\ & \dfrac{2}{x} &=& \dfrac{7}{6} \\\\ & \dfrac{x}{2} &=& \dfrac{6}{7} \\\\ & \mathbf{x} &=& \mathbf{\dfrac{12}{7}} \\ \hline \end{array} \)

In trapezoid \(PQRS\), \(\overline{PQ} \parallel \overline{RS}\).
Let \(X\) be the intersection of diagonals \(\overline{PR}\) and \(\overline{QS}\).
The area of triangle \(PQX\) is \(20\), and the area of triangle \(RSX\) is \(45\).
Find the area of trapezoid \(PQRS\).

\(\begin{array}{|rcll|} \hline \text{Let Area }A_1 &=& [PQX] \\ A_1 &=& 20 \\\\ \text{Let Area }A_2 &=& [RSX] \\ A_2 &=& 45 \\\\ \text{Let Area }A_3 &=& [PXS] \\\\ \text{Let Area }A_4 &=& [QXR] \\\\ \text{Let Area }A_5 &=& [PSR] \\ &=& A_2+A_3 \\\\ \text{Let Area }A_6 &=& [QSR] \\ A_6 &=& A_2 +A_4 \\ \hline \mathbf{A_5} &=& \mathbf{A_6} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{A_5} &=& \mathbf{A_6} \\ A_2+A_3 &=& A_2 +A_4 \\ \mathbf{A_3 }&=& \mathbf{A_4} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A_3 = \dfrac{SX*h_1}{2} && A_1 = \dfrac{QX*h_1}{2} \\ \mathbf{\dfrac{A_3}{A_1} }&=& \mathbf{\dfrac{SX}{QX}} \\\\ A_2 = \dfrac{SX*h_2}{2} && A_4 = \dfrac{QX*h_2}{2} \\ \mathbf{\dfrac{A_2}{A_4} }&=& \mathbf{\dfrac{SX}{QX}} \\\\ \dfrac{SX}{QX}=\mathbf{\dfrac{A_3}{A_1} }&=&\mathbf{\dfrac{A_2}{A_4} } \quad | \quad A_4 = A_3 \\\\ \dfrac{A_3}{A_1} &=& \dfrac{A_2}{A_3} \\\\ A_3^2 &=& A_1A_2 \\ A_3^2 &=& 20*45 \\ A_3^2 &=& 900 \\ \mathbf{A_3} &=& \mathbf{30} \\\\ A_4 &=& A_3 \\ \mathbf{A_4} &=& \mathbf{30} \\ \hline \end{array}\)

\(\text{The area of trapezoid $PQRS = A_1+A_2+A_3+A_4 = 20+45+30+30=\mathbf{125}$ }\)

​ I need Help

There are 38 boys, so \(38 - 18 = 20\) boys rest

There are 22 girls,  so \(22 - 9 = 13\) girls rest

a+b=13 b+c=24 c+a=19 what is the value of a+b+c.

\(\begin{array}{|lrcll|} \hline (1) & a+b=13 \\ (2) & b+c=24 \\ (3) & c+a=19 \\\\ (1)+(2)+(3): &(a+b)+(b+c)+(c+a) &=& 13+24+19 \\ & (a+b+c)+(a+b+c) &=& 13+24+19 \\ & 2(a+b+c) &=& 13+24+19 \\\\ & a+b+c &=& \dfrac{13+24+19}{2} \\\\ & a+b+c &=& \dfrac{56}{2} \\\\ & \mathbf{a+b+c} &=& \mathbf{28} \\ \hline \end{array}\)

A regular dodecahedron \(P_1 P_2 P_3 \dotsb P_{12}\) is inscribed in a circle with radius \(1\).
Compute  \((P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2\).
(The sum includes all terms of the form \((P_i P_j)^2\), where \(1 \le i < j \le 12\).

\(\small{ \begin{array}{|lrcll|} \hline \mathbf{\text{Compute}} \\ \hline &s=(P_1 P_2)^2 + (P_1 P_3)^2+(P_1 P_4)^2+(P_1 P_5)^2+(P_1 P_6)^2+(P_1 P_7)^2+(P_1 P_8)^2+(P_1 P_9)^2+(P_1 P_{10})^2+(P_1 P_{11})^2+(P_1 P_{12})^2 \\ & +(P_2 P_3)^2+(P_2 P_4)^2+(P_2 P_5)^2+(P_2 P_6)^2+(P_2 P_7)^2+(P_2 P_8)^2+(P_2 P_9)^2+(P_2 P_{10})^2+(P_2 P_{11})^2+(P_2 P_{12})^2 \\ & +(P_3 P_4)^2+(P_3 P_5)^2+(P_3 P_6)^2+(P_3 P_7)^2+(P_3 P_8)^2+(P_3 P_9)^2+(P_3 P_{10})^2+(P_3 P_{11})^2+(P_3 P_{12})^2 \\ & +(P_4 P_5)^2+(P_4 P_6)^2+(P_4 P_7)^2+(P_4 P_8)^2+(P_4 P_9)^2+(P_4 P_{10})^2+(P_4 P_{11})^2+(P_4 P_{12})^2 \\ & +(P_5 P_6)^2+(P_5 P_7)^2+(P_5 P_8)^2+(P_5 P_9)^2+(P_5 P_{10})^2+(P_5 P_{11})^2+(P_5 P_{12})^2 \\ & +(P_6 P_7)^2+(P_6 P_8)^2+(P_6 P_9)^2+(P_6 P_{10})^2+(P_6 P_{11})^2+(P_6 P_{12})^2 \\ & +(P_7 P_8)^2+(P_7 P_9)^2+(P_7 P_{10})^2+(P_7 P_{11})^2+(P_7 P_{12})^2 \\ & +(P_8 P_9)^2+(P_8 P_{10})^2+(P_8 P_{11})^2+(P_8 P_{12})^2 \\ & +(P_9 P_{10})^2+(P_9 P_{11})^2+(P_9 P_{12})^2 \\ & +(P_{10} P_{11})^2+(P_{10} P_{12})^2 \\ & +(P_{11} P_{12})^2 \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline (P_1 P_2)=(P_2 P_3)=(P_3 P_4)=(P_4 P_5)=(P_5 P_6)=(P_6 P_7)=(P_7 P_8) \\ =(P_8 P_9)=(P_9 P_{10})=(P_{10} P_{11})=(P_{11} P_{12})=(P_1 P_{12})= 2*\sin(15^\circ) \\\\ (P_1 P_2)=(P_2 P_4)=(P_3 P_5)=(P_4 P_6)=(P_5 P_7)=(P_6 P_8)=(P_7 P_9) \\ =(P_8 P_{10})=(P_9 P_{11})=(P_{10} P_{12})=(P_{1} P_{11})=(P_2 P_{12})= 2*\sin(30^\circ) \\\\ (P_1 P_4)=(P_2 P_5)=(P_3 P_6)=(P_4 P_7)=(P_5 P_8)=(P_6 P_9)=(P_7 P_{10}) \\ =(P_8 P_{11})=(P_9 P_{12})=(P_{1} P_{10})=(P_{2} P_{11})=(P_3 P_{12})= 2*\sin(45^\circ) \\\\ (P_1 P_5)=(P_2 P_6)=(P_3 P_7)=(P_4 P_8)=(P_5 P_9)=(P_6 P_{10})=(P_7 P_{11}) \\ =(P_8 P_{12})=(P_1 P_{9})=(P_{2} P_{10})=(P_{3} P_{11})=(P_5 P_{12})= 2*\sin(60^\circ) \\\\ (P_1 P_6)=(P_2 P_7)=(P_3 P_8)=(P_4 P_9)=(P_5 P_{10})=(P_6 P_{11})=(P_7 P_{12}) \\ =(P_1 P_{8})=(P_2 P_{9})=(P_{3} P_{10})=(P_{4} P_{11})=(P_5 P_{12})= 2*\sin(75^\circ) \\\\ (P_1 P_7)=(P_2 P_8)=(P_3 P_9)=(P_4 P_{10})=(P_5 P_{11})=(P_6 P_{12})= 2*\sin(90^\circ)=2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline s&=& 12*(2*\sin(15^\circ))^2 + 12*(2*\sin(30^\circ))^2+12*(2*\sin(45^\circ))^2 \\ && +12*(2*\sin(60^\circ))^2 +12*(2*\sin(75^\circ))^2+6*(2*\sin(90^\circ))^2 \\\\ s&=& 48\sin^2(15^\circ) + 48\sin^2(30^\circ)+48\sin^2(45^\circ) \\ && +48\sin^2(60^\circ) + 48\sin^2(75^\circ)+24 \\\\ s&=& 48\left( \sin^2(15^\circ) + \sin^2(30^\circ)+\sin^2(45^\circ)+\sin^2(60^\circ) + \sin^2(75^\circ) \right) +24 \\\\ s&=& 48\left( \sin^2(15^\circ) + \sin^2(30^\circ)+\sin^2(45^\circ)+\cos^2(30^\circ) + \cos^2(15^\circ) \right) +24 \\\\ s&=& 48\left( \sin^2(15^\circ) + \cos^2(15^\circ) + \sin^2(30^\circ)+\cos^2(30^\circ)+\sin^2(45^\circ)\right) +24 \\\\ s&=& 48\left( 2+\sin^2(45^\circ)\right) +24 \quad | \quad \sin(45^\circ)=\dfrac{\sqrt{2}}{2} \\\\ s&=& 48\left( 2+\dfrac{1}{2}\right) +24 \\\\ s&=& \dfrac{48*5}{2} +24 \\\\ s&=& 24*5 +24 \\\\ s&=& 24*6 \\\\ \mathbf{s}&=& \mathbf{144} \\\\ \mathbf{s} &=& \mathbf{12^2} \\ \hline \end{array}\)

The sum is \(\mathbf{12^2 = 144}\)