heureka

45!/(45-5)!5!

\(\begin{array}{|rcll|} \hline && \frac{45!}{(45-5)!5!} \\ &=& \dbinom{45}{5} \\ &=& \frac{45}{5}\cdot \frac{44}{4}\cdot \frac{43}{3}\cdot \frac{42}{2}\cdot \frac{41}{1} \\ &=& \frac{45}{15}\cdot \frac{44}{4}\cdot 43\cdot \frac{42}{2}\cdot \frac{41}{1} \\ &=& 3\cdot 11\cdot 43\cdot 21\cdot 41 \\ &=& \mathbf{1221759} \\ \hline \end{array}\)

xy+4x=2880

xy-12y=2160

x and y in both equations is the same

\(\begin{array}{|lrcll|} \hline (1) & xy+4x &=& 2880 \\ (2) & xy-12y &=& 2160 \\ \hline \\ (1)-(2): & xy+4x -(xy-12y)&=& 2880-2160 \\ & xy+4x -xy+12y &=& 720 \\ & 4x+12y &=& 720 \quad & | \quad : 4 \\ & x+3y &=& 180 \\ (3) & x &=& 180-3y \\ \hline \end{array}\)

(3) in (2):

\(\begin{array}{|rcll|} \hline (2) & xy-12y &=& 2160 \\ & y\cdot (x-12) &=& 2160 \quad & | \quad x = 180-3y \\ & y\cdot (180-3y-12) &=& 2160 \\ & 180y-3y^2-12y &=& 2160 \\ & -3y^2+168y &=& 2160 \quad & | \quad : (-3) \\ & y^2-56y &=&-720 \\ & y^2-56y +720 &=& 0 \\ & (y-36)(y-20) &=& 0 \\\\ & y_1 = 36 &\text{or}& \quad y_2 = 20 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (3) & x &=& 180-3y \\ & x_1 &=& 180-3\cdot 36 \\ & x_1 &=& 72 \\\\ & x_2 &=& 180-3\cdot 20 \\ & x_2 &=& 120 \\ \hline \end{array}\)

Solutions: (72,36) and (120,20)

50! /(45! x5!)

\(\begin{array}{|rcll|} \hline && \frac{50!}{45!5!} \\ &=& \dbinom{50}{5} \\ &=& \frac{50}{5}\cdot \frac{49}{4}\cdot \frac{48}{3}\cdot \frac{47}{2}\cdot \frac{46}{1} \\ &=& \frac{50}{5}\cdot 49\cdot \frac{48}{12}\cdot \frac{47}{1}\cdot \frac{46}{2} \\ &=&10\cdot 49\cdot 4\cdot 47\cdot 23 \\ &=& \mathbf{2118760} \\ \hline \end{array}\)

(1/2)^-3 = ?

Formula:

\(\begin{array}{|rcll|} \hline \left(\dfrac{a}{b} \right)^{-n} &=& \left(\dfrac{b}{a} \right)^n \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \left(\dfrac{1}{2} \right)^{-3} \\ &=& \left(\dfrac{2}{1} \right)^3 \\ &=& 2^3 \\ &=& 8 \\ \hline \end{array}\)

Ist diese Formel zur Berechnung von Primzahlen korrekt?

p*(p-2)+(p-2)^2+p

"p" muss dabei eine Primzahl größer als 2 sein, da man sonst immer nur 2 oder keine Primzahl als Ergebnis bekommen würde.

Diese Formel ist nicht korrekt.

Für z.B. p= 17 ergibt sich:

\(\begin{array}{|rcll|} \hline && p\cdot(p-2)+(p-2)^2+p \\ && 17\cdot (17-2) + (17-2)^2 + 17 \\ &=& 17\cdot 15 + 15^2 + 17 \\ &=& 255 + 225 + 17 \\ &=& 497 (= 7 \cdot 71 ) \\ \hline \end{array}\)

497 ist keine Primzahl. Ihre Faktoren sind 7 und 71.

What is the smallest positive n under 50,000 that satisfies the following:

n mod 43 = 22 

n mod 101 = 64 

n mod 211 = 30 . Thanks for help.

\(\begin{array}{rcll} n &\equiv& {\color{red}22} \pmod {{\color{green}43}} \\ n &\equiv& {\color{red}64} \pmod {{\color{green}101}} \\ n &\equiv& {\color{red}30} \pmod {{\color{green}211}} \\ \text{Let } m &=& 43\cdot 101\cdot 211 = 916373 \\ \end{array}\)

43 is a prime number, and 101 is a prime number, and 211 is a prime number.

Because 43 and 101 and 211 are relatively prim ( gcd(43,101,211) = 1! ) we can go on:

\(\small{ \begin{array}{lcl} n = \\ {\color{red}22} \cdot {\color{green}101\cdot 211} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}101\cdot 211) }^{\varphi({\color{green}43}) -1 } \mod {{\color{green}43}} ] }_{=\text{modulo inverse }(101\cdot 211) \pmod {43} } }_{=(101\cdot 211)^{42-1} \pmod {43}} }_{=(101\cdot 211)^{41} \pmod {43}} }_{=(21311\pmod{43})^{41} \pmod {43}} }_{=(26)^{41} \pmod {43}} }_{=5}\\ + {\color{red}64} \cdot {\color{green}43\cdot 211} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}43\cdot 211) }^{\varphi({\color{green}101}) -1} \mod {{\color{green}101}} ] }_{=\text{modulo inverse } (43\cdot 211) \pmod {101} } }_{=(43\cdot 211)^{100-1} \pmod {101}} }_{=(43\cdot 211)^{99} \pmod {101}} }_{=(9073\pmod{101})^{99} \pmod {101}} }_{=(84)^{99} \pmod {101}} }_{=95}\\ + {\color{red}30} \cdot {\color{green}43\cdot 101} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}43\cdot 101) }^{\varphi({\color{green}211}) -1 } \mod {{\color{green}211}} ] }_{=\text{modulo inverse } (43\cdot 101) \pmod {211} } }_{=(43\cdot 101)^{210-1} \mod {211}} }_{=(43\cdot 101)^{209} \mod {211}} }_{=(4343\pmod{211})^{209} \mod {211}} }_{=(123)^{209} \mod {211}} }_{=199} \\ + {\color{green}43}\cdot {\color{green}101}\cdot {\color{green}211} \cdot k \quad | \quad k\in Z \\ n = {\color{red}22} \cdot {\color{green}101\cdot 211} \cdot [5] + {\color{red}64} \cdot {\color{green}43\cdot 211} \cdot [95] + {\color{red}30} \cdot {\color{green}43\cdot 101} \cdot [199] + {\color{green}43}\cdot {\color{green}101}\cdot {\color{green}211} \cdot k \\ n = 2344210+ 55163840 + 25927710 + {\color{green}43}\cdot {\color{green}101}\cdot {\color{green}211} \cdot k \\ n = 83435760 + {\color{green}916373}\cdot k \quad | \quad k\in Z \\\\ n_{min} = 83435760 \pmod {916373 } \\ n_{min} = 45817 \\\\ \mathbf{n} \mathbf{=} \mathbf{45817 + 916373 \cdot k \quad | \quad k\in Z} \end{array} }\)

The smallest positive n under 50,000 is 45817

Marie works at a toy store, and makes different amounts of money depending oon what is purchased.

1 toy box set of wooden blocks costs $5,

1 doll house costs $16.99,

2 books costs $3.69,

1 toy box set of dollhouse accessories is $8.50, and a box set of

5 suffed animals is $13.99.

The tax is 20%.

If Marie sold 15 dollhouses, 12 dollhouse accessory sets, 7 wooden block sets, 29 books, and 20 stuffed animal sets, how much did she make, if you include tax?

\(\begin{array}{|rcll|} \hline && 1.2\cdot ( 15 \cdot $16.99 + 12 \cdot $8.50 + 7 \cdot $5 +29 \cdot \frac{$3.69}{2} +20 \cdot \frac{$13.99}{5} ) \\ &=& 1.2\cdot $501.315 \\ &=& $601.578 \\ \hline \end{array} \)

She makes $601.578 include tax.

What is the sum to this question:
A hiker walked 3 1/6 miles yesterday and 2 1/2 miles today.

\(\begin{array}{|rcll|} \hline && 3 \frac{1}{6} \ \text{miles}+ 2 \frac{1}{2}\ \text{miles} \\ &=& 3+2 +\frac{1}{6} + \frac{1}{2} \\ &=& 5 +\frac{1}{6} + \frac{1}{2} \\ &=& 5 +\frac{1}{6} + \frac{1}{2}\cdot \frac{3}{3} \\ &=& 5 +\frac{1}{6} + \frac{3}{6} \\ &=& 5 +\frac{4}{6} \\ &=& 5 +\frac{2}{3} \\ &=& 5 \frac{2}{3}\ \text{miles} \\ &=& 5.\bar{6}\ \text{miles} \\ \hline \end{array}\)

18/4 divided by 16/17

\(\begin{array}{|rcll|} \hline && \dfrac{ \frac{18}{4} } { \frac{16}{17} } \\\\ &=& \dfrac{ \frac{9}{2} } { \frac{16}{17} } \\\\ &=& \frac{9}{2} \cdot \frac{17}{16} \\\\ &=& \frac{9\cdot 17}{2\cdot 16} \\\\ &=& \frac{153}{32} \\\\ &=& 4\frac{25}{32} \\\\ &=& 4.78125 \\ \hline \end{array} \)

At what time, t, does the object hit the ground? s(t)=t^2−4t−12

\(\begin{array}{|rcll|} \hline s(t) =t^2-4t-12 &=& 0 \\ t^2-4t-12 &=& 0 \\ (t-6)(t+2) &=& 0 \\\\ t = 6 &\text{or}& t = -2 \quad (\text{no solution, because t < 0})\\ \hline \end{array} \)

At time t = 6 the object does hit the ground.