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heureka

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 #2
avatar+26367 
+2

Four people put their hats on the table as they arrive.
When they leave, each person picks up one hat.
It happens so no-one has picked up his own hat.
In how many ways can this have happened?

 

 

This have happened in 9 ways.

 

Formula:

see link: https://dd233cdb-a-62cb3a1a-s-sites.googlegroups.com/site/vravi68/2003JIAMNSVR.pdf?attachauth=ANoY7crVxJ9vuMj0rFmoEN_zsclHPQHKIpvsEFaZoYCp0RA8or4qPKuffYEwM1lsziada9eoD7zmTsB44rENxa_nvA3t_A_-rcUnBvaclgqWMMH7ohYxfRnND2BVnpVeNXOM5d7kYV4W_EYJ1-FUBBaKLWWq3-plEZg2-A56zBzCDjZYmg_PZzW3hwkte4ipwpRfxSCtM_ybHJv4sBAsQfF_GEID5SlzDQ%3D%3D&attredirects=0

 

\(\text{The number of permutations with all the elements $\\$ displaced from their original position for $n = 4$ is:}\)
\(\begin{array}{|rcll|} \hline && ^4C_2(4-2)! -\ ^4C_3(4-3)! +\ ^4C_4(4-4)! \\\\ &=& ^4C_2\cdot 2! -\ ^4C_3\cdot 1! +\ ^4C_4\cdot 0! \\\\ &=& 6\cdot 2 - 4\cdot 1 + 1\cdot 1 \\\\ &=& 12 - 4 + 1 \\\\ &\mathbf{=}&\mathbf{ 9 } \\ \hline \end{array}\)

 

laugh

30 мая 2018 г.
 #6
avatar+26367 
+2

Thank you, Melody!

 

laugh

29 мая 2018 г.
 #4
avatar+26367 
+2

Find an integer 

\(x \)

such that

\(0 \leq x < 527\ \text{and} \ x^{37} \equiv 3 \pmod{527}.\)

0 \leq x < 527\ \text{and} \ x^{37} \equiv 3 \pmod{527}.

 

\(\begin{array}{|lrcll|} \hline & x^{37} &\equiv& 3 \pmod{527} \\ \text{or}& x^{37} &=& 3 + n\cdot 527 \quad & | \quad 527 = 17\cdot 31 \\ & x^{37} &=& 3 + n\cdot 17\cdot 31 \\\\ \pmod{17}: & x^{37} &\equiv& 3 \pmod{17} \qquad (1) \\ \pmod{31}: & x^{37} &\equiv& 3 \pmod{31} \qquad (2) \\ \hline \end{array}\)

 

\(\text{By Fermat's Little Theorem $a^{p-1}\equiv1\pmod{p}$ :}\)

\(\begin{array}{|lrcll|} \hline a=x \quad p = 17 : & x^{17-1}\equiv 1\pmod{17} \\ & \mathbf{x^{16}\equiv 1\pmod{17}} \\\\ a=x \quad p = 31 : & x^{31-1}\equiv 1\pmod{31} \\ & \mathbf{x^{30}\equiv 1\pmod{31}} \\ \hline \end{array}\)

 

\(\text{reduce $x^{37}$ to $x^5$ and $x^7$ :}\)

\(\begin{array}{|rcll|} \hline x^{37} \pmod{17} &\equiv& \underbrace{x^{16}}_{=1}\cdot \underbrace{x^{16}}_{=1}\cdot x^{5} \pmod{17} \quad & | \quad x^{16} \equiv 1\pmod{17} \\ x^{37} \pmod{17} &\equiv& x^{5} \pmod{17} \equiv 3 \pmod{17} \quad & | \quad \text{ see $(1)$ } \\ \mathbf{ x^{5}} & \equiv & \mathbf{ 3 \pmod{17} } \qquad (3) \\ \\ \hline \\ x^{37} \pmod{31} &\equiv& \underbrace{x^{30}}_{=1}\cdot x^{7} \pmod{31} \quad & | \quad x^{30} \equiv 1\pmod{31} \\ x^{37} \pmod{31} &\equiv& x^{7} \pmod{31} \equiv 3 \pmod{31} \quad & | \quad \text{ see $(2)$ } \\ \mathbf{ x^{7} } & \equiv & \mathbf{ 3 \pmod{31} } \qquad (4) \\ \hline \end{array}\)

 

\(\text{reduce $x^{5}$ to $x$ :} \)

\(\begin{array}{|lrcll|} \hline & x^{16} &\equiv& 1\pmod{17} \quad & | \quad \cdot x \\ & x^{16}x &\equiv& x\pmod{17} \\ & x^{17}&\equiv& x\pmod{17} \quad & | \quad \text{if $17 \pmod{16} \equiv 1 $} \\\\ \text{or} & x^5 &\equiv& 3 \pmod{17} \quad & | \quad \text{raise to the power s } \\ & x^{5s} &\equiv& x \pmod{17} \\ \text{if} & 5s &\equiv& 1 \pmod{16} \\\\ \text{solve }s : & \\ & s &\equiv& 5^{-1} \pmod{16} \\ & s &\equiv& 5^{\phi(16)-1} \pmod{16} \quad & | \quad 5 \text{ and } 16 \text{ are co-prime!} \\ & s &\equiv& 5^{8-1} \pmod{16} \quad & | \quad \phi(16)= 8 \\ & s &\equiv& 5^{7} \pmod{16} \\ & s &\equiv& 13 \pmod{16} \\\\ & x &\equiv& x^{5s} \pmod{17} \\ & &\equiv& x^{5\cdot 13} \pmod{17} \\ & &\equiv& (x^{5})^{13} \pmod{17} \quad & | \quad x^5 \equiv 3 \pmod{17} \\ & &\equiv& 3^{13} \pmod{17} \\ & \mathbf{x} & \mathbf{\equiv}& \mathbf{ 12 \pmod{17} } \\ \hline \end{array}\)

 

\(\text{reduce $x^{7}$ to $x$ :}\)

\(\begin{array}{|lrcll|} \hline & x^{30} &\equiv& 1\pmod{31} \quad & | \quad \cdot x \\ & x^{30}x &\equiv& x\pmod{31} \\ & x^{31}&\equiv& x\pmod{31} \quad & | \quad \text{if $31 \pmod{30} \equiv 1 $} \\\\ \text{or} & x^7 &\equiv& 3 \pmod{31} \quad & | \quad \text{raise to the power s } \\ & x^{7s} &\equiv& x \pmod{31} \\ \text{if} & 7s &\equiv& 1 \pmod{30} \\ \text{solve }s : & \\ & s &\equiv& 7^{-1} \pmod{30} \\ & s &\equiv& 7^{\phi(30)-1} \pmod{30} \quad & | \quad 7 \text{ and } 30 \text{ are co-prime!} \\ & s &\equiv& 7^{8-1} \pmod{30} \quad & | \quad \phi(30)= 8 \\ & s &\equiv& 7^{7} \pmod{30} \\ & s &\equiv& 13 \pmod{30} \\\\ & x &\equiv& x^{7s} \pmod{31} \\ & &\equiv& x^{7\cdot 13} \pmod{31} \\ & &\equiv& (x^{7})^{13} \pmod{31} \quad & | \quad x^7 \equiv 3 \pmod{31} \\ & &\equiv& 3^{13} \pmod{31} \\ & \mathbf{x} & \mathbf{\equiv}& \mathbf{ 24 \pmod{17} } \\ \hline \end{array}\)


\(\text{finally solve the system for $x$:}\)

\(\begin{array}{|lrcl|rl|rl|} \hline (1) & x &\equiv& 12 \pmod{17} \\ (2) & x &\equiv& 24 \pmod{31} \\ \\ \hline \\ &&& 17 \text{ and } 31 \text{ are co-prime!} \\ & x &=& 12 \cdot 31 \cdot 31^{-1} \pmod{17} & &31^{-1}\pmod{17}& &17^{-1}\pmod{31}\\ & &+& 24 \cdot 17 \cdot 17^{-1} \pmod{31} & \equiv&31^{\phi(17)-1}\pmod{17}& \equiv & 17^{\phi(31)-1}\pmod{31}\\ & &+& 17 \cdot 31 n \qquad n \in Z & \equiv & 31^{16-1}\pmod{17} & \equiv & 17^{30-1}\pmod{31}\\ & & & & \equiv & 31^{15}\pmod{17} & \equiv & 17^{29}\pmod{31}\\ & & & & \equiv & 14^{15}\pmod{17} & \equiv & (17^{2})^{14}\cdot 17\pmod{31}\\ & & & & \equiv & (14^2)^7\cdot 14 \pmod{17} & \equiv & 10^{14}\cdot 17 \pmod{31}\\ & & & & \equiv & 9^7\cdot 14 \pmod{17} & \equiv & (10^2)^{7}\cdot 17 \pmod{31}\\ & & & & \equiv & 11 \pmod{17} & \equiv & 7^{7}\cdot 17 \pmod{31}\\ & & & & & & \equiv & 11 \pmod{31}\\ & x &=& 12\cdot 31 \cdot 11 \\ & &+& 24 \cdot 17 \cdot 11 \\ & &+& 17 \cdot 31 n \\ & x &=& 4092+4488+527n \\ & x &=& \underbrace{8580}_{=148 \pmod {527 }}+527n \\ & x &=& 148 +527n \qquad 0 \leq x < 527 \qquad n=0!\\ & \mathbf{x} & \mathbf{=} & \mathbf{148} \\ \hline \end{array}\)

 

 

laugh

29 мая 2018 г.
 #2
avatar+26367 
+1

A rock is dropped from a bridge, When it's 1m from the water it takes 0.05 seconds to hit the water.

Calculate the height of the bridge.

 

\(\begin{array}{|lrcll|} \hline (1)& h &=& \frac{g}{2}t^2 \\ & t &=& \sqrt{\frac{2h}{g}} \\\\ (2)& h-1 &=& \frac{g}{2}(t-0.05)^2 \quad & | \quad t=\sqrt{\frac{2h}{g}}\\ & h-1 &=& \frac{g}{2} \left(\sqrt{\frac{2h}{g}}-0.05 \right)^2 \\ & h-1 &=& \frac{g}{2} \left(\frac{2h}{g} -2\cdot 0.05 \cdot \sqrt{\frac{2h}{g}} + 0.05^2 \right) \\ & h-1 &=& \frac{g}{2} \cdot \frac{2h}{g} -\frac{g}{2} \cdot 2\cdot 0.05 \cdot \sqrt{\frac{2h}{g}} + \frac{g}{2} \cdot 0.05^2 \\ & h-1 &=& h - 0.05\cdot g \cdot \sqrt{\frac{2h}{g}} + 0.05^2 \cdot \frac{g}{2} \\ & -1 &=& - 0.05\cdot g \cdot \sqrt{\frac{2h}{g}} + 0.05^2 \cdot \frac{g}{2} \\ & 0.05\cdot g \cdot \sqrt{\frac{2h}{g}} &=& 1+ 0.05^2 \cdot \frac{g}{2} \quad | \quad \text{square both sides} \\ & 0.05^2\cdot g^2 \cdot \frac{2h}{g} &=& \left(1+ 0.05^2 \cdot \frac{g}{2} \right)^2 \\ & 0.05^2\cdot g \cdot 2h &=& \left(1+ 0.05^2 \cdot \frac{g}{2} \right)^2 \\\\ & \mathbf{ h } & \mathbf{=} & \mathbf{ \dfrac{ \left(1+ 0.05^2 \cdot \frac{g}{2} \right)^2} {2\cdot 0.05^2\cdot g } } \\ \hline \end{array}\)

 

\(\text{Let $g = \frac{9.81\ m}{s^2}$ }\\ \text{$h$ the height of the bridge}: \)

 

\(\begin{array}{|rcll|} \hline h & = & \dfrac{ \left(1+ 0.05^2 \cdot \frac{g}{2} \right)^2} {2\cdot 0.05^2\cdot g } \\\\ & = & \dfrac{ \left(1+ 0.05^2 \cdot \frac{9.81}{2} \right)^2} {2\cdot 0.05^2\cdot 9.81 } \\\\ & = & \dfrac{ \left(1+ 0.05^2 \cdot \frac{9.81}{2} \right)^2} {2\cdot 0.05^2\cdot 9.81 } \\\\ & = & \dfrac{ 1.0122625^2} {0.04905 } \\\\ & = & \dfrac{ 1.02467536891} {0.04905 } \\\\ & = & 20.8904254619 \\ \hline \end{array}\)

 

The height of the bridge is  \(\approx 20.89\ m\)

 

laugh

28 мая 2018 г.
 #20
avatar+26367 
+1

Thank you Melody!

 

laugh

28 мая 2018 г.