heureka

Find the sum of the orders mod 83 over all elements of the \(set \{1,2,3,\ldots,82\}\).

(If multiple elements have the same order, add that term multiple times.)

 ord83(1)=1
 ord83(2)=82
 ord83(3)=41
 ord83(4)=41
 ord83(5)=82
 ord83(6)=82
 ord83(7)=41
 ord83(8)=82
 ord83(9)=41
 ord83(10)=41
 ord83(11)=41
 ord83(12)=41
 ord83(13)=82
 ord83(14)=82
 ord83(15)=82
 ord83(16)=41
 ord83(17)=41
 ord83(18)=82
 ord83(19)=82
 ord83(20)=82
 ord83(21)=41
 ord83(22)=82
 ord83(23)=41
 ord83(24)=82
 ord83(25)=41
 ord83(26)=41
 ord83(27)=41
 ord83(28)=41
 ord83(29)=41
 ord83(30)=41
 ord83(31)=41
 ord83(32)=82
 ord83(33)=41
 ord83(34)=82
 ord83(35)=82
 ord83(36)=41
 ord83(37)=41
 ord83(38)=41
 ord83(39)=82
 ord83(40)=41
 ord83(41)=41
 ord83(42)=82
 ord83(43)=82
 ord83(44)=41
 ord83(45)=82
 ord83(46)=82
 ord83(47)=82
 ord83(48)=41
 ord83(49)=41
 ord83(50)=82
 ord83(51)=41
 ord83(52)=82
 ord83(53)=82
 ord83(54)=82
 ord83(55)=82
 ord83(56)=82
 ord83(57)=82
 ord83(58)=82
 ord83(59)=41
 ord83(60)=82
 ord83(61)=41
 ord83(62)=82
 ord83(63)=41
 ord83(64)=41
 ord83(65)=41
 ord83(66)=82
 ord83(67)=82
 ord83(68)=41
 ord83(69)=41
 ord83(70)=41
 ord83(71)=82
 ord83(72)=82
 ord83(73)=82
 ord83(74)=82
 ord83(75)=41
 ord83(76)=82
 ord83(77)=41
 ord83(78)=41
 ord83(79)=82
 ord83(80)=82
 ord83(81)=41
 ord83(82)=2

 sum = 4923

A country has three denominations of coins, worth 7, 10, and 53 units of value.

What is the maximum number of units of currency which one cannot have if they are only carrying these three kinds of coins?

The maximum number of units of currency which one cannot have if they are only carrying these three kinds of coins is 46

see McNugget-Problem

Find the order of 15 mod 257.

1. 257 is a prime and
2. \(257-1 = 2^8\) is a power of a prime, so
3. the order of 257 must also be a power of two. The order is \(2^k\) for some \(k= 0,1,2,\ldots, 8\)

Test:

\(\begin{array}{|r|r|l|} \hline k & 2^k \\ \hline 0 & 1 & 15^1 \equiv 15 \pmod{257} \\ 1 & 2 & 15^2 \equiv 225 \pmod{257} \\ 2 & 4 & 15^4 \equiv 253 \pmod{257} \\ 3 & 8 & 15^8 \equiv 16 \pmod{257} \\ 4 & 16 & 15^{16} \equiv 256 \pmod{257} \\ 5 & \color{red}32& 15^{32} \equiv {\color{red}1} \pmod{257} \\ \hline \end{array}\)

The order of 15 mod 257 is 32

Let \(z\) and \(w\) be complex numbers satisfying \(|z| = 5\), \(|w| = 2\), and \(z\overline{w} = 6+8i\). 

Find the value of \(|z+w|^2\), \(|zw|^2\), \(|z-w|^2\),  \(\left|\dfrac{z}{w}\right|^2\) .

\(\begin{array}{|rcll|} \hline && \mathbf{|z+w|^2} \\ &=& |z|^2+|w|^2+2*\Re{(z\overline{w})} \\ &=& 5^2+2^2+2*6 \quad | \quad \Re{(z\overline{w})} = \Re{(6+8i)} = 6 \\ &=& \mathbf{41} \\ \hline && \mathbf{|zw|^2} \\ &=&|z|^2|w|^2 \\ &=& 5^2*2^2 \\ &=& 25*4 \\ &=& \mathbf{100} \\ \hline && \mathbf{|z-w|^2} \\ &=& |z|^2+|w|^2-2*\Re{(z\overline{w})} \\ &=& 5^2+2^2-2*6 \quad | \quad \Re{(z\overline{w})} = \Re{(6+8i)} = 6 \\ &=& \mathbf{17} \\ \hline && \mathbf{\left|\dfrac{z}{w}\right|^2} \\ &=& \dfrac{|z|^2}{|w|^2} \\ &=& \dfrac{5^2}{2^2} \\ &=& \dfrac{25}{4} \\ &=& \mathbf{6.25} \\ \hline \end{array}\)

1)

A, B, C, D, E, F, G, H, I, J all represent a different integer from 0 to 9.

If ABCDE-FGHIJ=66995, what is the value of each letter.

(ABCDE and FGHIJ are 5 digits integers, not multiplication)

\(ABCDE=83497 \\ FGHIJ=16502 \\ ABCDE-FGHIJ=66995\)

Beim versuch das im Titel genannte Problem zu lösen bin ich auf diese Seite gekommen.
Der Rechner spuckte mit auf meine Frage nach

\(\large \lim \limits_{n\to\infty} \left(\dfrac{1}{n!}\right)^{\dfrac{1}{n}}\)

liebenswerter weise

\(e^{-\dfrac{\ln(\infty!)} {\infty} } \)

aus.
Jetzt ist es schön und gut diese Lösung zu haben,

allerdings verstehe ich nicht ganz wie man ohne magische Taschenrechner darauf kommen würde.

\(\begin{array}{|rcll|} \hline && \mathbf{ \lim \limits_{n\to\infty} \left(\dfrac{1}{n!}\right)^{\dfrac{1}{n}} } \\\\ &=& \lim \limits_{n\to\infty} \left( e\right)^{\ln\left( \left(\dfrac{1}{n!}\right)^{\dfrac{1}{n}} \right) } \quad | \quad \text{Basiswechsel nach e} \\\\ &=& \lim \limits_{n\to\infty} \left( e\right)^{\dfrac{1}{n} \ln\left(\dfrac{1}{n!} \right) } \\\\ &=& \lim \limits_{n\to\infty} \left( e\right)^{\dfrac{1}{n} \Big(\ln(1) - \ln\left( n!\right) \Big) } \quad | \quad \ln(1) = 0 \\\\ &=& \lim \limits_{n\to\infty} \left( e\right)^{\dfrac{1}{n} \Big(0 - \ln\left( n!\right) \Big) } \\\\ &=& \lim \limits_{n\to\infty} \left( e\right)^{\dfrac{1}{n} \Big(-\ln\left( n!\right) \Big) } \\\\ &=& \lim \limits_{n\to\infty} \left( e\right)^{ \Big(-\ln\left( \left(n!\right)^n \right) \Big) } \\\\ &=& \lim \limits_{n\to\infty} \dfrac{1}{\left( e\right)^{ \Big(\ln\left( \left(n!\right)^n \right) \Big) }} \\\\ &=& \dfrac{1}{ e^{\infty}} \\\\ &=& \mathbf{0} \\ \hline \end{array}\)

What is unit digit of 7^2018

\(\begin{array}{|rcll|} \hline && \mathbf{{\color{red}7}^{2018} \pmod{10}}\\ &&&\boxed{{\color{red}7^2} \pmod{10} \\ \equiv 49 \pmod{10}\\ \equiv 9 \pmod{10} \\ \equiv 9-10 \pmod{10}\\ \equiv {\color{green}-1} \pmod{10}} \\ &\equiv& \left({\color{green}-1}\right)^{1009} \pmod{10} \\ &\equiv& -1 \pmod{10} \\ &\equiv& -1+10 \pmod{10} \\ &\equiv& \mathbf{9 \pmod{10}} \\ \hline \end{array} \)

The unit digit of \(\mathbf{7^{2018}}\) is 9

What is unit digit of 7^2018

\(\begin{array}{|rcll|} \hline && \mathbf{{\color{red}7}^{2018} \pmod{10}}\\ &&&\boxed{ {\color{red}7} \pmod{10} \\ \equiv 7-10 \pmod{10} \\ \equiv {\color{green}-3} \pmod{10} } \\ &\equiv& \left({\color{green}-3}\right)^{2018} \pmod{10} \\ &\equiv& (-1)^{2018}3^{2018} \pmod{10} \\ &\equiv& \mathbf{3^{2018} \pmod{10}} \\ &&&\boxed{{\color{blue}3^2} \pmod{10} \\ \equiv 9 \pmod{10} \\ \equiv 9-10 \pmod{10}\\ \equiv {\color{brown}-1} \pmod{10}} \\ &\equiv& \left({\color{blue}3^2}\right)^{1009} \pmod{10} \\ &\equiv& \left({\color{brown}-1}\right)^{1009} \pmod{10} \\ &\equiv& -1 \pmod{10} \\ &\equiv& -1+10 \pmod{10} \\ &\equiv& \mathbf{9 \pmod{10}} \\ \hline \end{array} \)

The unit digit of \(\mathbf{7^{2018}}\) is 9

This question here is giving me a hard time can any help me?

1)

which angle in the quadrilateral with vertices A(2,-4),, B(-5,-2),C(-4,2) and D(4,0) is a right angle?

\(\begin{array}{|rcll|} \hline \overline{CD}^2 = 8^2+2^2 = 68 \\ \overline{CB}^2 = 4^2+1^2 = 17 \\ \overline{BD}^2 = 9^2+2^2 = 85 \\ \hline \end{array}\)

Pythagoras:

\(\begin{array}{|rcll|} \hline \overline{CD}^2+\overline{CB}^2 &\overset{?}{=}& \overline{BD}^2 \quad | \quad \text{If true, then C is a right angle!} \\ 68+17 &\overset{?}{=}& 85 \\ 85 &=& 85 \quad | \quad \text{So C is a right angle!} \\ \hline \end{array}\)

use the properties of this term tto find the fifteenth term of the geometric sequence

\(9x,\ x^2,\ 2xy,\ 8y,\ \ldots\)

I assume

\(\text{Let $r$ the common ratio }\\ \text{Let $a_1 = 9x$ } \\ \text{Let $a_2 = x^2 $ } \\ \text{Let $a_3 = 2xy $ } \\ \text{Let $a_4 = 8y $ } \\ \text{Let $a_{15} = \ ? $ } \)

\(\begin{array}{|rcll|} \hline r &=& \dfrac{a_2}{a_1} \\ &=& \dfrac{x^2}{9x} \\\\ \mathbf{r} &=& \mathbf{\dfrac{x}{9}} \qquad (1) \\ \mathbf{x^2} &=& \mathbf{9xr} \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline r &=& \dfrac{a_3}{a_2} \\ &=& \dfrac{2xy}{x^2} \\\\ \mathbf{2xy} &=& \mathbf{x^2r} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline r &=& \dfrac{a_4}{a_3} \\ &=& \dfrac{8y}{2xy} \\\\ \mathbf{2xy} &=& \mathbf{\dfrac{8y}{r} } \qquad (4) \\ \hline \end{array}\)

\(\mathbf{(3)=(4)}\)

\(\begin{array}{|rcll|} \hline 2xy = x^2r &=& \dfrac{8y}{r} \\ x^2r &=& \dfrac{8y}{r} \quad | \quad \mathbf{x^2=9xr} \qquad (2) \\ 9xr^2 &=& \dfrac{8y}{r} \\ \mathbf{r^3} &=& \mathbf{\dfrac{8y}{9x} } \qquad (5)\quad | \quad x\ne 0! \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline r^3 = \left(\dfrac{x}{9}\right)^3 &=& \dfrac{8y}{9x} \quad | \quad \mathbf{r=\dfrac{x}{9}} \qquad (1), \qquad \mathbf{r^3=\dfrac{8y}{9x} } \qquad (5) \\ \left(\dfrac{x}{9}\right)^3 &=& \dfrac{8y}{9x} \\ x^4 &=& 9^2\cdot 8y \\ x^4 &=& 648y \\ \mathbf{y} &=& \mathbf{\dfrac{x^4}{648} } \qquad (6) \\ \hline \end{array}\)

\(\text{Let $a_1 = a = 9x$ } \\ \text{Let $a_2 = ar = 9xr $ } \\ \text{Let $a_3 = ar^2 = 9xr^2 $ } \\ \text{Let $a_4 = ar^3 $ } \\ \text{Let $a_{15} = ar^{14}=9xr^{14} $ } \)

\(\mathbf{x=\ ? }\)

\(\begin{array}{|rcll|} \hline a_3 &=& 9xr^2 \quad | \quad a_3 = 2xy \\ 2xy &=& 9xr^2 \quad | \quad \mathbf{r=\dfrac{x}{9}} \qquad (1) \\ 2xy &=& 9x\left(\dfrac{x}{9}\right)^2 \\ 2xy &=& \dfrac{x^3}{9} \\ x^3 &=& 18xy \quad | \quad \mathbf{y=\dfrac{x^4}{648} } \qquad (6) \\ x^3 &=& 18x\dfrac{x^4}{648} \\ x^3 &=& \dfrac{18x^5}{648} \\ x^2 &=& \dfrac{648}{18} \\ \mathbf{x^2} &=& \mathbf{36} \qquad \mathbf{x=\pm6} \\ \hline \end{array}\)

\(\mathbf{y=\ ? }\)

\(\begin{array}{|rcll|} \hline \mathbf{y} &=& \mathbf{\dfrac{x^4}{648} } \quad | \quad \mathbf{x^2=36} \\ y &=& \dfrac{36^2}{648} \\ \mathbf{y} &=& \mathbf{2} \\ \hline \end{array}\)

1. geometric sequence

\(x=6,\ y=2,\ r=\dfrac{x}{9}=\dfrac{6}{9}=\dfrac{2}{3} \)

\(\begin{array}{|rcll|} \hline a_1 &=& 54 \\ a_2 &=& 36 \\ a_3 &=& 24 \\ a_4 &=& 16 \\\\ a_{15} &=& 54\cdot \left( \dfrac{2}{3} \right)^{14} \\\\ &=& \dfrac{54}{3^3}\cdot \dfrac{2^{14}}{3^{11}} \\\\ &=& 2\cdot \dfrac{2^{14}}{3^{11}} \\\\ &=& \dfrac{2^{15}}{3^{11}} \\\\ &=& \dfrac{32768}{177147} \\\\ \mathbf{a_{15}} &=& \mathbf{ 0.18497631910 } \\ \hline \end{array}\)

2. geometric sequence

\(x=-6,\ y=2,\ r=\dfrac{x}{9}=\dfrac{-6}{9}=-\dfrac{2}{3}\)

\(\begin{array}{|rcll|} \hline a_1 &=& -54 \\ a_2 &=& 36 \\ a_3 &=& -24 \\ a_4 &=& 16 \\\\ a_{15} &=& -54\cdot \left( -\dfrac{2}{3} \right)^{14} \\\\ &=& -\dfrac{54}{3^3}\cdot \dfrac{2^{14}}{3^{11}} \\\\ &=& -2\cdot \dfrac{2^{14}}{3^{11}} \\\\ &=& - \dfrac{2^{15}}{3^{11}} \\\\ &=& -\dfrac{32768}{177147} \\\\ \mathbf{a_{15}} &=& \mathbf{ -0.18497631910 } \\ \hline \end{array}\)