heureka

Triangles ABC and ABD are isosceles with

$AB=AC=BD$,
and$\overline{BD}$ intersects$\overline{AC}$ at $E$.
If $\overline{BD}\perp\overline{AC}$, then what is the value of $\angle C+\angle D$?

$\angle D = BDA = DAB \\ \angle C = BCA = CBA \\ \angle \alpha = DBA\\ \angle \beta = BAC \\ \angle \alpha +\angle \beta = 90^{\circ}$

$\begin{array}{|rcll|} \hline \angle \alpha &=& 180^{\circ} - 2\times \angle D \\ \angle \beta&=& 180^{\circ} - 2\times \angle C \\ \angle \alpha +\angle \beta &=& 180^{\circ} - 2\times \angle D+180^{\circ} - 2\times \angle C \quad & | \quad \angle \alpha +\angle \beta = 90^{\circ}\\ 90^{\circ} &=& 180^{\circ} - 2\times \angle D+180^{\circ} - 2\times \angle C \\ 90^{\circ} &=& 360^{\circ} - 2\times (\angle C + \angle D) \\ 2\times(\angle C + \angle D) &=& 360^{\circ} - 90^{\circ} \\ 2\times(\angle C + \angle D) &=& 270^{\circ} \\ \mathbf{ \angle C + \angle D } & \mathbf{=}& \mathbf{135^{\circ}} \\ \hline \end{array}$

The numbers 1,2,3,4,5,6,7,8,9 are arranged in a list so that each number
is either greater than all the numbers that come before it or
is less than all the numbers that come before it.
For example, 4,5,6,3,2,7,1,8,9 is one such list:

notice that (for instance) the 6 is greater than all the numbers that come before it,
and the 2 is less than all the numbers that come before it.
How many are such lists of the numbers 1,2,3,4,5,6,7,8,9 possible?
 

$\begin{array}{|rrrr|} \hline \begin{array}{r|r} 1.& 123456789 \\ 2.& 213456789 \\ 3.& 231456789 \\ 4.& 234156789 \\ 5.& 234516789 \\ 6.& 234561789 \\ 7.& 234567189 \\ 8.& 234567819 \\ 9.& 234567891 \\ 10.& 321456789 \\ 11.& 324156789 \\ 12.& 324516789 \\ 13.& 324561789 \\ 14.& 324567189 \\ 15.& 324567819 \\ 16.& 324567891 \\ 17.& 342156789 \\ 18.& 342516789 \\ 19.& 342561789 \\ 20.& 342567189 \\ 21.& 342567819 \\ 22.& 342567891 \\ 23.& 345216789 \\ 24.& 345261789 \\ 25.& 345267189 \\ 26.& 345267819 \\ 27.& 345267891 \\ 28.& 345621789 \\ 29.& 345627189 \\ 30.& 345627819 \\ 31.& 345627891 \\ 32.& 345672189 \\ 33.& 345672819 \\ 34.& 345672891 \\ 35.& 345678219 \\ 36.& 345678291 \\ 37.& 345678921 \\ 38.& 432156789 \\ 39.& 432516789 \\ 40.& 432561789 \\ 41.& 432567189 \\ 42.& 432567819 \\ 43.& 432567891 \\ 44.& 435216789 \\ 45.& 435261789 \\ 46.& 435267189 \\ 47.& 435267819 \\ 48.& 435267891 \\ 49.& 435621789 \\ 50.& 435627189 \\ 51.& 435627819 \\ 52.& 435627891 \\ 53.& 435672189 \\ 54.& 435672819 \\ 55.& 435672891 \\ 56.& 435678219 \\ 57.& 435678291 \\ 58.& 435678921 \\ 59.& 453216789 \\ 60.& 453261789 \\ 61.& 453267189 \\ 62.& 453267819 \\ 63.& 453267891 \\ 64.& 453621789 \\ \end{array} & \begin{array}{r|r} 65.& 453627189 \\ 66.& 453627819 \\ 67.& 453627891 \\ 68.& 453672189 \\ 69.& 453672819 \\ 70.& 453672891 \\ 71.& 453678219 \\ 72.& 453678291 \\ 73.& 453678921 \\ 74.& 456321789 \\ 75.& 456327189 \\ 76.& 456327819 \\ 77.& 456327891 \\ 78.& 456372189 \\ 79.& 456372819 \\ 80.& 456372891 \\ 81.& 456378219 \\ 82.& 456378291 \\ 83.& 456378921 \\ 84.& 456732189 \\ 85.& 456732819 \\ 86.& 456732891 \\ 87.& 456738219 \\ 88.& 456738291 \\ 89.& 456738921 \\ 90.& 456783219 \\ 91.& 456783291 \\ 92.& 456783921 \\ 93.& 456789321 \\ 94.& 543216789 \\ 95.& 543261789 \\ 96.& 543267189 \\ 97.& 543267819 \\ 98.& 543267891 \\ 99.& 543621789 \\ 100.& 543627189 \\ 101.& 543627819 \\ 102.& 543627891 \\ 103.& 543672189 \\ 104.& 543672819 \\ 105.& 543672891 \\ 106.& 543678219 \\ 107.& 543678291 \\ 108.& 543678921 \\ 109.& 546321789 \\ 110.& 546327189 \\ 111.& 546327819 \\ 112.& 546327891 \\ 113.& 546372189 \\ 114.& 546372819 \\ 115.& 546372891 \\ 116.& 546378219 \\ 117.& 546378291 \\ 118.& 546378921 \\ 119.& 546732189 \\ 120.& 546732819 \\ 121.& 546732891 \\ 122.& 546738219 \\ 123.& 546738291 \\ 124.& 546738921 \\ 125.& 546783219 \\ 126.& 546783291 \\ 127.& 546783921 \\ 128.& 546789321 \\ \end{array} & \begin{array}{r|r} 129.& 564321789 \\ 130.& 564327189 \\ 131.& 564327819 \\ 132.& 564327891 \\ 133.& 564372189 \\ 134.& 564372819 \\ 135.& 564372891 \\ 136.& 564378219 \\ 137.& 564378291 \\ 138.& 564378921 \\ 139.& 564732189 \\ 140.& 564732819 \\ 141.& 564732891 \\ 142.& 564738219 \\ 143.& 564738291 \\ 144.& 564738921 \\ 145.& 564783219 \\ 146.& 564783291 \\ 147.& 564783921 \\ 148.& 564789321 \\ 149.& 567432189 \\ 150.& 567432819 \\ 151.& 567432891 \\ 152.& 567438219 \\ 153.& 567438291 \\ 154.& 567438921 \\ 155.& 567483219 \\ 156.& 567483291 \\ 157.& 567483921 \\ 158.& 567489321 \\ 159.& 567843219 \\ 160.& 567843291 \\ 161.& 567843921 \\ 162.& 567849321 \\ 163.& 567894321 \\ 164.& 654321789 \\ 165.& 654327189 \\ 166.& 654327819 \\ 167.& 654327891 \\ 168.& 654372189 \\ 169.& 654372819 \\ 170.& 654372891 \\ 171.& 654378219 \\ 172.& 654378291 \\ 173.& 654378921 \\ 174.& 654732189 \\ 175.& 654732819 \\ 176.& 654732891 \\ 177.& 654738219 \\ 178.& 654738291 \\ 179.& 654738921 \\ 180.& 654783219 \\ 181.& 654783291 \\ 182.& 654783921 \\ 183.& 654789321 \\ 184.& 657432189 \\ 185.& 657432819 \\ 186.& 657432891 \\ 187.& 657438219 \\ 188.& 657438291 \\ 189.& 657438921 \\ 190.& 657483219 \\ 191.& 657483291 \\ 192.& 657483921 \\ \end{array} & \begin{array}{r|r} 193.& 657489321 \\ 194.& 657843219 \\ 195.& 657843291 \\ 196.& 657843921 \\ 197.& 657849321 \\ 198.& 657894321 \\ 199.& 675432189 \\ 200.& 675432819 \\ 201.& 675432891 \\ 202.& 675438219 \\ 203.& 675438291 \\ 204.& 675438921 \\ 205.& 675483219 \\ 206.& 675483291 \\ 207.& 675483921 \\ 208.& 675489321 \\ 209.& 675843219 \\ 210.& 675843291 \\ 211.& 675843921 \\ 212.& 675849321 \\ 213.& 675894321 \\ 214.& 678543219 \\ 215.& 678543291 \\ 216.& 678543921 \\ 217.& 678549321 \\ 218.& 678594321 \\ 219.& 678954321 \\ 220.& 765432189 \\ 221.& 765432819 \\ 222.& 765432891 \\ 223.& 765438219 \\ 224.& 765438291 \\ 225.& 765438921 \\ 226.& 765483219 \\ 227.& 765483291 \\ 228.& 765483921 \\ 229.& 765489321 \\ 230.& 765843219 \\ 231.& 765843291 \\ 232.& 765843921 \\ 233.& 765849321 \\ 234.& 765894321 \\ 235.& 768543219 \\ 236.& 768543291 \\ 237.& 768543921 \\ 238.& 768549321 \\ 239.& 768594321 \\ 240.& 768954321 \\ 241.& 786543219 \\ 242.& 786543291 \\ 243.& 786543921 \\ 244.& 786549321 \\ 245.& 786594321 \\ 246.& 786954321 \\ 247.& 789654321 \\ 248.& 876543219 \\ 249.& 876543291 \\ 250.& 876543921 \\ 251.& 876549321 \\ 252.& 876594321 \\ 253.& 876954321 \\ 254.& 879654321 \\ 255.& 897654321 \\ 256.& 987654321 \\ \end{array}\\ \hline \end{array}$

The answer is 256

Computing...

1.

$\text{Numbers}:\\ \begin{array}{rrrrrrr} 31 & 30 & 29 & 28 & 27 & 26 & \\ 32 & 13 & 12 & 11 & 10 & 25 & \\ \ldots & 14 & 3 & 2 & 9 & 24 & \\ \ldots & 15 & 4 & 1 & 8 & 23 & \\ \ldots & 16 & 5 & 6 & 7 & 22 & \\ \ldots & 17 & 18 & 19 & 20 & 21 & \\ \end{array}$

2.

$\text{Coordinates of the lattice points}:\\ \begin{array}{rrrrrrr} \text{Start at $(0, 0) = 1$ }\\\\ (-3,3) & (-2,3) & (-1,3) & (0,3) & (1,3) & (2,3) & \\ (-3,2) & (-2,2) & (-1,2) & (0,2) & (1,2) & (2,2) & \\ \ldots & (-2,1) & (-1,1) & (0,1) & (1,1) & (2,1) & \\ \ldots & (-2,0) & (-1,0) & (0,0) & (1,0) & (2,0) & \\ \ldots & (-2,-1) & (-1,-1) & (0,-1) & (1,-1) & (2,-1) & \\ \ldots & (-2,-2) & (-1,-2) & (0,-2) & (1,-2) & (2,-2) & \\ \end{array}$

3.

$\text{Algorithem to get the lattice point coordinates:} \\ \begin{array}{|r|r|} \hline \text{lattice point number} & \text{coordinate }(x,y) \\ \hline 1.& (0,0) \\ 2.& (0,1) \\ 3.& (-1,1) \\ 4.& (-1,0) \\ 5.& (-1,-1) \\ 6.& (0,-1) \\ 7.& (1,-1) \\ 8.& (1,0) \\ 9.& (1,1) \\ 10.& (1,2) \\ 11.& (0,2) \\ 12.& (-1,2) \\ 13.& (-2,2) \\ 14.& (-2,1) \\ 15.& (-2,0) \\ 16.& (-2,-1) \\ 17.& (-2,-2) \\ 18.& (-1,-2) \\ 19.& (0,-2) \\ 20.& (1,-2) \\ 21.& (2,-2) \\ 22.& (2,-1) \\ 23.& (2,0) \\ 24.& (2,1) \\ 25.& (2,2) \\ 26.& (2,3) \\ 27.& (1,3) \\ 28.& (0,3) \\ 29.& (-1,3) \\ 30.& (-2,3) \\ 31.& (-3,3) \\ 32.& (-3,2) \\ \ldots \\ 1843.& (21,15) \\ \ldots \\ 2017.& (22,14) \\ \ldots \\ 2018.& (22,15) \\ \ldots \\ 2019.& (22,16) \\ \ldots \\ 2201.& (23,15) \\ \ldots \\ \hline \end{array}$

4.

short algorithm to get (x,y) the index in a grid-net of the spiral in c++:

    int n_max = 2818;
    int y_coordinate = 0;
    int x_coordinate = 0;
    for(int i = 0; i < n_max; ++i) {
        // output Number i+1
        // output x_coordinate
        // output y_coordinate
        if(abs(y_coordinate) <= abs(x_coordinate) && (y_coordinate != x_coordinate || y_coordinate >= 0))
            y_coordinate += ((x_coordinate >= 0) ? 1 : -1);
        else
            x_coordinate += ((y_coordinate >= 0) ? -1 : 1);
    }

Find the lattice point coordinate of 2018.

We find (22,15).

The four numbers directly adjacent to the number “2018” in
this spiral are (21,15), (22,14), (22,16), (23,15).

The Numbers are 1843, 2017, 2019, 2201.

4.  The numbers 1, 2, 3, 4, . . . are written starting with 1, spiraling outward
in a counterclockwise direction, as shown below.

What is the sum of the four numbers directly adjacent to the number “2018” in this spiral?

.................. 26
13 12 11 10 25
14 3 2 9 24
15 4 1 8 23
16 5 6 7 22
17 18 19 20 21

$\begin{array}{cccc} \text{Start at $(0, 0) = 1$ }\\\\ && \boxed{2019\\\tiny{(22,16)}} \\ && \uparrow \\ & \boxed{1843\\\tiny{(21,15)}} \leftarrow & \boxed{2018\\\tiny{(22,15)}} & \rightarrow \boxed{2201\\\tiny{(23,15)}} \\ & & \downarrow \\ & & \boxed{2017\\\tiny{(22,14)}} \\ \end{array}$

$1843+2017+2019+2201 = 8080$

How many 9 step paths are there from E to G which pass through F?

We can simplify:

$\begin{array}{|rcll|} \hline && \dfrac{4!}{3!1!}\times \dfrac{5!}{2!3!} \\\\ &=& \dfrac{3!4}{3!1!}\times \dfrac{3!\times 4 \times 5}{2!3!} \\\\ &=& \dfrac{ 4}{ 1!}\times \dfrac{4 \times 5}{2! } \\\\ &=& \dfrac{ 4}{1}\times \dfrac{4 \times 5}{2} \\\\ &=& 4\times 2 \times 5 \\\\ &=& 4\times 10 \\ &=& 40 \\ \hline \end{array}$

There are 40
9 step paths are there from E to G which pass through F.

In general you can solve 2-D Pathways:

what does b  represent?

$\begin{array}{|lrcll|} \hline &y &=& ax^2+bx+c \\ &y' &=& 2ax+b \\ x=0: &y' &=& b \\ \hline \end{array}$

so b is the slope at x = 0.

In the diagram below, we have DE = 2EC and AB = DC = 20. Find the length of FG.

$\text{Let $DC = 20$ } \\ \text{Let $DE = \dfrac23\cdot 20$ } \\ \text{Let $EC = \dfrac13\cdot 20$ } \\ \text{Let $FG = x + EC $ or $x=FG-EC $ }\\ \text{Let $BG = p $ } \\ \text{Let $GC = q $ }$

1. intercept theorem

$\begin{array}{|rcll|} \hline \dfrac{q}{x} &=& \dfrac{p}{DE-x} \quad & \quad DE= \dfrac23\cdot 20 \\\\ \dfrac{q}{x} &=& \dfrac{p}{\dfrac23\cdot 20-x} \\\\ \mathbf{\dfrac{p}{q}} & \mathbf{=} & \mathbf{\dfrac{40-3x}{3x} \qquad (1)} \\ \hline \end{array}$

2. intercept theorem

$\begin{array}{|rcll|} \hline \dfrac{q}{DE-x} &=& \dfrac{p}{x+EC} \quad & \quad DE= \dfrac23\cdot 20 \qquad EC=\dfrac13\cdot 20 \\\\ \dfrac{q}{ \dfrac23\cdot 20-x} &=& \dfrac{p}{x+\dfrac13\cdot 20} \\\\ \mathbf{\dfrac{p}{q}} & \mathbf{=} & \mathbf{\dfrac{3x+20}{40-3x} \qquad (2)} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (1)=(2): & \mathbf{\dfrac{p}{q}} = \mathbf{\dfrac{40-3x}{3x} }&\mathbf{=}& \mathbf{\dfrac{3x+20}{40-3x}} \\\\ & \dfrac{40-3x}{3x} & = & \dfrac{3x+20}{40-3x} \\\\ & (40-3x)^2 & = & 3x(3x+20) \\\\ & 1600-240x+\not{9x^2} & = & \not{9x^2} +60x \\\\ & 1600-240x & = & 60x \\\\ & 300x &=& 1600 \quad & | \quad : 300 \\\\ & x &=& \dfrac{1600}{300} \\\\ & x &=& \dfrac{16}{3} \\\\ & FG &=& x + EC \qquad EC=\dfrac{20}{3} \\\\ & &=& \dfrac{16}{3} + \dfrac{20}{3} \\\\ & &=& \dfrac{36}{3} \\\\ & \mathbf{FG} &\mathbf{=} & \mathbf{12} \\ \hline \end{array}$

Vollständige Induktion

$\large{(a+b)^n\ge a^n+b^n}$

Ich habe probleme den Induktionschritt n+1 anzuwenden. Kann mir jemand helfen?

Induktionsanfang:
$n = 1:$  linke Seite: $(a+b)^1 = a+b$
               rechte Seite: $a^1+b^1 = a+b$
Für $n=1$ sind beide Seiten gleich und die Aussage ist wahr!

Die Induktionsannahme $(I.A.)$ lautet:

 $(a + b)^n \ge a^n + b^n$

Induktionsschluss:

$\begin{array}{|lrcll|} \hline n+1:\\ \text{linke Seite:} \\ && & (a+b)^{n+1} \\ &&=& (a+b)^n(a+b)^1 \\ &&=& (a+b)(a+b)^n \\ &&\overset{I.A.}{\ge}& (a+b)(a^n+b^n) \\ &&=& a^{n+1}+b^{n+1}+ab^n+ba^n \\\\ \text{rechte Seite:} \\ &&&a^{n+1}+b^{n+1}\\\\ \text{Ergebnis:}\\ && & (a+b)^{n+1} \ge a^{n+1}+b^{n+1}+ab^n+ba^n \ge a^{n+1}+b^{n+1}\ \checkmark \\ \hline \end{array}$

What is the smallest positive integer that will satisfy the following congruences:

N mod 105 = 104,

N mod 111 = 110,

N mod 121 = 111,

N mod 122 = 111

Thanks for any help.

$\begin{array}{|l|cll|} \hline \begin{array}{|rcll|} \hline \mathbf{ n } & \mathbf{\equiv} & \mathbf{104 \pmod{105}} \\ \mathbf{ n } & \mathbf{\equiv} & \mathbf{110 \pmod{111}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n -104 &=& 105x \\ & n -110 &=& 111y \\\\ \Rightarrow & n &=& 104+105x \\ & n &=& 110+111y \\\\ & n=104+105x&=& 110+111y \\ & 104+105x&=& 110+111y \\ & 105x-111y&=& 6 \quad & | \quad : 3 \\ & \mathbf{35x-37y}& \mathbf{=}& \mathbf{2} \qquad x,y \in Z \\ &\rightarrow \quad x &=& -1+37b^{^1)} \qquad b \in Z \\\\ & n &=& 104+105x \\ & n &=& 104+105(-1+37b) \\ & n &=& 104-105+105\cdot 37b \\ & n &=& -1+ 3885b \\ & \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \qquad (1)\\ \end{array}\\ \hline \end{array}$

$\begin{array}{|l|cll|} \hline \begin{array}{|rcll|} \hline \mathbf{ n } & \mathbf{\equiv} & \mathbf{111 \pmod{121}} \\ \mathbf{ n } & \mathbf{\equiv} & \mathbf{111 \pmod{122}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n -111 &=& 121x \\ & n -111 &=& 122y \\\\ \Rightarrow & n &=& 111+121x \\ & n &=& 111+122y \\\\ & n=111x+121y&=& 111+122y \\ & 111x+121y&=& 111+122y \\\\ & \mathbf{121x-122y}&\mathbf{=}& \mathbf{0} \qquad x,y \in Z \\ &\rightarrow \quad x &=& 122a^{^2)} \qquad a \in Z \\\\ & n &=& 111+121x \\ & n &=& 111+121(122a) \\ & n &=& 111+121\cdot 122a \\ & n &=& 111+14762a \\ & \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \qquad (2)\\ \end{array} \\ \hline \end{array}$

After reducing, we have two formulas:

$\begin{array}{|rcll|} \hline \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \qquad &(1)\\ \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \qquad &(2)\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \\ \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n +1 &=& 3885x \\ & n -111 &=& 14762y \\\\ \Rightarrow & n &=& -1+3885x \\ & n &=& 111+14762y \\\\ & n=-1+3885x&=& 111+14762y \\ & -1+3885x&=& 111+14762y \\ & \mathbf{3885x-14762y}& \mathbf{=}& \mathbf{112} \qquad x,y \in Z \\ &\rightarrow \quad x &=&3378+14762g^{^3)} \qquad g \in Z \\\\ & n &=& -1+3885x \\ & n &=& -1+3885\cdot (3378+14762g) \\ & n &=& -1+3885\cdot 3378+ 3885\cdot 14762g \\ & \mathbf{n} &\mathbf{=}& \mathbf{13123529 + 57350370g \qquad g \in Z } \\ \end{array}$

The smallest positive integer is  $\mathbf{13\ 123\ 529}$

Proof:

$\begin{array}{|rcll|} \hline \mathbf{13\ 123\ 529} &\text{mod } 105 =& 104\\ \mathbf{13\ 123\ 529} &\text{mod } 111 =& 110\\ \mathbf{13\ 123\ 529} &\text{mod } 121 =& 111 \\ \mathbf{13\ 123\ 529} &\text{mod } 122 =& 111 \\ \hline \end{array}$

$\begin{array}{lcll} \hline ^1) \\ \text{Solve of the diophantine equation $35x - 37y = 2$ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 35x &=& 2 + 37y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 2 + 37y } {35}} \\ &=& \dfrac{ 2 +35y+2y } {35} \\ &=& \dfrac{ 35y + 2 +2y } {35} \\ &=& \dfrac{35y}{35} + \dfrac{ 2 +2y } {35} \\ x &=& y + \dfrac{ 2 +2y} {35} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ 2 + 2y } {35} \\ & 35a &=& 2 + 2y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{rcll} 2y &=& -2 + 35a \\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -2 + 35a } {2}} \\ &=& \dfrac{ -2 + 34a+a } {2} \\ &=& \dfrac{ 34a - 2 +a } {2} \\ &=& -\dfrac{-2}{2} +\dfrac{34a}{2} + \dfrac{ a } {2} \\ y &=& -1+17a+ \dfrac{ a } {2} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &b &=& \dfrac{a} {2} \\ & 2b &=& a \\ \end{array} \\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ 2b } \\ \end{array} \end{array}$

$\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -2 + 35a } {2}} \quad & | \quad \mathbf{a = 2b }\\ & = & \dfrac{-2 + 35\cdot (2b) } {2} \\ \mathbf{y} & \mathbf{=} & \mathbf{-1+35b} \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 2 + 37y } {35}} \quad & | \quad \mathbf{y = -1+35b }\\ & = & \dfrac{ 2 + 37\cdot (-1+35b ) } {11} \\ \mathbf{x} & \mathbf{=} & \mathbf{-1 + 37b } \\ \end{array}$

$\begin{array}{lcll} \hline ^2) \\ \text{Solve of the diophantine equation $121x - 122y = 0 $ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 121x &=& 122y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 122y } {121}} \\ &=& \dfrac{ 121y+y } {121} \\ &=& \dfrac{121y}{121} + \dfrac{y} {121} \\ x &=& y + \dfrac{ y} {121} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{y} {121} \\ & 121a &=& y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{y} &\mathbf{=}& \mathbf{ 121a } \\ \end{array} \end{array}$

$\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 122y } {121}} \quad & | \quad \mathbf{y=121a }\\ & = & \dfrac{122\cdot (121a) } {121} \\ \mathbf{x} & \mathbf{=} & \mathbf{122a} \\ \end{array}$

$\begin{array}{lcll} \hline ^3) \\ \text{Solve of the diophantine equation $3885x - 14762y = 112 $ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 3885x &=& 112 + 14762y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 112 + 14762y} {3885}} \\ &=& \dfrac{ 112 + 11655y + 3107y } {3885} \\ &=& \dfrac{11655y + 112 + 3107y }{3885} \\ &=& \dfrac{11655y }{3885} + \dfrac{112 + 3107y} {3885} \\ x &=& 3y + \dfrac{ 112 + 3107y } {3885} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{112 + 3107y } {3885} \\ & 3885a &=& 112 + 3107y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{rcll} 3107y &=& -112 + 3885a \\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -112 + 3885a} {3107}} \\ &=& \dfrac{ -112 + 3107a + 778a } {3107} \\ &=& \dfrac{3107a - 112 + 778a }{3107} \\ &=& \dfrac{3107a }{3107} + \dfrac{- 112 + 778a} {3107} \\ y &=& 3a + \dfrac{ 112 + 3107a } {3107} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &b &=& \dfrac{-112 + 778a } {3107} \\ & 3107b &=& -112 + 778a \\ \end{array} \\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{rcll} 778a &=& 112 + 3107b \\ \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 112 + 3107b} {778}} \\ &=& \dfrac{ 112 + 2334b + 773b } {778} \\ &=& \dfrac{2334b +112 + 773b }{778} \\ &=& \dfrac{2334b }{3107} + \dfrac{112 + 773b} {778} \\ a &=& 3b + \dfrac{ 112 + 773b} {778} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &c &=& \dfrac{112 + 773b } {778} \\ & 778c &=& 112 + 773b \\ \end{array} \\ \text{The variable with the smallest coefficient is $b$. The equation is transformed after $b$: }\\ \begin{array}{rcll} 773b &=& -112 + 778c\\ \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ -112 + 778c} {773}} \\ &=& \dfrac{ -112 + 773c + 5c } {773} \\ &=& \dfrac{773c -112 + 5c }{773} \\ &=& \dfrac{773c }{773} + \dfrac{-112 + 5c } {773} \\ b &=& c + \dfrac{ -112 + 5c} {773} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &d &=& \dfrac{-112 + 5c } {773} \\ & 773d &=& -112 + 5c \\ \end{array} \\ \text{The variable with the smallest coefficient is $c$. The equation is transformed after $c$: }\\ \begin{array}{rcll} 5c &=& 112 + 773d\\ \mathbf{c} &\mathbf{=}& \mathbf{\dfrac{ 112 + 773d } {5}} \\ &=& \dfrac{ 110+2 + 770d + 3d } {5} \\ &=& \dfrac{110+770d+2 + 5d }{5} \\ &=& \dfrac{110 }{5}+\dfrac{770d }{5} + \dfrac{2 + 3d } {5} \\ c &=& 22 + 154d + \dfrac{2 + 3d } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &e &=& \dfrac{ 2 + 3d } {5} \\ & 5e &=& 2 + 3d \\ \end{array} \\ \text{The variable with the smallest coefficient is $d$. The equation is transformed after $d$: }\\ \begin{array}{rcll} 3d &=& -2 + 5e\\ \mathbf{d} &\mathbf{=}& \mathbf{\dfrac{ -2 + 5e } {3}} \\ &=& \dfrac{ -2 + 3e + 2e } {3} \\ &=& \dfrac{3e -2 + 2e}{3} \\ &=& \dfrac{3e }{3} + \dfrac{-2 + 2e} {3} \\ d &=& e + \dfrac{-2 + 2e } {3} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &f &=& \dfrac{ -2 + 2e } {3} \\ & 3f &=& -2 + 2e \\ \end{array} \\ \text{The variable with the smallest coefficient is $e$. The equation is transformed after $e$: }\\ \begin{array}{rcll} 2e &=& 2 + 3f\\ \mathbf{e} &\mathbf{=}& \mathbf{\dfrac{2 + 3f } {2}} \\ &=& \dfrac{ 2+2f+f } {2} \\ &=& \dfrac{2 }{2} + \dfrac{2f }{2} + \dfrac{f} {2} \\ e &=& 1 + f + \dfrac{f} {2} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &g &=& \dfrac{ f} {2} \\ & 2g &=& f \\ \end{array} \\ \text{The variable with the smallest coefficient is $f$. The equation is transformed after $f$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{f} &\mathbf{=}& \mathbf{ 2g } \\ \end{array} \end{array}$

$\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{e} &\mathbf{=}& \mathbf{\dfrac{ 2 + 3f } {2}} \quad & | \quad \mathbf{f=2g }\\ & = & \dfrac{2 + 3\cdot 2g } {2} \\ \mathbf{e} & \mathbf{=} & \mathbf{1 + 3g } \\\\ \mathbf{d} &\mathbf{=}& \mathbf{\dfrac{ -2 + 5e } {3}} \quad & | \quad \mathbf{e=1 + 3g }\\ & = & \dfrac{-2 + 5\cdot (1 + 3g) } {2} \\ \mathbf{d} & \mathbf{=} & \mathbf{1 + 5g } \\\\ \mathbf{c} &\mathbf{=}& \mathbf{\dfrac{ 112 + 773d } {5}} \quad & | \quad \mathbf{d=1 + 5g }\\ & = & \dfrac{112 + 773\cdot (1 + 5g) } {2} \\ \mathbf{c} & \mathbf{=} & \mathbf{177 + 773g } \\\\ \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ -112 + 778c } {773}} \quad & | \quad \mathbf{c=177 + 773g }\\ & = & \dfrac{-112 + 778\cdot (177 + 773g) } {773} \\ \mathbf{b} & \mathbf{=} & \mathbf{ 178 + 778g } \\\\ \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 112 + 3107b } {778}} \quad & | \quad \mathbf{b=178 + 778g }\\ & = & \dfrac{112 + 3107\cdot (178 + 778g) } {778} \\ \mathbf{a} & \mathbf{=} & \mathbf{ 711 + 3107g } \\\\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -112 + 3885a } {3107}} \quad & | \quad \mathbf{a=711 + 3107g }\\ & = & \dfrac{-112 + 3885\cdot (711 + 3107g) } {3107} \\ \mathbf{y} & \mathbf{=} & \mathbf{ 889 + 3885g } \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 112 + 14762y } {3885}} \quad & | \quad \mathbf{y=889 + 3885g }\\ & = & \dfrac{112 + 14762\cdot (889 + 3885g) } {3885} \\ \mathbf{x} & \mathbf{=} & \mathbf{ 3378 + 14762g } \\ \end{array}$

What effect does replacing x with  x−4 have on the graph for the function f(x)?

f(x)=|x−6|+2

The graph is shifted 4 units left.

The graph is shifted 4 units right.

The graph is shifted 4 units up.

The graph is shifted 4 units down.