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 #2
avatar+18956 
0

2)

H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH
intersect the circumcircle of traingle ABC at A prime, B prime and C prime.
We know angle AHB : angle BHC : angle CHA = 9 : 10 : 11.
Find angle AprimeBprimeCprime in degrees.
Picture: https://latex.artofproblemsolving.com/6/1/1/6119f02c59ea35f11dceab068d51811cace24c1d.png

 

\(\text{Let $\angle A'B'C' = \angle CAB = x$ } \\ \text{Let $\angle ACB= \alpha $ } \\ \text{Let $\angle BAC= \beta $ } \\ \text{Let $\angle CBA= \gamma $ } \\ \text{Let $\angle AHB = \gamma + \beta $} \\ \text{Let $\angle BHC = \alpha + \gamma $} \\ \text{Let $\angle CHA = \beta + \alpha $} \\ \text{Let $ \mathbf{x = 180^{\circ}-2 \times\gamma }$} \)

 

In triangle ABC we set the angles \( \alpha, \beta \text{ and } \gamma\)

 

further:

\(\mathbf{ \large{x =180^{\circ} - 2 \times \gamma} } \)

 

 

\(\begin{array}{|rcrcrcrcrcr|} \hline \angle AHB &:& \angle BHC &:& \angle CHA &=& 9 &:& 10 &:& 11 \\ \gamma + \beta &:& \alpha + \gamma &:& \beta + \alpha &=& 9 &:& 10 &:& 11 \\ \hline \end{array}\)

 

\(\begin{array}{rclrclrcl} \dfrac{\alpha + \gamma }{\beta + \alpha} &=& \dfrac{10}{11} \qquad & \qquad \dfrac{\gamma + \beta}{\beta + \alpha} &=& \dfrac{9}{11} \\\\ \dfrac{\alpha + \gamma +\gamma + \beta }{\beta + \alpha} &=& \dfrac{10+9}{11} \\\\ \dfrac{\alpha + \beta + 2\times\gamma }{\alpha + \beta } &=& \dfrac{19}{11} \quad &| \quad \alpha + \beta + \gamma = 180^{\circ} \\\\ & & \quad &| \quad \alpha + \beta = 180^{\circ} - \gamma \\\\ \dfrac{180^{\circ} - \gamma + 2\times\gamma }{180^{\circ} - \gamma } &=& \dfrac{19}{11} \\\\ \dfrac{180^{\circ} + \gamma }{180^{\circ} - \gamma } &=& \dfrac{19}{11} \\\\ 11\times(180^{\circ} + \gamma ) &=&19\times(180^{\circ} - \gamma ) \\ 11\times 180^{\circ} + 11\gamma &=&19\times 180^{\circ} - 19\gamma \\ 30\gamma &=&8\times 180^{\circ}\\ \mathbf{\gamma} &\mathbf{=}& \mathbf{48^{\circ}} \\ \end{array} \)

 

\(\begin{array}{|rcll|} \hline \angle A'B'C' = x &=& 180^{\circ} - 2 \times \gamma \\ \angle A'B'C' &=& 180^{\circ} - 2 \times \gamma \\ &=& 180^{\circ} - 2 \times 48^{\circ} \\ &=& 180^{\circ} - 96^{\circ} \\ &=& \mathbf{84^{\circ}} \\ \hline \end{array} \)

 

\(\text{The angle $A'B'C'$ in degrees is $\mathbf{84^{\circ}}$ } \)

 

\(\begin{array}{rcll} \dfrac{\alpha+\gamma}{\alpha+\beta} &=& \dfrac{10}{11} \quad & | \quad \alpha+ \beta=180^{\circ}- \gamma = 180^{\circ}-84^{\circ} = 132^{\circ} \\\\ \dfrac{\alpha+48^{\circ} }{132^{\circ}} &=& \dfrac{10}{11} \\\\ \alpha &=& \dfrac{10\times 132^{\circ}}{11} -48^{\circ} \\\\ \mathbf{ \alpha } & \mathbf{=} & \mathbf{72^{\circ}} \\ \end{array}\)

 

\(\begin{array}{rcll} \alpha+\beta &=& 132^{\circ} \\ \beta &=& 132^{\circ}-\alpha \\ \beta &=& 132^{\circ}-72^{\circ} \\ \mathbf{ \beta } & \mathbf{=} & \mathbf{60^{\circ}} \\ \end{array}\)

 

\(\begin{array}{|rcll|} \hline \angle AHB = \gamma + \beta = 108^{\circ} \\ \angle BHC = \alpha + \gamma = 120^{\circ} \\ \angle CHA = \beta + \alpha = 132^{\circ} \\ \hline \end{array} \)

 

 

laugh

heureka 22 févr. 2018
 #2
avatar+18956 
+2

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

 

laugh

heureka 20 févr. 2018
 #7
avatar+18956 
+2

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.

 

laugh

heureka 20 févr. 2018
 #4
avatar+18956 
+1
heureka 16 févr. 2018
 #1
avatar+18956 
+1

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}\)

 

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

 

 

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heureka 15 févr. 2018