Menu Close

Determine-all-solutions-in-the-integers-of-the-following-Diophantine-equations-a-56x-72y-40-b-24x-138y-18-c-221x-35y-11-

Tinku Tara 0 Comments
Pin




Question Number 136830 by bramlexs22 last updated on 26/Mar/21
Determine all solutions in the integers of the following Diophantine equations: (a)56x+72y=40 (b)24x+138y=18 (c)221x+35y=11
$$ \\ $$Determine all solutions in the integers of the following Diophantine equations:
(a)56x+72y=40
(b)24x+138y=18
(c)221x+35y=11
Answered by floor(10²Eta[1]) last updated on 26/Mar/21
(a)56x+72y=40 ⇒7x+9y=5 7x+9y=5 (look mod 3) x≡2(mod 3)⇒x=3a+2 7(3k+2)+9y=5 y=−1−(7/3)a⇒3∣a⇒a=3k x=9k+2, y=−1−7k (b)24x+138y=18 ⇒4x+23y=3 (look mod 2) y≡1(mod 2)⇒y=2a+1 4x+23(2a+1)=3 x=−5−((23)/2)a⇒2∣a⇒a=2k x=−5−23k, y=4k+1 (c)221x+35y=11 (look mod 5) x≡1(mod 5)⇒x=5a+1 221(5a+1)+35y=11 y=−((221)/7)a−6⇒7∣a⇒a=7k x=35k+1, y=−221k−6
$$\left(\mathrm{a}\right)\mathrm{56x}+\mathrm{72y}=\mathrm{40} \\ $$$$\Rightarrow\mathrm{7x}+\mathrm{9y}=\mathrm{5} \\ $$$$\mathrm{7x}+\mathrm{9y}=\mathrm{5}\:\:\left(\mathrm{look}\:\mathrm{mod}\:\mathrm{3}\right) \\ $$$$\mathrm{x}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{3}\right)\Rightarrow\mathrm{x}=\mathrm{3a}+\mathrm{2} \\ $$$$\mathrm{7}\left(\mathrm{3k}+\mathrm{2}\right)+\mathrm{9y}=\mathrm{5} \\ $$$$\mathrm{y}=−\mathrm{1}−\frac{\mathrm{7}}{\mathrm{3}}\mathrm{a}\Rightarrow\mathrm{3}\mid\mathrm{a}\Rightarrow\mathrm{a}=\mathrm{3k} \\ $$$$\mathrm{x}=\mathrm{9k}+\mathrm{2},\:\mathrm{y}=−\mathrm{1}−\mathrm{7k} \\ $$$$\left(\mathrm{b}\right)\mathrm{24x}+\mathrm{138y}=\mathrm{18} \\ $$$$\Rightarrow\mathrm{4x}+\mathrm{23y}=\mathrm{3}\:\left(\mathrm{look}\:\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{y}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right)\Rightarrow\mathrm{y}=\mathrm{2a}+\mathrm{1} \\ $$$$\mathrm{4x}+\mathrm{23}\left(\mathrm{2a}+\mathrm{1}\right)=\mathrm{3} \\ $$$$\mathrm{x}=−\mathrm{5}−\frac{\mathrm{23}}{\mathrm{2}}\mathrm{a}\Rightarrow\mathrm{2}\mid\mathrm{a}\Rightarrow\mathrm{a}=\mathrm{2k} \\ $$$$\mathrm{x}=−\mathrm{5}−\mathrm{23k},\:\mathrm{y}=\mathrm{4k}+\mathrm{1} \\ $$$$\left(\mathrm{c}\right)\mathrm{221x}+\mathrm{35y}=\mathrm{11}\:\left(\mathrm{look}\:\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{x}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right)\Rightarrow\mathrm{x}=\mathrm{5a}+\mathrm{1} \\ $$$$\mathrm{221}\left(\mathrm{5a}+\mathrm{1}\right)+\mathrm{35y}=\mathrm{11} \\ $$$$\mathrm{y}=−\frac{\mathrm{221}}{\mathrm{7}}\mathrm{a}−\mathrm{6}\Rightarrow\mathrm{7}\mid\mathrm{a}\Rightarrow\mathrm{a}=\mathrm{7k} \\ $$$$\mathrm{x}=\mathrm{35k}+\mathrm{1},\:\mathrm{y}=−\mathrm{221k}−\mathrm{6} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *