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x-4-mod-5-x-3-mod-4-




Question Number 134267 by bramlexs22 last updated on 01/Mar/21
 { ((x≡ 4 (mod 5))),((x≡ 3 (mod 4 ))) :}
$$\begin{cases}{\mathrm{x}\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{5}\right)}\\{\mathrm{x}\equiv\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{4}\:\right)}\end{cases} \\ $$
Answered by EDWIN88 last updated on 01/Mar/21
 { ((x=5k+4)),((x=4ℓ+3)) :} ⇔ 5k+4 = 4ℓ+3  ℓ = ((5k+1)/4) ; (k,ℓ) = (3,4), (7,9),(11,14),...,  (4λ−1, 5λ−1) ⇒then x=5(4λ−1)+4  x = 20λ−1 , λ∈ Z = 19, 39, 59, 79,...
$$\begin{cases}{\mathrm{x}=\mathrm{5k}+\mathrm{4}}\\{\mathrm{x}=\mathrm{4}\ell+\mathrm{3}}\end{cases}\:\Leftrightarrow\:\mathrm{5k}+\mathrm{4}\:=\:\mathrm{4}\ell+\mathrm{3} \\ $$$$\ell\:=\:\frac{\mathrm{5k}+\mathrm{1}}{\mathrm{4}}\:;\:\left(\mathrm{k},\ell\right)\:=\:\left(\mathrm{3},\mathrm{4}\right),\:\left(\mathrm{7},\mathrm{9}\right),\left(\mathrm{11},\mathrm{14}\right),…, \\ $$$$\left(\mathrm{4}\lambda−\mathrm{1},\:\mathrm{5}\lambda−\mathrm{1}\right)\:\Rightarrow\mathrm{then}\:\mathrm{x}=\mathrm{5}\left(\mathrm{4}\lambda−\mathrm{1}\right)+\mathrm{4} \\ $$$$\mathrm{x}\:=\:\mathrm{20}\lambda−\mathrm{1}\:,\:\lambda\in\:\mathbb{Z}\:=\:\mathrm{19},\:\mathrm{39},\:\mathrm{59},\:\mathrm{79},… \\ $$

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