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Question Number 185 by 123456 last updated on 25/Jan/15

solve 10x≡25(mod 15)

$$\mathrm{solve}\:\mathrm{10x}\equiv\mathrm{25}\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$

Answered by mreddy last updated on 14/Dec/14

gcd(10,15)=5 and 5 divided 25 so   there are 5 solutions  Solutions are given by equation  10x+15y=25  particular solution x=1, y=1  general solution x=1+3k, y=1−2k  x=1, 4, 7, 10, 13 (mod 15)

$$\mathrm{gcd}\left(\mathrm{10},\mathrm{15}\right)=\mathrm{5}\:\mathrm{and}\:\mathrm{5}\:\mathrm{divided}\:\mathrm{25}\:\mathrm{so}\: \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{5}\:\mathrm{solutions} \\ $$$$\mathrm{Solutions}\:\mathrm{are}\:\mathrm{given}\:\mathrm{by}\:\mathrm{equation} \\ $$$$\mathrm{10x}+\mathrm{15y}=\mathrm{25} \\ $$$$\mathrm{particular}\:\mathrm{solution}\:\mathrm{x}=\mathrm{1},\:\mathrm{y}=\mathrm{1} \\ $$$$\mathrm{general}\:\mathrm{solution}\:\mathrm{x}=\mathrm{1}+\mathrm{3k},\:\mathrm{y}=\mathrm{1}−\mathrm{2k} \\ $$$$\mathrm{x}=\mathrm{1},\:\mathrm{4},\:\mathrm{7},\:\mathrm{10},\:\mathrm{13}\:\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$

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