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Question Number 115674 by naka3546 last updated on 27/Sep/20

Solve  for  x        x  ≡  1  mod  3        x  ≡  4  mod  5        x  ≡  6  mod  7

Solveforxx1mod3x4mod5x6mod7

Answered by floor(10²Eta[1]) last updated on 27/Sep/20

x≡1(mod3)⇒x=3a+1  3a+1≡4(mod5)⇒3a≡3(mod5)⇒a≡1(mod5)  a=5b+1⇒x=3(5b+1)+1=15b+4  15b+4≡6(mod7)⇒15b≡b≡2(mod7)  b=7c+2⇒x=15(7c+2)+4  ⇒x=105c+34 ∀ c∈Z

x1(mod3)x=3a+13a+14(mod5)3a3(mod5)a1(mod5)a=5b+1x=3(5b+1)+1=15b+415b+46(mod7)15bb2(mod7)b=7c+2x=15(7c+2)+4x=105c+34cZ

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