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Question Number 115674 by naka3546 last updated on 27/Sep/20
Solveforxx≡1mod3x≡4mod5x≡6mod7
Answered by floor(10²Eta[1]) last updated on 27/Sep/20
x≡1(mod3)⇒x=3a+13a+1≡4(mod5)⇒3a≡3(mod5)⇒a≡1(mod5)a=5b+1⇒x=3(5b+1)+1=15b+415b+4≡6(mod7)⇒15b≡b≡2(mod7)b=7c+2⇒x=15(7c+2)+4⇒x=105c+34∀c∈Z
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