Resolver o seguinte sistema: x ≡ 1 (mod 2) x ≡ 2 (mod 3) x ≡ 5...

2*3*7=42  

x ≡ 1 (mod 2) ==> vezes 21 ==> 21x= 21 (mod 42) (i)

x ≡ 2 (mod 3) ==>vezes 14 ==>  14x= 28 (mod 42)  (ii)

x ≡ 5 (mod 7) ==>vezes 6 ==>  6x=30 (mod 42)    (iii)

subtraindo (i)-(ii)-(iii)

x =21 (mod 42)- 28 (mod 42)-30 (mod 42)

x= -37 (mod 42)

x=5