If a > 1, and a | b , then 3b+1 mod a =
b(mod a)=3b(mod a)=0b(mod\ a)=3b(mod\ a)=0b(mod a)=3b(mod a)=0
(3b+1)(mod a)=1(3b+1)(mod\ a)=1(3b+1)(mod a)=1
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