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A.Avinash_Goutham
 2 years ago
Best ResponseYou've already chosen the best response.0u start like a=kn + b........

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.3\(a≡b \pmod n \implies a=n.k+b \) where \( k\in \mathbb{N} \) \(c≡d \pmod n \implies c=n.p+d \) where \( p\in \mathbb{N} \) \(ac = nk+bnpd = n(kp) +(bd)\), where \( (kp)\in \mathbb{N} \). Hence \((a−c)≡(b−d) \pmod n \) QED!

A.Avinash_Goutham
 2 years ago
Best ResponseYou've already chosen the best response.0c= pn+d ac=n(kp)+bd :P......m late

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.3Similarly we can show that \( (a+c)≡(b+d) \pmod n \) and more generally \( (pa+qc)≡(pb+qd) \pmod n \)

Hollywood_chrissy
 2 years ago
Best ResponseYou've already chosen the best response.1so basically we need to show that difference between a and c will be a multiple of n
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