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Question Number 206999 by efronzo1 last updated on 03/May/24

Prove that 6^(20) −1 = 0 (mod 7)

Provethat620−1=0(mod7)

Commented by Frix last updated on 03/May/24

Prove that n^(2k) −1=0(mod(n+1)): n^(2k) −1=(n+1)Σ_(j=1) ^(2k) ((−1)^(j+1) n^(2k−j) ) With n=6∧k=10 6^(20) −1=7Σ_(j=1) ^(20) ((−1)^(j+1) 6^(20−j) )

Provethatn2k−1=0(mod(n+1)):n2k−1=(n+1)∑2kj=1((−1)j+1n2k−j)Withn=6∧k=10620−1=7∑20j=1((−1)j+1620−j)

Answered by Rasheed.Sindhi last updated on 03/May/24

6^(20) −1≡0(mod 7) (−1)^(20) −1≡(mod 7) [∵ 6≡−1(mod 7)] 1−1≡0(mod 7) 0≡0(mod 7) Hence proved

620−1≡0(mod7)(−1)20−1≡(mod7)[∵6≡−1(mod7)]1−1≡0(mod7)0≡0(mod7)Henceproved

Answered by A5T last updated on 03/May/24

6^6 ≡1(mod 7)⇒6^(20) =(6^6 )^3 6^2 ≡6^2 ≡1(mod 7) ⇒6^(20) −1≡0(mod 7)

66≡1(mod7)⇒620=(66)362≡62≡1(mod7)⇒620−1≡0(mod7)

Answered by Rasheed.Sindhi last updated on 03/May/24

6^(20) −1≡0(mod 7) 6^5 −1)(6^5 +1)(6^(10) +1)≡0(mod 7) 6^5 −1)(7776+1)(6^(10) +1)≡0(mod 7) 6^5 −1)(7777)(6^(10) +1)≡0(mod 7) 6^5 −1)(0)(6^(10) +1)≡0(mod 7) 0≡0(mod 7) Proved

620−1≡0(mod7)65−1)(65+1)(610+1)≡0(mod7)65−1)(7776+1)(610+1)≡0(mod7)65−1)(7777)(610+1)≡0(mod7)65−1)(0)(610+1)≡0(mod7)0≡0(mod7)Proved

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