Prove-that-6-20-1-0-mod-7-

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

Prove that 6^{20} - 1 = 0 (\mod 7)

Commented by Frix last updated on 03/May/24

Prove that n^{2 k} - 1 = 0 (\mod (n + 1)) :

n^{2 k} - 1 = (n + 1) \underset{j = 1}{\sum^{2 k}} ({(- 1)}^{j + 1} n^{2 k - j})

With n = 6 \land k = 10

6^{20} - 1 = 7 \underset{j = 1}{\sum^{20}} ({(- 1)}^{j + 1} 6^{20 - j})

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

6^{20} - 1 \equiv 0 (m o d 7)

{(- 1)}^{20} - 1 \equiv (m o d 7) [∵ 6 \equiv - 1 (m o d 7)]

1 - 1 \equiv 0 (m o d 7)

0 \equiv 0 (m o d 7)

H e n c e p r o v e d

Answered by A5T last updated on 03/May/24

6^{6} \equiv 1 (m o d 7) \Rightarrow 6^{20} = {(6^{6})}^{3} 6^{2} \equiv 6^{2} \equiv 1 (m o d 7)

\Rightarrow 6^{20} - 1 \equiv 0 (m o d 7)

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

6^{20} - 1 \equiv 0 (m o d 7)

6^{5} - 1) (6^{5} + 1) (6^{10} + 1) \equiv 0 (m o d 7)

6^{5} - 1) (7776 + 1) (6^{10} + 1) \equiv 0 (m o d 7)

6^{5} - 1) (7777) (6^{10} + 1) \equiv 0 (m o d 7)

6^{5} - 1) (0) (6^{10} + 1) \equiv 0 (m o d 7)

0 \equiv 0 (m o d 7)

Proved