6-2020-8-2020-mod-49-Show-your-elegant-workings-please-Thanks-a-lot-

Question Number 121608 by naka3546 last updated on 10/Nov/20

(6^{2020} + 8^{2020}) m o d 49 ?

S h o w y o u r e l e g a n t w o r k i n g s, p l e a s e .

T h a n k s a l o t .

Answered by mr W last updated on 10/Nov/20

6^{2020} + 8^{2020}

= {(7 - 1)}^{2020} + {(7 + 1)}^{2020}

= {(1 - 7)}^{2020} + {(1 + 7)}^{2020}

= \underset{k = 0}{\sum^{2020}} C_{k}^{2020} {(- 1)}^{k} 7^{k} + \underset{k = 0}{\sum^{2020}} C_{k}^{2020} 7^{k}

= \underset{k = 0, 2, \dots}{\sum^{2020}} 2 C_{k}^{2020} 7^{k}

= 2 + 2 \underset{k = 1}{\sum^{1010}} C_{2 k}^{2020} 7^{2 k}

= 2 + 2 \underset{k = 1}{\sum^{1010}} C_{2 k}^{2020} 49^{k}

\Rightarrow (6^{2020} + 8^{2020}) m o d 49 = 2

Answered by Rasheed.Sindhi last updated on 10/Nov/20

\equiv A n O ther W ay \equiv

^{∙} 6^{7} \equiv - 1 (m o d 49)

2020 = 7 \times 288 + 4

{(6^{7})}^{288} {(6)}^{4} \equiv {(- 1)}^{144} {(- 1)}^{4} (m o d 49)

6^{2020} \equiv 1 (m o d 49) \dots \dots \dots \dots \dots \dots . . A

^{∙} 8^{7} \equiv 1 (m o d 49)

{(8^{7})}^{288} {(8)}^{4} \equiv {(1)}^{288} {(1)}^{4} (m o d 49)

8^{2020} \equiv 1 (m o d 49) \dots \dots \dots \dots \dots . . B

A + B :

6^{2020} + 8^{2020} \equiv 2 (m o d 49)