(6^(2020) + 8^(2020) ) mod 49 ? Show your elegant workings , please. Thanks a lot.


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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

	$6^{2020} + 8^{2020}$ $= {(7 - 1)}^{2020} + {(7 + 1)}^{2020}$ $= {(1 - 7)}^{2020} + {(1 + 7)}^{2020}$ $= \underset{k = 0}{\sum^{2020}} C_{k}^{2020} {(- 1)}^{k} 7^{k} + \underset{k = 0}{\sum^{2020}} C_{k}^{2020} 7^{k}$ $= \underset{k = 0, 2, . . .}{\sum^{2020}} 2 C_{k}^{2020} 7^{k}$ $= 2 + 2 \underset{k = 1}{\sum^{1010}} C_{2 k}^{2020} 7^{2 k}$ $= 2 + 2 \underset{k = 1}{\sum^{1010}} C_{2 k}^{2020} 49^{k}$ $\Rightarrow (6^{2020} + 8^{2020}) m o d 49 = 2$
	Answered by Rasheed.Sindhi last updated on 10/Nov/20

	$\equiv A n O ther W ay \equiv$ $^{∙} 6^{7} \equiv - 1 (m o d 49)$ $2020 = 7 \times 288 + 4$ ${(6^{7})}^{288} {(6)}^{4} \equiv {(- 1)}^{144} {(- 1)}^{4} (m o d 49)$ $6^{2020} \equiv 1 (m o d 49) . . . . . . . . . . . . . . . . . . . . A$ $^{∙} 8^{7} \equiv 1 (m o d 49)$ ${(8^{7})}^{288} {(8)}^{4} \equiv {(1)}^{288} {(1)}^{4} (m o d 49)$ $8^{2020} \equiv 1 (m o d 49) . . . . . . . . . . . . . . . . . B$ $A + B :$ $6^{2020} + 8^{2020} \equiv 2 (m o d 49)$
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