6^(273) +8^(273) :49 prove the division


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Question Number 98448 by Quvonchbek last updated on 14/Jun/20

$6^{273} + 8^{273} : 49 p r o v e t h e d i v i s i o n$
Commented by Rasheed.Sindhi last updated on 14/Jun/20

$49 ∣ 2^{273} (3^{273} + 4^{273})$ $∵ 49 ∤ 2^{273} ∴ 49 ∣ (3^{273} + 4^{273})$ $O r 3^{273} + 4^{273} \equiv 0 (m o d 49)$
Commented by mr W last updated on 14/Jun/20

$d o w e n e e d t o p r o v e t h a t$ $3^{273} + 4^{273} \equiv 0 (m o d 49)$ $h e r e s i r ?$
Commented by Rasheed.Sindhi last updated on 14/Jun/20

$Yes sir, If I understood the question .$
Commented by 1549442205 last updated on 14/Jun/20

$3^{273} + 4^{273} = 27^{91} + 64^{91} = 91 . (27^{90} - 27^{89} . 64 + 27^{88} . 64^{2} - . . . - 27 . 64^{89} + 64^{90})$ $Since 91 = 13.7, it is enough to prove that$ $A = (27^{90} - 27^{89} . 64 + 27^{88} . 64^{2} - . . . - 27 . 64^{89} + 64^{90}) ⋮ 7$ $Since 64^{k} = {(63 + 1)}^{k} = 1 (\mod 7) and$ $27^{90 - k} = {(28 - 1)}^{90 - k} = {(- 1)}^{90 - k} (\mod 7),$ $so 27^{90 - k} . 64^{k} = {(- 1)}^{90 - k} . = {(- 1)}^{90 - k} (\mod 7) (k = 0, 1 . . . 90)$ $it follows that A = 1 - 1 + 1 - 1 + . . . - 1 + 1 = 1 (\mod 7)$ $because A has 91 terms$ $Therefore A {isn}^{'} t divisible by 7 and so$ $3^{273} + 4^{273} {isn}^{'} t divisible by 49$ $it is only divisible by 7$ $I answered the question of Mr . W$
Commented by mr W last updated on 14/Jun/20

$t h a n k s s i r!$ $b u t 3^{273} + 4^{273} i s d i v i s i b l e b y 49 .$
	Answered by Rasheed.Sindhi last updated on 14/Jun/20

	$6^{273} + 8^{273} \equiv 0 (m o d 49)$ $\Rightarrow 2^{273} (3^{273} + 4^{273}) \equiv 0 (m o d 49)$ $\Rightarrow (3^{273} + 4^{273}) \equiv 0 (m o d 49) [∵ 49 ∤ 2^{273}]$ $^{∙} 4^{21} \equiv 1 (m o d 49)$ ${(4^{21})}^{13} \equiv {(1)}^{13} (m o d 49)$ $4^{273} \equiv 1 (m o d 49) . . . . . . . . . . . . (i)$ $^{∙} 3^{42} \equiv 1 (m o d 49)$ ${(3^{42})}^{6} \equiv {(1)}^{6} (m o d 49)$ $3^{252} \equiv 1 (m o d 49)$ $3^{252} . 3^{21} \equiv 1 . 3^{21} \equiv 48 (m o d 49)$ $3^{273} \equiv 48 (m o d 49) . . . . . . . . . . . (i i)$ $(i) + (i i) : 3^{273} + 4^{273} \equiv 49 \equiv 0 (m o d 49)$ $3^{273} + 4^{273} \equiv 0 (m o d 49)$ $P r o v e d$
	Commented by mr W last updated on 14/Jun/20

	$g r e a t!$
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