32^(32^(32) ) ≡ r (mod 7) r = ?


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Question Number 187619 by BaliramKumar last updated on 19/Feb/23

$32^{32^{32}} \equiv r (m o d 7)$ $r = ?$
	Answered by SEKRET last updated on 19/Feb/23

	${(32)}^{32^{32}} \equiv r \mod (7) r = ?$ ${(4 \cdot 7 + 4)}^{32^{32}}$ $4^{1} \equiv 4 \mod (7)$ $4^{2} \equiv 2 \mod (7)$ $4^{3} \equiv 1 \mod (7)$ $4^{4} \equiv 4 \mod (7) \to T = 3$ $32^{32} \equiv x \mod (3)$ ${(3 \cdot 10 + 2)}^{32}$ $2^{1} \equiv 2 \mod (3)$ $2^{2} \equiv 1 \mod (3)$ $2^{3} \equiv 2 \mod (3) \to T_{2} = 2$ $32 \equiv 0 \mod (2)$ $32^{1} \equiv 4 \mod (7)$ $32^{32^{32}} = 4 \mod (7)$ $A B D U L A Z I Z A B D U V A L I Y E V$
	Commented by BaliramKumar last updated on 20/Feb/23

	$T h a n k s S i r$
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