32-32-32-r-mod-7-r-

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

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