Question-183737

Question Number 183737 by Michaelfaraday last updated on 29/Dec/22

Answered by Frix last updated on 29/Dec/22

\mod (2222^{6 k}; 7) = 1

\mod (2222^{6 k + 1}; 7) = 3

\mod (2222^{6 k + 2}; 7) = 2

\mod (2222^{6 k + 3}; 7) = 6

\mod (2222^{6 k + 4}; 7) = 4

\mod (2222^{6 k + 5}; 7) = 5

\mod (5555^{6 k}; 7) = 1

\mod (5555^{6 k + 1}; 7) = 4

\mod (5555^{6 k + 2}; 7) = 2

\mod (5555^{6 k + 3}; 7) = 1

\mod (5555^{6 k + 4}; 7) = 4

\mod (5555^{6 k + 5}; 7) = 2

5555 = 6 k + 5

2222 = 6 k + 2

\Rightarrow

2222^{5555} = 2222^{6 k + 5} \Rightarrow \mod (2222^{5555}; 7) = 5

5555^{2222} = 5555^{6 k + 2} \Rightarrow \mod (5555^{2222}; 7) = 2

5 + 2 = 7

Answered by mr W last updated on 29/Dec/22

2222^{5555} + 5555^{2222} m o d 7

= {(317 \times 7 + 3)}^{5555} + {(793 \times 7 + 4)}^{2222} m o d 7

= 3^{5555} + 4^{2222} m o d 7

= {(34 \times 7 + 5)}^{1111} + {(2 \times 7 + 2)}^{1111} m o d 7

= 5^{1111} + 2^{1111} m o d 7

= {(7 - 2)}^{1111} + 2^{1111} m o d 7

= - 2^{1111} + 2^{1111} m o d 7

= 0

Commented by Michaelfaraday last updated on 29/Dec/22

t h a n k s s i r