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

Answered by Frix last updated on 29/Dec/22

$$\mathrm{mod}({2222}^{6k};7)=1$$$$\mathrm{mod}({2222}^{6k+1};7)=3$$$$\mathrm{mod}({2222}^{6k+2};7)=2$$$$\mathrm{mod}({2222}^{6k+3};7)=6$$$$\mathrm{mod}({2222}^{6k+4};7)=4$$$$\mathrm{mod}({2222}^{6k+5};7)=5$$$$$$$$\mathrm{mod}({5555}^{6k};7)=1$$$$\mathrm{mod}({5555}^{6k+1};7)=4$$$$\mathrm{mod}({5555}^{6k+2};7)=2$$$$\mathrm{mod}({5555}^{6k+3};7)=1$$$$\mathrm{mod}({5555}^{6k+4};7)=4$$$$\mathrm{mod}({5555}^{6k+5};7)=2$$$$$$$$5555=6k+5$$$$2222=6k+2$$$$$$$$\Rightarrow$$$${2222}^{5555}={2222}^{6k+5}\Rightarrow \mathrm{mod}({2222}^{5555};7)=5$$$${5555}^{2222}={5555}^{6k+2}\Rightarrow \mathrm{mod}({5555}^{2222};7)=2$$$$5+2=7$$

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

$${2222}^{5555}+{5555}^{2222}mod7$$$$={(317\times 7+3)}^{5555}+{(793\times 7+4)}^{2222}mod7$$$$=3^{5555}+4^{2222}mod7$$$$={(34\times 7+5)}^{1111}+{(2\times 7+2)}^{1111}mod7$$$$=5^{1111}+2^{1111}mod7$$$$={(7-2)}^{1111}+2^{1111}mod7$$$$=-2^{1111}+2^{1111}mod7$$$$=0$$

Commented by Michaelfaraday last updated on 29/Dec/22

$$thankssir$$

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