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Question Number 82991 by ahmadshahhimat775@gmail.com last updated on 26/Feb/20

Commented by mr W last updated on 26/Feb/20

$${11}^{800}mod9$$$$=2^{800}mod9$$$$=7^{200}mod9$$$$=4^{100}mod9$$$$=7^{50}mod9$$$$=4^{25}mod9$$$$=4\times 7^{12}mod9$$$$=4\times 4^{6}mod9$$$$=4\times 7^{3}mod9$$$$=4\times 7\times 4mod9$$$$=4\times 1mod9$$$$=4$$

Commented by mr W last updated on 26/Feb/20

$$why{don}^{\prime }tyoutypethequestioninstead$$$$ofpostinganimage?$$

Answered by TANMAY PANACEA last updated on 27/Feb/20

$$\frac{{(9+2)}^{800}}{9}=\frac{9^{800}+800c_1{9}^{799}2+...+2^{800}}{9}=\frac{\mathit{f}\left(9\right)+2^{800}}{9}$$$$\frac{\mathit{f}\left(9\right)+{\left(2^8\right)}^{100}}{9}=\frac{\mathit{f}\left(9\right)+{\left(256\right)}^{100}}{9}=\frac{f\left(9\right)+{(28\times 9+4)}^{100}}{9}$$$$=\frac{f\left(9\right)+g\left(9\right)+4^{100}}{9}=\frac{f\left(9\right)+g\left(9\right)+{\left(4^4\right)}^{25}}{9}=\frac{f\left(9\right)+g\left(9\right)+{(252+4)}^{25}}{9}$$$$=\frac{f\left(9\right)+g\left(9\right)+h\left(9\right)+4^{25}}{9}$$$$=\frac{f\left(9\right)+g\left(9\right)+h\left(9\right)+4\times {\left(4^3\right)}^8}{9}$$$$=\frac{f\left(9\right)+g\left(9\right)+h\left(9\right)+4\times {(63+1)}^8}{9}$$$$=\frac{f\left(9\right)+g\left(9\right)+h\left(9\right)+4[k\left(9\right)+1]}{9}\left[note63isdivisibleby9\right]$$$$andk\left(9\right)={(63+1)}^8=\left({63}^8\right)+8c_1{\left(63\right)}^7+...+1$$$$\mathit{s}\mathit{o}$$$$=\frac{f\left(9\right)+g\left(9\right)+h\left(9\right)+4k\left(9\right)+4}{9}\mathit{r}\mathit{e}\mathit{m}\mathit{a}\mathit{i}\mathit{n}\mathit{d}\mathit{e}\mathit{r}=4$$

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